Maharashtra Board

Maharashtra Board Class 9 Maths Solutions Chapter 5 Linear Equations in Two Variables Problem Set 5

Question 1.
Choose the correct alternative answers for the following questions.

i. If 3x + 5y = 9 and 5x + 3y = 7, then what is the value of x + y ?
(A) 2
(B) 16
(C) 9
(D) 7
Answer:
(A) 2

ii. ‘When 5 is subtracted from length and breadth of the rectangle, the perimeter becomes 26’. What is the mathematical form of the statement ?
(A) x – y = 8
(B) x + y = 8
(C) x + y = 23
(D) 2x + y = 21
Answer:
(C) x + y = 23

iii. Ajay is younger than Vijay by 5 years. Sum of their ages is 25 years. What is Ajay’s age?
(A) 20 years
(B) 15 years
(C) 10 years
(D) 5 years
Answer:
(C) 10 years

Hints:
i. Adding the given equations,
3x+ 5y = 9
5x + 3y = 7
8x + 8y = 16
∴ x + y = 2 .. [Dividing both sides by 8]

ii. Let the length of the rectangle be ‘x’ and that of breadth be ‘y’.
Perimeter of rectangle = 2[(x – 5) + (y – 5)]
∴ 26 = 2(x + y – 10)
∴ x + y – 10 = 13
∴ x + y = 23

iii. Let the age of Ajay bex years.
∴ x + (x + 5) = 25
∴ 2x = 20
∴ x = 10 years

Question 2.
Solve the following simultaneous equations.
i. 2x + y = 5 ; 3x – y = 5
ii. x – 2y = -1 ; 2x – y = 7
iii. x + y = 11 ; 2x – 3y = 7
iv. 2x + y = -2 ; 3x – y = 7
V. 2x – y = 5 ; 3x + 2y = 11
vi. x – 2y – 2 ; x + 2y = 10
Solution:
ii. 2x + y = 5 …(i)
3x – y = 5 …(ii)
Adding equations (i) and (ii),
2x + y = 5
+ 3x – y = 5
5x = 10
∴ x = 10/5
∴ x = 2
Substituting x = 2 in equation (i),
2(2) + y = 5
4 + y = 5
∴ y = 5 – 4 = 1
∴ (2, 1) is the solution of the given equations.

ii. x – 2y = -1
∴x = 2y – 1 … .(i)
∴ 2x – y = 7 ….(ii)
Substituting x = 2y – 1 in equation (ii),
2(2y – 1) – y = 7
∴ 4y – 2 – y = 7
∴ 3y = 7 + 2
∴ 3y = 9
∴ y = 9/3
∴ y = 3
Substituting y = 3 in equation (i),
x = 2y – 1
∴ x = 2(3) – 1
∴ x = 6 – 1 = 5
∴ (5, 3) is the solution of the given equations.

iii. x + y = 11
∴ x = 11 – y …(i)
2x – 3y = 7 …….(ii)
Substituting x = 11 -y in equation (ii),
2(11 – y) – 3y = 7
∴ 22 – 2y – 3y = 1
∴ 22 – 5y = 7
∴ 22 – 7 = 5y
∴ 15 = 5y
∴ y = \(\frac { 15 }{ 5 }\)
∴ y = 3
Substituting y = 3 in equation (i),
x = 11 – y
∴ x = 11 – 3 = 8
∴ (8, 3) is the solution of the given equations.

iv. 2x + y = -2 …(i)
3x – y = 7 …(ii)
Adding equations (i) and (ii),
2x + y = -2
+ 3x – y = l
5x = 5
∴ x = \(\frac { 5 }{ 5 }\)
∴ x = 1
Substituting x = 1 in equation (i),
2x + y = -2
∴ 2(1) +y = -2
2 + y = -2
∴ y = – 2 – 2
∴ y = -4
∴ (1, -4) is the solution of the given equations.

v. 2x – y = 5
∴ -y = 5 – 2x
∴ y = 2x – 5 …(i)
3x + 2y = 11 ……(ii)
Substituting y = 2x – 5 in equation (ii),
3x + 2(2x – 5) = 11
∴ 3x + 4x- 10= 11
∴ 7x = 11 + 10
∴ 7x = 21
∴ x = \(\frac { 21 }{ 7 }\)
∴ x = 3
Substituting x = 3 in equation (i),
y = 2x – 5
∴ y = 2(3) – 5
∴ y = 6 – 5 = 1
∴(3,1) is the solution of the given equations.

vi. x – 2y = -2
∴ x = 2y – 2 …(i)
x + 2y = 10 …..(ii)
Substituting x = 2y – 2 in equation (ii),
2y – 2 + 2y = 10
∴ 4y = 10 + 2
∴ 4y= 12
∴ y = \(\frac { 12 }{ 7 }\)
∴ y = 3
Substituting y = 3 in equation (i),
x = 2y – 2
∴ x = 2(3) – 2
∴ x = 6 – 2 = 4
∴ (4, 3) is the solution of the given equations.

Question 3.
By equating coefficients of variables, solve the following equations. [3 Marks each]
i. 3x – 4y = 7 ; 5x + 2y = 3
ii. 5x + ly= 17 ; 3x – 2y = 4
iii. x – 2y = -10 ; 3x – 3y = -12
iv. 4x+y = 34 ; x + 4y = 16
Solution:
i. 3x – 4y = 7 …(i)
5x + 2y = 3 ….(ii)
Multiplying equation (ii) by 2,
10x + 4y = 6 …(iii)
Adding equations (i) and (iii),

∴ x = 1
Substituting x = 1 in equation (i),
3x – 4y = 7
∴ 3(1) – 4y = 7
∴ 3 – 4y = 7
∴ 3 – 7 = 4y
∴ -4 = 4y
∴ y = \(\frac { -4 }{ 4 }\)
∴ y = -1
∴ (1, -1) is the solution of the given equations.

ii. 5x + 7y = 17 …(i)
3x – 2y = 4 ….(ii)
Multiplying equation (i) by 2,
10x + 14y = 34 …(iii)
Multiplying equation (ii) by 7,
21x – 14y = 28 …..(iv)
Adding equations (iii) and (iv),

∴ x = 2
Substituting x = 2 in equation (ii),
3x – 2y = 4
∴ 3(2) – 2y = 4
∴ 6 – 2y = 4
∴ 6 – 4 = 2y
∴ 2 = 2y
∴ y = \(\frac { 2 }{ 2 }\)
∴ y = 1
∴ (2,1) is the solution of the given equations.

iii. x – 2y = -10 ….(i)
3x – 5y = -12 …….(ii)
Multiplying equation (i) by 3,
3x – 6y = -30 …(iii)
Subtracting equation (ii) from (iii),

∴ y = 18
Substituting y = 18 in equation (i),
x – 2y = -10
∴ x – 2(18) = -10
∴ x – 36 = -10
∴ x = -10 + 36 = 26
∴ (26, 18) is the solution of the given equations.

iv. 4x + y = 34 …(i)
x + 4y = 16 …… (ii)
Multiplying equation (i) by 4,
16x + 4y = 136 …(iii)
Subtracting equation (ii) from (iii),

x = 8
Substituting x = 8 in equation (i),
4x + y = 34
∴ 4(8) + y = 34
∴ 32 + y = 34
∴ y = 34 – 32 = 2
∴ (8, 2) is the solution of the given equations.

Question 4.
Solve the following simultaneous equations.

Solution:
i. \(\frac{x}{3}+\frac{y}{4}=4\)
Multiplying both sides by 12,
4x + 3y = 48 …(i)
\(\frac{x}{2}-\frac{y}{4}=1\)
Multiplying both sides by 8,
4x – 2y = 8 …..(ii)
Subtracting equation (ii) from (i),

∴ y = 8
Substituting y = 8 in equation (ii),
4x – 2y = 8
∴ 4x – 2(8) = 8
∴ 4x – 16 = 8
∴ 4x = 8+ 16
∴ 4x = 24
∴ x = \(\frac { 24 }{ 4 }\)
∴ x = 6
∴ (6, 8) is the solution of the given equations.

ii. \(\frac { x }{ 3 }\) + 5y = 13
Multiplying both sides by 3,
x + 15y = 39 …(i)
2x + \(\frac { y }{ 2 }\) =19
Multiplying both sides by 2,
4x + y = 38 …….(ii)
Multiplying equation (i) by 4,
4x + 60y = 156 …(iii)
Subtracting equation (ii) from (iii),
4x + 60y =156 4x + y= 38

∴ y = 2
Substituting y = 2 in equation (i),
x + 15y = 39
∴ x+ 15(2) = 39
∴ x + 30 = 39
∴ x = 39 – 30 = 9
∴ (9,2) is the solution of the given equations.

iii. \(\frac { 2 }{ x }\) + \(\frac { 3 }{ y }\) = 13
Multiplying both sides by 5,

Question 5.
A two digit number is 3 more than 4 times the sum of its digits. If 18 is added to this number, the sum is equal to the number obtained by interchanging the digits. Find the number.
Solution:
Let the digit in unit’s place be ‘x’ and the digit in ten’s place be ‘y’.

According to the first condition,
a two digit number is 3 more than 4 times the sum of its digits.
10y + x = 4(x + y) + 3
∴ 10y + x = 4x + 4y + 3
∴ x – 4x + 10y – 4y = 3
∴ – 3x + 6y = 3
Dividing both sides by -3,
x – 2y = -1 …(i)
According to the second condition,
if 18 is added to the number, the sum is equal to the number obtained by interchanging the digits.
10y + x + 18= 10x + y
∴ x – 10x + 10y – y = -18
∴ – 9x + 9y = -18
Dividing both sides by – 9,
x – y = 2 ……(ii)
Subtracting equation (ii) from (i),

∴ y = 3
Substituting y = 3 in equation (ii),
x – y = 2
∴ x – 3 = 2
∴ x = 2 + 3 = 5
∴ Original number = 10y + x
= 10(3) + 5
= 30 + 5
= 35
The required number is 35.

Question 6.
The total cost of 8 books and 5 pens is ₹ 420 and the total cost of 5 books and 8 pens is ₹321. Find the cost of 1 book and 2 pens.
Solution:
Let the cost of one book be ₹ x and the cost of one pen be ₹ y.
According to the first condition,
the total cost of 8 books and 5 pens is ₹ 420.
∴ 8x + 5y = 420 …(i)
According to the second condition, the total cost of 5 books and 8 pens is ₹ 321.
5x + 8y = 321 ….(ii)
Multiplying equation (i) by 5,
40x + 25y = 2100 …(iii)
Multiplying equation (ii) by 8,
40x + 64y = 2568 … (iv)
Subtracting equation (iii) from (iv),

∴ y = 12
Substituting y = 12 in equation (i),
8x + 5y = 420
∴ 8x + 5(12) = 420
∴ 8x + 60 = 420
∴ 8x = 420 – 60
∴ 8x = 360
∴ x = \(\frac { 360 }{ 8 }\)
∴ x = 45
Cost of 1 book and 2 pens = x + 2y
= 45 + 2(12)
= 45 + 24
= ₹69
∴ The cost of 1 book and 2 pens is ₹69.

Question 7.
The ratio of incomes of two persons is 9 : 7. The ratio of their expenses is 4 : 3. Every person saves ₹ 200, find the income of each.
Solution:
Let the income of first person be ₹ x and that of second person be ₹ y.
According to the first condition,
the ratio of their incomes is 9 : 7.
∴ \(\frac { x }{ y }\) = \(\frac { 9 }{ 7 }\)
∴ 7x = 9y
∴ 7x – 9y = 0 …….(i)
Each person saves ₹ 200.
Expenses of first person = Income – Saving = x – 200
Expenses of second person = y – 200
According to the second condition,
the ratio of their expenses is 4 : 3
∴ \(\frac { x – 200 }{ y – 200 }\) = \(\frac { 4 }{ 3 }\)
∴ 3(x – 200) = 4(y – 200)
∴ 3x – 600 = 4y – 800
∴ 3x – 4y = – 800 + 600
∴ 3x – 4y = -200 …(ii)
Multiplying equation (i) by 4,
28x-36y =0 …(iii)
Multiplying equation (ii) by 9,
27x-36y = -1800 …(iv)
Subtracting equation (iv) from (iii),

Substituting x = 1800 in equation (i),
7x – 9y = 0
∴ 7(1800) – 9y = 0
∴ 9y = 7 x 1800
∴ y = \(\frac { 7 \times 1800 }{ 9 }\)
y = 7 x 200
∴ y = 1400
∴ The income of first person is ₹ 1800 and that of second person is ₹ 1400.

Question 8.
If the length of a rectangle is reduced by 5 units and its breadth is increased by 3 units, then the area of the rectangle is reduced by 9 square units. If length is reduced by 3 units and breadth is increased by 2 units, then the area of rectangle will increase by 67 square units. Then find the length and breadth of the rectangle.
Solution:
Let the length of the rectangle be ‘x’ units and the breadth of the rectangle be ‘y’ units.
Area of the rectangle = xy sq. units
length of the rectangle is reduced by 5 units
∴ length = x – 5
breadth of the rectangle is increased by 3 units
∴ breadth = y + 3
area of the rectangle is reduced by 9 square units
∴ area of the rectangle = xy – 9
According to the first condition,
(x – 5)(y + 3) = xy – 9
∴ xy + 3x – 5y – 15 = xy – 9
∴ 3x – 5y = -9 + 15
∴ 3x – 5y = 6 …(i)
length of the rectangle is reduced by 3 units
∴ length = x – 3
breadth of the rectangle is increased by 2 units
∴ breadth = y + 2
area of the rectangle is increased by 67 square units
∴ area of the rectangle = xy + 61
According to the second condition,
(x – 3)(y + 2) = xy + 67
∴ xy + 2x – 3y – 6 = xy + 67
∴ 2x – 3y = 67 + 6
∴ 2x – 3y = 73 …(ii)
Multiplying equation (i) by 3,
9x – 15y = 18 . ..(iii)
Multiplying equation (ii) by 5,
10x – 15y = 365 …(iv)
Subtracting equation (iii) from (iv), 10x- 15y= 365 9x-15y= 18

Substituting x = 347 in equation (ii),
2x – 3y = 73
∴ 2(347) – 3y = 73
∴ 694 – 73 = 3y
∴ 621 = 3y
∴ y = \(\frac { 621 }{ 3 }\)
∴ y = 207
∴ The length and breadth of rectangle are 347 units and 207 units respectively.

Question 9.
The distance between two places A and B on a road is 70 kilometres. A car starts from A and the other from B. If they travel in the same direction, they will meet in 7 hours. If they travel towards each other they will meet in 1 hour, then find their speeds.
Solution:
Let the speed of the car starting from A (first car) be ‘x’ km/hr and that starting from B (second car) be ‘y’ km/hr. (x > y)
According to the first condition,
Distance covered by the first car in 7 hours = 7x km
Distance covered by the second car in 7 hours = 7y km

If the cars are travelling in the same direction, 7x – 7y = 70
Dividing both sides by 7,
x – y = 10 …(i)
According to the second condition,
Distance covered by the first car in
1 hour = x km
Distance covered by the second car in 1 hour = y km

If the cars are travelling in the opposite direction
x + y = 70 …(ii)
Adding equations (i) and (ii),

∴ x = 40
Substituting x = 40 in equation (ii), x + y = 70
∴ 40 +y = 70
∴ y = 70 – 40 = 30
∴ The speed of the cars starting from places A and B are 40 km/hr and 30 km/hr respectively.

Question 10.
The sum of a two digit number and the number obtained by interchanging its digits is 99. Find the number.
Solution:
Let the digit in unit’s place be ‘x’ and the digit in ten’s place be ‘y’.

According to the given condition,
the sum of a two digit number and the number
obtained by interchanging its digits is 99.
∴ 10y + x + 10x +y = 99
∴ 11x + 11y = 99
Dividing both sides by 11,
x + y = 9
If y = 1, then x = 8
If y = 2, then x = 7
If y = 3, then x = 6 and so on.
∴ The number can be 18, 27, 36, … etc.

Maharashtra Board Class 9 Maths Chapter 5 Linear Equations in Two Variables Practice Set 5 Intext Questions and Activities

Question 1.
On the glasses of following spectacles, write numbers such that (Textbook pg. no. 82)
i. Their sum is 42 and difference is 16.

ii. Their sum is 37 and difference is 11.

iii. Their sum is 54 and difference is 20.

iv. Their sum is … and difference is … .

Answer:
ii. x + y = 37 and x – y = 11
∴ x = 24, y = 13
iii. x +y = 54 and x – y = 20
∴ x = 37, y =17

Question 2.
There are instructions written near the arrows in the following diagram. From this information form suitable equations and write in the boxes indicated by arrows. Select any two equations from these boxes and find their solutions. Also verify the solutions. By taking one pair of equations at a time, how many pairs can be formed ? Discuss the solutions for these pairs. (Textbook pg. no. 92)

Answer:
Here, if we take a pair of any two equations, we get following 6 pairs.

1. equation (i) and (ii)
2. equation (i) and (iii)
3. equation (i) and (iv)
4. equation (ii) and (iii)
5. equation (ii) and (iv)
6. equation (iii) and (iv)

Solution of each pair given above is (21, 15).
Here, all four equations are of same rectangle. By solving any two equations simultaneously, we get length and breadth of the rectangle.

Question 3.
Find the function.

∴ Given function = \(\frac { 6 }{ 14 }\)
Verify the answer obtained. (Textbook pg. no. 92)
Answer:
For the fraction\(\frac { 6 }{ 14 }\), if the numerator is multiplied by 3 and 3 is subtracted from the denominator, we get fraction \(\frac { 18 }{ 11 }\).
Similarly, for the fraction \(\frac { 6 }{ 14 }\), if the numerator is increased by 8 and the denominator is doubled, we get fraction \(\frac { 1 }{ 2 }\).

Maharashtra Board Class 9 Maths Solutions

Maharashtra State Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.2

Question 1.
Two circles having radii 3.5 cm and 4.8 cm touch each other internally. Find the distance between their centres.
Solution:
Let the two circles having centres P and Q touch each other internally at point R.
Here, QR = 3.5 cm, PR = 4.8 cm

The two circles touch each other internally.
∴ By theorem of touching circles,
P – Q – R
PQ = PR – QR
= 4.8 – 3.5
= 1.3 cm
[The distance between the centres of circles touching internally is equal to the difference in their radii]

Question 2.
Two circles of radii 5.5 cm and 4.2 cm touch each other externally. Find the distance between their centres.
Solution:
Let the two circles having centres P and R touch each other externally at point Q.
Here, PQ = 5.5 cm, QR = 4.2 cm

The two circles touch each other externally.
∴ By theorem of touching circles,
P – Q – R
PR = PQ + QR
= 5.5 + 4.2
= 9.7 cm
[The distance between the centres of the circles touching externally is equal to the sum of their radii]

Question 3.
If radii of two circles are 4 cm and 2.8 cm. Draw figure of these circles touching each other
i. externally
ii. internally.
Solution:
i. Circles touching externally:

ii. Circles touching internally:

Question 4.
In the adjoining figure, the circles with centres P and Q touch each other at R A line passing through R meets the circles at A and B respectively. Prove that –

i. seg AP || seg BQ,
ii. ∆APR ~ ∆RQB, and
iii. Find ∠RQB if ∠PAR = 35°.
Solution:
The circles with centres P and Q touch each other at R.
∴ By theorem of touching circles,
P – R – Q
i. In ∆PAR,
seg PA = seg PR [Radii of the same circle]
∴ ∠PRA ≅ ∠PAR (i) [Isosceles triangle theorem]
Similarly, in ∆QBR,
seg QR = seg QB [Radii of the same circle]
∴ ∠RBQ ≅ ∠QRB (ii) [Isosceles triangle theorem]
But, ∠PRA ≅ ∠QRB (iii) [Vertically opposite angles]
∴ ∠PAR ≅ ∠RBQ (iv) [From (i) and (ii)]
But, they are a pair of alternate angles formed by transversal AB on seg AP and seg BQ.
∴ seg AP || seg BQ [Alternate angles test]
ii. In ∆APR and ∆RQB,
∠PAR ≅ ∠QRB [From (i) and (iii)]
∠APR ≅ ∠RQB [Alternate angles]
∴ ∆APR – ∆RQB [AA test of similarity]
iii. ∠PAR = 35° [Given]
∴ ∠RBQ = ∠PAR= 35° [From (iv)]
In ∆RQB,
∠RQB + ∠RBQ + ∠QRB = 180° [Sum of the measures of angles of a triangle is 180°]
∴ ∠RQB + ∠RBQ + ∠RBQ = 180° [From (ii)]
∴ ∠RQB + 2 ∠RBQ = 180°
∴ ∠RQB + 2 × 35° = 180°
∴ ∠RQB + 70° = 180°
∴ ∠RQB = 110°

Question 5.
In the adjoining figure, the circles with centres A and B touch each other at E. Line l is a common tangent which touches the circles at C and D respectively. Find the length of seg CD if the radii of the circles are 4 cm, 6 cm.

Construction : Draw seg AF ⊥ seg BD.
Solution:
i. The circles with centres A and B touch each other at E. [Given]
∴ By theorem of touching circles,
A – E – B

∴ ∠ACD = ∠BDC = 90° [Tangent theorem]
∠AFD = 90° [Construction]
∴ ∠CAF = 90° [Remaining angle of ꠸AFDC]
∴ ꠸AFDC is a rectangle. [Each angle is of measure 900]
∴ AC = DF = 4 cm [Opposite sides of a rectangle]
Now, BD = BF + DF [B – F – C]
∴ 6 = BF + 4 BF = 2 cm
Also, AB = AE + EB
= 4 + 6 = 10 cm
[The distance between the centres of circles touching externally is equal to the sum of their radii]

ii. Now, in ∆AFB, ∠AFB = 90° [Construction]
∴ AB² = AF² + BF² [Pythagoras theorem]
∴ 10² = AF² + 2²
∴ 100 = AF² + 4
∴ AF² = 96
∴ AF = \(\sqrt { 96 }\) [Taking square root of both sides]
= \(\sqrt{16 \times 6}\)
= 4 \(\sqrt { 6 }\) cm
But, CD = AF [Opposite sides of a rectangle]
∴ CD = 4 \(\sqrt { 6 }\) cm

Question 1.
Take three collinear points X – Y – Z as shown in figure.
Draw a circle with centre X and radius XY. Draw another circle with centre Z and radius YZ.
Note that both the circles intersect each other at the single point Y. Draw a line / through point Y and perpendicular to seg XZ. What is line l (Textbook pg. no. 56)
Solution:

Line l is a common tangent of the two circles.

Question 2.
Take points Y – X – Z as shown in the figure. Draw a circle with centre Z and radius ZY.
Also draw a circle with centre X and radius XY. Note that both the circles intersect each other at the point Y.
Draw a line l perpendicular to seg YZ through point Y. What is line l? (Textbook pg. no. 56)
Solution:

Line l is a common tangent of the two circles.

If two circles in the same plane intersect with a line in the plane in only one point, they are said to be touching circles and the line is their common tangent.

The point common to the circles and the line is called their common point of contact.

1. Circles touching externally:
For circles touching externally, the distance between their centres is equal to sum of their radii, i.e. AB = AC + BC

2. Circles touching internally:
For circles touching internally, the distance between their centres is equal to difference of their radii,
i. e. AB = AC – BC

Question 3.
The circles shown in the given figure are called externally touching circles. Why? (iexthook pg. no. 57)

Answer:
Circles with centres R and S lie in the same plane and intersect with a line l in the plane in one and only one point T [R – T – S].
Hence the given circles are externally touching circles.

Question 4.
The circles shown in the given figure are called internally touching circles, why? (Textbook pg. no. 57)

Answer:
Circles with centres N and M lie in the same plane and intersect with a line p in the plane in one and only one point T [K – N – M].
Hence, the given circles are internally touching circles.

Question 5.
In the given figure, the radii of the circles with centres A and B are 3 cm and 4 cm respectively. Find
i. d(A,B) in figure (a)
ii. d(A,B) in figure (b) (Textbook pg. no. 57)

Solution:
i. Here, circle with centres A and B touch each other externally at point C.
∴ d(A, B) = d(A, C) + d(B ,C)
= 3 + 4
∴ d(A,B) = 7 cm
[The distance between the centres of circles touching externally is equal to the sum of their radii]
ii. Here, circle with centres A and 13 touch each other internally at point C.
∴ d(A, B) = d(A, C) – d(B, C)
= 4 – 3
∴ d(A,B) = 1 cm
[The distance between the centres of circles touching internally is equal to the difference in their radii]

Maharashtra Board Class 10 Maths Solutions

Maharashtra State Board Class 10 Maths Solutions Chapter 3 Circle Practice Set 3.1

Question 1.
In the adjoining figure, the radius of a circle with centre C is 6 cm, line AB is a tangent at A. Answer the following questions.
i. What is the measure of ∠CAB? Why?
ii. What is the distance of point C from line AB? Why?
iii. d(A, B) = 6 cm, find d(B, C).
iv. What is the measure of ∠ABC? Why?

Solution:
i. line AB is the tangent to the circle with centre C and radius AC. [Given]
∴ ∠CAB = 90° (i) [Tangent theorem]
ii. seg CA ⊥ line AB [From (i)]
radius = l(AC) = 6 cm
∴ The distance of point C from line AB is 6 cm.
iii. In ∆CAB, ∠CAB = 90° [From (i)]
∴ BC² = AB² + AC² . [Pythagoras theorem]

= 6² + 6²
= 2 × 6²
∴ BC = \(\sqrt{2 \times 6^{2}}\) [Taking square root of both sides]
= 6 \(\sqrt { 2 }\) cm
∴ d(B, C) = 6 cm
iv. In ∆ABC,
AC = AB = 6cm
∴ ∠ABC = ∠ACB [Isosceles triangle theorem]
Let ∠ABC = ∠ACB =x
In ∆ABC,
∠CAB + ∠ABC + ∠ACB = 180° [Sum of the measures of angles of a triangle is 180°]
∴ 90° + x + x = 180°
∴ 90 + 2x = 180°
∴ 2x = 180°- 90°
∴ x = \(\frac{90^{\circ}}{2}\)
∴ x = 45°
∴ ∠ABC = 45°

Question 2.
In the adjoining figure, O is the centre of the circle. From point R, seg RM and seg RN are tangent segments touching the circle at M and N. If (OR) = 10 cm and radius of the circle = 5 cm, then
i. What is the length of each tangent segment?
ii. What is the measure of ∠MRO?
iii. What is the measure of ∠MRN?

Solution:
seg RM and seg RN are tangents to the circle with centre O. [Given]
∴ ∠OMR = ∠ONR = 90° [Tangent theorem]

i. In ∆OMR, ∠OMR = 90°
∴ OR² = OM² + RM² [Pythagoras theorem]
∴ 10² = 5² + RM²
∴ 100 = 25 + RM²
∴ RM² = 75
∴ RM = \(\sqrt { 75 }\) [Taking square root of both sides]
∴ RM = RN [Tangent segment theorem]
Length of each tangent segment is 5 \(\sqrt { 3 }\) cm.
ii. In ∆RMO,
∠OMR = 90° [Tangent theorem]
OM = 5 cm and OR = 10 cm
∴ OM = \(\frac { 1 }{ 2 } \) OR
∴ ∠MRO = 30° (i) [Converse of 30° – 60° – 90° theorem]
Similarly, ∠NRO = 30°
iii. But, ∠MRN = ∠MRO + ∠NRO [Angle addition property]
= 30° + 30° [From (i)]
∴ ∠MRN = 60°

Question 3.
Seg RM and seg RN are tangent segments of a circle with centre O. Prove that seg OR bisects ∠MRN as well as ∠MON.

Solution:
Proof:
In ∆OMR and ∆ONR,
seg RM ≅ seg RN [Tangent segment theorem]
seg OM ≅ seg ON [Radii of the same circle]
seg OR ≅ seg OR [Common side]
∴ ∆OMR ≅ ∆ONR [SSS test of congruency]
{∴ ∠MRO ≅ ∠NRO
∠MOR ≅ ∠NOR } [c.a.c.t.]
∴ seg OR bisects ∠MRN and ∠MON.

Question 4.
What is the distance between two parallel tangents of a circle having radius 4.5 cm? Justify your answer.
Solution:
Let the lines PQ and RS be the two parallel tangents to circle at M and N respectively. Through centre O, draw line AB || line RS.

OM = ON = 4.5 [Given]
line AB || line RS [Construction]
line PQ || line RS [Given]
∴ line AB || line PQ || line RS
Now, ∠OMP = ∠ONR = 90° (i) [Tangent theorem]
For line PQ || line AB,
∠OMP = ∠AON = 90° (ii) [Corresponding angles and from (i)]
For line RS || line AB,
∠ONR = ∠AOM = 90° (iii) [Corresponding angles and from (i)]
∠AON + ∠AOM = 90° + 90° [From (ii) and (iii)]
∴ ∠AON + ∠AOM = 180°
∴ ∠AON and ∠AOM form a linear pair.
∴ ray OM and ray ON are opposite rays.
∴ Points M, O, N are collinear. (iv)
∴ MN = OM + ON [M – O – N, From (iv)]
∴ MN = 4.5 + 4.5
∴ MN = 9 cm
∴ Distance between two parallel tangents PQ and RS is 9 cm.

Question 1.
In the adjoining figure, seg QR is a chord of the circle with centre O. P is the midpoint of the chord QR. If QR = 24, OP = 10, find radius of the circle. To find solution of the problem, write the theorems that are useful. Using them, solve the problem. (Textbook pg. no. 48)

Solution:
Theorems which are useful to find solution:
i. The segment joining the centre of a circle and the midpoint of a chord is perpendicular to the chord.
ii. In a right angled triangle, sum of the squares of the perpendicular sides is equal to square of its hypotenuse.
QP = \(\frac { 1 }{ 2 } \) (QR) [P is the midpoint of chord QR]

\(\frac { 1 }{ 2 } \) × 24 = 12 units
Also, seg OP ⊥ chord QR [The segment joining centre of a circle and midpoint of a chord is perpendicular to the chord]
In ∆OPQ, ∠OPQ = 90°
∴ OQ² = OP² + QP² [Pythagoras theorem]
= 10² + 12²
= 100 + 144
= 244
∴ OQ = \(\sqrt { 244 }\) = 2\(\sqrt { 61 }\) units.
∴ The radius of the circle is 2\(\sqrt { 61 }\) units.

Question 2.
In the adjoining figure, M is the centre of the circle and seg AB is a diameter, seg MS ⊥ chord AD, seg MT ⊥ chord AC, ∠DAB ≅ ∠CAB.
i. Prove that: chord AD ≅ chord AC.
ii. To solve this problem which theorems will you use?
a. The chords which are equidistant from the centre are equal in length.
b. Congruent chords of a circle are equidistant from the centre.

iii. Which of the following tests of congruence of triangles will be useful?
a. SAS b. ASA c. SSS d. AAS e. Hypotenuse-side test.
Using appropriate test and theorem write the proof of the above example. (Textbook pg. no, 48)
Solution:
Proof:
i. ∠DAB ≅ ∠CAB [Given]
∴ ∠SAM ≅ ∠TAM (i) [A – S – D, A – M – B, A -T – C]
In ∆SAM and ∆TAM,
∠SAM ≅ ∠TAM [From (i)]
∠ASM ≅ ∠ATM [Each angle is of measure 90°]
seg AM ≅ seg AM [Common side]
∴ ∆SAM ≅ ∆TAM [AAS [SAA] test of congruency]
∴ side MS ≅ side MT [c.s.c.t]
But, seg MS ⊥ chord AD [Given]
seg MT ⊥chord AC
∴ chord AD ≅ chord AC [Chords of a circle equidistant from the centre are congruent]
ii. Theorem used for solving the problem:
The chords which are equidistant from the centre are equal in length.
iii. Test of congruency useful in solving the above problem is AAS ISAAI test of congruency.

Question 3.
i. Draw segment AB. Draw perpendicular bisector l of the segment AB. Take point P on the line l as centre,
PA as radius and draw a circle. Observe that the circle passes through point B also. Find the reason.
ii. Taking any other point Q on the line l, if a circle is drawn with centre Q and radius QA, will it pass through B? Think.
iii. How many such circles can be drawn, passing through A and B? Where will their centres lie? (Textbook pg. no. 49)
Solution:

i. Draw the circle with centre P and radius PA.
line l is the perpendicular bisector of seg AB.
Every point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment.
∴ PA = PB … [Perpendicular bisector theorem]
∴ PA = PB = radius
∴ The circle with centre P and radius PA passes through point B.

ii. The circle with any other point Q and radius QA is drawn.
QA = QB = radius … [Perpendicular bisector theorem]
∴ The circle with centre Q and radius QA passes through point B.

iii. We can draw infinite number of circles passing through A and B.
All their centres will lie on the perpendicular bisector of AB (i.e., line l)

Question 4.
i. Take any three non-collinear points. What should be done to draw a circle passing through all these points? Draw a circle through these points.
ii. Is it possible to draw one more circle passing through these three points? Think of it. (Textbook pg. no. 49)
Solution:

i. Let points A, B, C be any three non collinear points.
Draw the perpendicular bisector of seg AB (line l).
∴ Points A and B are equidistant from any point of line l ….(i)[Perpendicular bisector theorem]
Draw the perpendicular bisector of seg BC (line m) to intersect line l at point P.
∴ Points B and C are equidistant from any point of line m ….(ii) [Perpendicular bisector theorem]
∴ PA = PB …[From (i)]
PB = PC … [From (ii)]
∴ PA = PB = PC = radius
∴ With PA as radius the required circle is drawn through points A, B, C.
ii. It is not possible to draw more than one circle passing through these three points.

Question 5.
Take 3 collinear points D, E, F. Try to draw a circle passing through these points. If you are not able to draw a circle, think of the reason. (Textbook pg. no. 49)
Solution:

Let D, E, F be the collinear points.
The perpendicular bisector of DE and EF drawn (i.e., line l and line m) do not intersect at a common point.
∴ There is no single common point (centre of circle) from which a circle can be drawn passing through points D, E and F.
Hence, we cannot draw a circle passing through points D, E and F.

Question 6.
Which theorem do we use in proving that hypotenuse is the longest side of a right angled triangle? (Textbook pg. no. 52)

Solution:
In ∆ABC,
∠ABC = 90°
∴ ∠BAC < 90° and ∠ACB < 90° [Given]
∴ ∠ABC > ∠BAC and ∠ABC > ∠ACB
∴ AC > BC and AC > AB [Side opposite to greater angle is greater]
∴ Hypotenuse is the longest side in right angled triangle.
We use theorem, If two angles of a triangle are not equal, then the side opposite to the greater angle is greater than the side opposite to the smaller angle.

Question 7.
Theorem: Tangent segments drawn from an external point to a circle are congruent
Draw radius AP and radius AQ and complete the following proof of the theorem.
Given: A is the centre of the circle.
Tangents through external point D touch the circle at the points P and Q.
To prove: seg DP ≅ seg DQ
Construction: Draw seg AP and seg AQ.

Solution:
Proof:
In ∆PAD and ∆QAD,
seg PA ≅ [segQA] [Radii of the same circle]
seg AD ≅ seg AD [Common side]

∠APD = ∠AQD = 90° [Tangent theorem]
∴ ∆PAD = ∆QAD [By Hypotenuse side test]
∴ seg DP = seg DQ [c.s.c.t]

Maharashtra Board Class 10 Maths Solutions

Maharashtra State Board Class 9 Maths Solutions Chapter 5 Linear Equations in Two Variables Practice Set 5.1

Question 1.
By using variables x and y form any five linear equations in two variables.
Answer:
The general form of a linear equation in two variables x and y is ax + by + c = 0,
where a, b, c are real numbers and a ≠ 0, b ≠ 0.
Five linear equations in two variables are as follows:
i. 3x + 4y – 12 = 0
ii. 3x – 4y + 12 = 0
iii. 5x + 5y – 6 = 0
iv. 7x + 12y – 11 = 0
v. x – y + 5 = 0

Question 2.
Write five solutions of the equation x + y = 1.
Answer:
i. x = 1, y = 6
ii. x = -1, y = 8
iii. x = 5, y = 2
iv. x = 0, y = 7
v. x = 10, y = -3

Question 3.
Solve the following sets of simultaneous equations.
i. x + y = 4 ; 2x – 5y = 1
ii. 2x + y = 5 ; 3x – y = 5
iii. 3x – 5y = 16; x – 3y= 8
iv. 2y – x = 0; 10x + 15y = 105
v. 2x + 3y + 4 = 0; x – 5y = 11
vi. 2x – 7y = 7; 3x + y = 22
Solution:
i. Substitution Method:
x + y = 4
∴ x = 4 – y …(i)
2x – 5y = 1 ……(ii)
Substituting x = 4 – y in equation (ii),
2(4 – y) – 5y = 1
∴ 8 – 2y – 5y = 1
∴ 8 – 7y = 1
∴ 8 – 1 = 7y
∴ 7 = 7y
∴ y = \(\frac { 7 }{ 7 }\)
∴ y = 1
Substituting y = 1 in equation (i),
x = 4 – 1 = 3
∴ (3,1) is the solution of the given equations.

Alternate method:
Elimination Method:
x + y = 4 …(i)
2x – 5y = 1 ……(ii)
Multiplying equation (i) by 5,
5x + 5y = 20 … (iii)
Adding equations (ii) and (iii),
2x – 5y = 1
+ 5x + 5y = 20
7 = 21
∴ x = \(\frac { 21 }{ 7 }\)
∴ x = 3
Substituting x = 3 in equation (i),
3 + y = 4
∴ y = 4 – 3 = 1
(3,1) is the solution of the given equations.

ii. 2x + y = 5 …(i)
3x – y = 5 …(ii)
Adding equations (i) and (ii),
2x + y = 5
+ 3x – y = 5
5x = 10
∴ x = \(\frac { 10 }{ 5 }\)
∴ x = 2
Substituting x = 2 in equation (i),
2(2) + y = 5
4 + y = 5
∴ y = 5 – 4 = 1
∴ (2, 1) is the solution of the given equations.

iii. 3x – 5y = 16 …(i)
x – 3y = 8
∴x = 8 + 3y …..(ii)
Substituting x = 8 + 3y in equation (i),
3(8 + 3y) – 5y = 16
24 + 9y- 5y = 16
∴4y= 16 – 24
∴ 4y = -8
∴ y = \(\frac { -8 }{ 4 }\)
y = -2
Substituting y = -2 in equation (ii),
x = 8 + 3 (-2)
∴ x = 8 – 6 = 2
∴ (2, -2) is the solution of the given equations.

iv. 2y – x = 0
∴ x = 2y …(i)
10x + 15y = 105 …(ii)
Substituting x = 2y in equation (ii),
10(2y) + 15y = 105
∴ 20y + 15y = 105
∴ 35y = 105
∴ y = \(\frac { 105 }{ 35 }\)
∴ y = 3
Substituting y = 3 in equation (i),
x = 2y
∴ x = 2(3) = 6
∴ (6, 3) is the solution of the given equations.

v. 2x + 3y + 4 = 0 …(i)
x – 5y = 11
∴x = 11 + 5y …(ii)
Substituting x = 11 + 5y in equation (i),
2(11 +5y) + 3y + 4 = 0
∴ 22 + 10y + 3y + 4 = 0
∴ 13y + 26 = 0
∴ 13y = -26
∴y = \(\frac { -26 }{ 13 }\)
∴ y = -2
Substituting y = -2 in equation (ii),
x = 11 + 5y
∴ x = 11 + 5(-2)
∴ x = 11 – 10 = 1
∴ (1, -2) is the solution of the given equations.

vi. 2x – 7y = 7 …(i)
3x + y = 22
∴ y = 22 – 3x ……(ii)
Substituting y = 22 – 3x in equation (i),
2x – 7(22 – 3x) = 7
∴ 2x – 154 + 21x = 7
∴ 23x = 7 + 154
∴ 23x = 161
∴ x = \(\frac { 161 }{ 23 }\)
∴ x = 7
Substituting x = 7 in equation (ii),
y = 22 – 3x
∴ y = 22 – 3(7)
∴ 7 = 22 -21= 1
∴ (7, 1) is the solution of the given equations.

Question 1.
Solve the following equations. (Textbook pg. no. 80)
i. m + 3 = 5
ii. 3y + 8 = 22
iii. \(\frac { x }{ 3 }\) = 2
iv. 2p = p + \(\frac { 4 }{ 9 }\)
Solution:
i. m + 3 = 5
m = 5 – 3
∴m = 2

ii. 3y + 8 = 22
∴ 3y = 22 – 8
∴ 3y = 14
∴ y = \(\frac { 14 }{ 9 }\)

iii. \(\frac { x }{ 3 }\) = 2
∴ x = 2 × 3
∴ x = 6

iv. 2p = p + \(\frac { 4 }{ 9 }\)
∴ 2p – p = \(\frac { 4 }{ 9 }\)
∴ p = \(\frac { 4 }{ 9 }\)

Question 2.
Which number should be added to 5 to obtain 14? (Textbook pg. no. 80)
Solution:
x + 5 = 14
∴ x = 14 – 5
x = 9
∴ 9 + 5 = 14

Question 3.
Which number should be subtracted from 8 to obtain 2? (Textbook pg. no. 80)
Solution:
8 – y = 2
∴ y = 8 – 2
∴ y = 6
∴ 8 – 6 = 2

Question 4.
x + y = 5 and 2x + 2y= 10 are two equations in two variables. Find live different solutions of x + y = 5, verify whether same solutions satisfy the equation 2x + 2y = 10 also. Observe both equations. Find the condition where two equations in two variables have all solutions in common. (Textbook pg. no. 82)
Solution:
Five solutions of x + y = 5 are given below:
(1,4), (2, 3), (3, 2), (4,1), (0, 5)
The above solutions also satisfy the equation 2x + 2y = 10.
∴ x + y = 5 …[Dividing both sides by 2]
∴ If the two equations are the same, then the two equations in two variables have all solutions common.

Question 5.
3x – 4y – 15 = 0 and y + x + 2 = 0. Can these equations be solved by eliminating x ? Is the solution same? (Textbook pg. no. 84)
Solution:
3x – 4y – 15 = 0
∴ 3x – 4y = 15 …(i)
y + x + 2 = 0
∴ x + y = -2 ……(ii)
Multiplying equation (ii) by 3,
3x + 3y = -6 …(iii)
Subtracting equation (iii) from (i),

∴ y = -3
Substituting y = -3 in equation (ii),
∴ x – 3 = -2
∴ x = – 2 + 3
∴ x = 1
∴ (x, y) = ( 1, -3)
Yes, the given equations can be solved by eliminating x. Also, the solution will remain the same.

Maharashtra Board Class 9 Maths Solutions

Maharashtra State Board Class 9 Maths Solutions Chapter 4 Ratio and Proportion Problem Set 4

Question 1.
Select the appropriate alternative answer for the following questions.

i . If 6 : 5 = y : 20, then what will be the value of y?
(A) 15
(B) 24
(C) 18
(D) 22.5
Answer:
(B) 24

ii. What is the ratio of 1 mm to 1 cm ?
(A) 1 : 100
(B) 10: 1
(C) 1 : 10
(D) 100: 1
Answer:
(C) 1 : 10

iii. The ages of Jatin, Nitin and Mohasin are 16, 24 and 36 years respectively. What is the ratio of Nitin’s age to Mohasin’s age ?
(A) 3 : 2
(B) 2 : 3
(C) 4 : 3
(D) 3 : 4
Answer:
(B) 2 : 3

iv. 24 bananas were distributed between Shubham and Anil in the ratio 3 : 5, then how many bananas did Shubham get?
(A) 8
(B) 15
(C) 12
(D) 9
Answer:
(D) 9

v. What is the mean proportional of 4 and 25?
(A) 6
(B) 8
(C) 10
(D) 12
Answer:
(C) 10

Hints:
i . \(\frac{6}{5}=\frac{y}{20}\)
∴ \(\quad y=\frac{6 \times 20}{5}=24\)

ii. \(\frac{1 \mathrm{mm}}{1 \mathrm{cm}}=\frac{1 \mathrm{mm}}{10 \mathrm{mm}}=\frac{1}{10}=1 : 10\)

iii. \(\frac{\text { Age of Nitin }}{\text { Age of Mohasin }}=\frac{24}{36}=\frac{12 \times 2}{12 \times 3}\)
\(\frac { 2 }{ 3 }\) = 2 : 3

iv. 3x + 5x = 24
∴ 8x = 24
∴ x = 3
∴ Number of bananas with Shubham = 3x = 9

v. x² = 4 x 25
∴ x² = 100
∴ x = 10

Question 2.
For the following numbers write the ratio of first number to second number in the reduced form. [1 Mark each]
i. 21,48
ii. 36,90
iii. 65,117
iv. 138,161
v. 114,133
Solution:
i. 21,48
\(\text { Ratio }=\frac{21}{48}=\frac{3 \times 7}{3 \times 16}=\frac{7}{16}=7 : 16\)
ii. 36,90
\(\text { Ratio }=\frac{36}{90}=\frac{18 \times 2}{18 \times 5}=\frac{2}{5}=2 : 5\)
iii. 65,117
\(\text { Ratio }=\frac{65}{117}=\frac{13 \times 5}{13 \times 9}=\frac{5}{9}=5 : 9\)
iv. 138,161
\(\text { Ratio }=\frac{138}{161}=\frac{23 \times 6}{23 \times 7}=\frac{6}{7}=6 : 7\)
v. 114,133
\(\text { Ratio }=\frac{114}{113}=\frac{19 \times 6}{19 \times 7}=\frac{6}{7}=6 : 7\)

Question 3.
Write the following ratios in the reduced form.
i. Radius to the diameter of a circle.
ii. The ratio of diagonal to the length of a rectangle, having length 4 cm and breadth 3 cm.
iii. The ratio of numbers denoting perimeter to area of a square, having side 4 cm.
Solution:
i. Radius to the diameter of a circle.
Let radius of the circle be r
then, diameter = 2 x radius = 2r
Ratio of radius to diameter of circle

∴ Ratio of radius to diameter of circle is 1 : 2.

ii. The ratio of diagonal to the length of a rectangle, having length 4 cm and breadth 3 cm.

Let □ ABCD be a rectangle.
In ∆ABC, ∠B = 90°
AC² = AB² + BC² … [Pythagoras theorem]
= 4² + 3² = 16 + 9
∴ AC² = 25
AC = 5 … [Taking square root on both side]
The ratio of diagonal to the length of a rectangle = \(\frac { AC }{ AB }\)
= \(\frac { 5 }{ 4 }\)
= 5 : 4
∴ The ratio of diagonal to the length of a rectangle is 5 : 4

iii. The ratio of numbers denoting perimeter to area of a square, having side 4 cm. side of square = 4cm
Perimeter of square = 4 x side = 4 x 4 = 16 cm
Area of square = (side)² = (4)² = 16 cm²
Ratio of numbers denoting perimeter to area of square

∴ The ratio of numbers denoting perimeter to area of a square is 1 : 1.

Question 4.
Check whether the following numbers are in continued proportion.
i. 2, 4, 8
ii. 1, 2, 3
iii. 9, 12, 16
iv. 3, 5, 8
Solution:
If a, b, c are in continued proportion then b² = ac.
i. 2, 4, 8
Let, a = 2, b = 4 and c = 8
Here, b² = 4² = 16
ac = 2 x 8 = 16
∴ b² = ac
∴ 2, 4,8 are in continued proportion.

ii. 1, 2, 3
Let, a = 1, b = 2 and c = 3
Here, b² = 2² = 4
ac = 1 x 3 = 3
∴ b² ≠ ac
∴ 1, 2,3 are not in continued proportion.

iii. 9, 12, 16
Let, a = 9, b = 12 and c = 16
Here, b² = 12² = 144
ac = 9 x 16 = 144
∴ b² = ac
∴ 9, 12, 16 are in continued proportion.

iv. 3, 5, 8
Let, a = 3, b = 5 and c = 8
Here, b² = 5² = 25
ac = 3 x 8 = 24
∴ b² ≠ ac
∴ 3, 5, 8 are not in continued proportion.

Question 5.
a, b, c are in continued proportion. If a = 3 and c = 27, then find b.
Solution:
a, b, c are in continued proportion. …[Given]
∴ b² = ac
∴ b² = 3 x 27 …[∵ a = 3 and c = 27]
∴ b² = 81
∴ b = 9 …[Taking square root of both sides]

Question 6.
Convert the following ratios into percentages.

Solution:
i. Let 37 : 500 = x%

∴ 37 : 500 = 7.4%

Question 7.
Write the ratio of first quantity to second quantity in the reduced form.
i. 1024 MB, 1.2 GB [(1024 MB = 1GB)]
ii. ₹ 17, ₹ 25 and 60 paise
iii. 5 dozen, 120 units
iv. 4 sq.m, 800 sq.cm
v. 1.5 kg, 2500 gm
Solution:
i. 1024 MB, 1.2 GB
1024 MB = 1 GB

ii. ₹ 17, ₹ 25 and 60 paise
₹ 17 = 17 x 100 paise = 1700 paise
₹ 25 and 60 paise = (25x 100) paise + 60 paise
= (2500 + 60) paise
= 2560 paise

iii. 5 dozen, 120 units
5 dozen = 5 x 12 units = 60 units

iv. 4 sq.m, 800 sq.cm
4 sq.m = 4 x 10000 sq.cm = 40000 sq.cm

v. 1.5 kg, 2500 gm
1.5 kg = 1.5 x 1000 gm = 1500gm

Question 8.
If \(\frac { a }{ b }\) = \(\frac { 2 }{ 3 }\), then find the values of the following expressions.

Solution:

Question 9.
If a, b, c, d are in proportion, then prove that

Solution:
a, b, c are in continued proportion. …[Given]

Question 10.
If a, b, c are ¡n continued proportion, then prove that

Solution:
a, b, c are in continued proportion. … [Given]
∴ \(\frac { a }{ b }\) = \(\frac { b }{ c }\)
Let \(\frac { a }{ b }\) = \(\frac { b }{ c }\) = k
∴ b = ck
∴ a = bk
= (ck)k . .. [From (j)]
a = ck² …(ii)

Question 11.
Solve:
\( \frac{12 x^{2}+18 x+42}{18 x^{2}+12 x+58}=\frac{2 x+3}{3 x+2}\)
Solution:


∴ 29(2x + 3) = 21 (3x + 2)
∴ 5 + 87= 63x + 42
∴ 87 – 42 = 63x – 58x
∴ 45 = 5x
∴ x = 9
∴ x = 9 is the solution of the given equation.

Question 12.
If \( \frac{2 x-3 y}{3 z+y}=\frac{z-y}{z-x}=\frac{x+3 z}{2 y-3 x}\) ,then prove that every ratio = \(\frac { x }{ y }\).
Solution:

Question 13.
If \(\frac{b y+c z}{b^{2}+c^{2}}=\frac{c z+a x}{c^{2}+a^{2}}=\frac{a x+b y}{a^{2}+b^{2}}\), then prove \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
Solution:

Question 1.
Take 5 pieces of card paper. Write the following statements, one on each paper.

a, b, c, d are positive numbers and \(\frac{a}{b}=\frac{c}{d}\) is given. Which of the above statements are true or false, write at the back of each card, if false explain why. (Textbook pg. no. 70)
Answer:
i. True
ii. True
iii. False
Here, numerator and denominator are multiplied by two different numbers a and b.
iv. False
Here, different numbers a and b are subtracted from numerator and denominator.
v. True

Question 2.
In the following activity, the values of a and b can be changed. That is by changing a : b we can create many examples. Teachers should give lot of practice to the students and encourage them to construct their own examples. (Textbook pg. no. 70)

Question 3.
Observe the political map of India from a Geography text book. Study the scale of this map.
From the given scale find the straight line distances between various cities like
i. New Delhi to Bengaluru
ii. Mumbai to Kolkata
iii. Jaipur to Bhuvaneshvar. (Textbook pg. no. 77)
[Students should attempt the above activity on their own.]

Maharashtra Board Class 9 Maths Solutions

Maharashtra State Board Class 10 Maths Solutions Chapter 2 Pythagoras Theorem Problem Set 2

Question 1.
Some questions and their alternative answers are given. Select the correct alternative. [1 Mark each]

i. Out of the following which is the Pythagorean triplet?
(A) (1,5,10)
(B) (3,4,5)
(C) (2,2,2)
(D) (5,5,2)
Answer: (B)
Hint: Refer Practice set 2.1 Q.1 (i)

ii. In a right angled triangle, if sum of the squares of the sides making right angle is 169, then what is the length of the hypotenuse?
(A) 15
(B) 13
(C) 5
(D) 12
Answer: (B)
Hint:

iii. Out of the dates given below which date constitutes a Pythagorean triplet?
(A) 15/08/17
(B) 16/08/16
(C) 3/5/17
(D) 4/9/15
Answer: (A)
Hint:

iv. If a, b, c are sides of a triangle and a² + b² = c², name the type of the triangle.
(A) Obtuse angled triangle
(B) Acute angled triangle
(C) Right angled triangle
(D) Equilateral triangle
Answer: (C)

v. Find perimeter of a square if its diagonal is 10\(\sqrt { 2 }\) cm.
(A) 10 cm
(B) 40\(\sqrt { 2 }\) cm
(C) 20 cm
(D) 40 cm
Answer: (D)
Hint:

vi. Altitude on the hypotenuse of a right angled triangle divides it in two parts of lengths 4 cm and 9 cm. Find the length of the altitude.
(A) 9 cm
(B) 4 cm
(C) 6 cm
(D) 2\(\sqrt { 6 }\)
Answer: (C)
Hint:

vii. Height and base of a right angled triangle are 24 cm and 18 cm find the length of its hypotenuse.
(A) 24 cm
(B) 30 cm
(C) 15 cm
(D) 18 cm
Answer: (B)
Hint:

viii. In ∆ABC, AB = 6\(\sqrt { 3 }\) cm, AC = 12 cm, BC = 6 cm. Find measure of ∠A.
(A) 30°
(B) 60°
(C) 90°
(D) 45°
Answer: (A)
Hint:

Question 2.
Solve the following examples.
i. Find the height of an equilateral triangle having side 2a.
ii. Do sides 7 cm, 24 cm, 25 cm form a right angled triangle? Give reason.
iii. Find the length of a diagonal of a rectangle having sides 11 cm and 60 cm.
iv. Find the length of the hypotenuse of a right angled triangle if remaining sides are 9 cm and 12 cm.
v. A side of an isosceles right angled triangle is x. Find its hypotenuse.
vi. In ∆PQR, PQ = \(\sqrt { 8 }\), QR = \(\sqrt { 5 }\), PR = \(\sqrt { 3 }\) . Is ∆PQR a right angled triangle? If yes, which angle is of 90°?
Solution:
i. Let ∆ABC be the given equilateral triangle.

∴ ∠B = 60° [Angle of an equilateral triangle]
Let AD ⊥BC, B – D – C.
In ∆ABD, ∠B = 60°, ∠ADB = 90°
∴ ∠BAD = 30° [Remaining angle of a triangle]
∴ ∆ABD is a 30° – 60° – 90° triangle.
∴ AD = \(\frac{\sqrt{3}}{2}\) AB [Side opposite to 60°]
= \(\frac{\sqrt{3}}{2}\) × 2a
= a\(\sqrt { 3 }\) units
The height of the equilateral triangle is a\(\sqrt { 3 }\) units.

ii. The sides of the triangle are 7 cm, 24 cm and 25 cm.
The longest side of the triangle is 25 cm.
∴ (25)² = 625
Now, sum of the squares of the remaining sides is,
(7)² + (24)² = 49 + 576
= 625
∴ (25)² = (7)² + (24)²
∴ Square of the longest side is equal to the sum of the squares of the remaining two sides.
∴ The given sides will form a right angled triangle. [Converse of Pythagoras theorem]

iii. Let ꠸ABCD be the given rectangle.
AB = 11 cm, BC = 60 cm

In ∆ABC, ∠B = 90° [Angle of a rectangle]
∴ AC² = AB² + BC² [Pythagoras theorem]
= 112 + 602
= 121 +3600
= 3721
∴ AC = \(\sqrt { 3721 }\) [Taking square root of both sides]
= 61 cm
The length of the diagonal of the rectangle is 61 cm.
∴ The length of the diagonal of the rectangle is 61 cm.

iv. Let ∆PQR be the given right angled triangle.
In ∆PQR, ∠Q = 90°

∴ PR² = PQ² + QR² [Pythagoras theorem]
= 9² + 12²
= 81 + 144
= 225
∴ PR = \(\sqrt { 225 }\) [Taking square root of both sides]
= 15 cm
∴ The length of the hypotenuse of the right angled triangle is 15 cm.

v. Let ∆PQR be the given right angled isosceles triangle.
PQ = QR = x.

In ∆PQR, ∠Q = 90° [Pythagoras theorem]
∴ PR² = PQ² + QR²
= x² + x²
= 2x²
∴ PR = \(\sqrt{2 x^{2}}\) [Taking square root of both sides]
= x \(\sqrt { 2 }\) units
∴ The hypotenuse of the right angled isosceles triangle is x \(\sqrt { 2 }\) units.
∴ The hypotenuse of the right angled isosceles triangle is x \(\sqrt { 2 }\) units.

vi. Longest side of ∆PQR = PQ = \(\sqrt { 8 }\)
∴ PQ² = (\(\sqrt { 8 }\))² = 8
Now, sum of the squares of the remaining sides is,

QR² + PR² = (\(\sqrt { 5 }\))² + (\(\sqrt { 3 }\))²
= 5 + 3
= 8
∴ PQ² = QR² + PR²
∴ Square of the longest side is equal to the sum of the squares of the remaining two sides.
∴ ∆PQR is a right angled triangle. [Converse of Pythagoras theorem]
Now, PQ is the hypotenuse.
∴∠PRQ = 90° [Angle opposite to hypotenuse]
∴ ∆PQR is a right angled triangle in which ∠PRQ is of 90°.

Question 3.
In ∆RST, ∠S = 90°, ∠T = 30°, RT = 12 cm, then find RS and ST.
Solution:
in ∆RST, ∠S = 900, ∠T = 30° [Given]
∴ ∠R = 60° [Remaining angle of a triangle]
∴ ∆RST is a 30° – 60° – 90° triangle.

∴ RS = \(\frac { 1 }{ 2 } \) RT [Side opposite to 30°]
= \(\frac { 1 }{ 2 } \) × 12 = 6cm
Also, ST = \(\frac{\sqrt{3}}{2}\) RT [Side opposite to 60°]
= \(\frac{\sqrt{3}}{2}\) × 12 = 6 \(\sqrt { 3 }\) cm
∴ RS = 6 cm and ST = 6 \(\sqrt { 3 }\) cm

Question 4.
Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq. cm.
Solution:
Let ꠸ABCD be the given rectangle.
BC = 16cm
Area of rectangle = length × breadth
Area of ꠸ABCD = BC × AB

∴ 192 = I6 × AB
∴ AB = \(\frac { 192 }{ 16 } \)
= 12cm
Now, in ∆ABC, ∠B = 90° [Angle of a rectangle]
∴ AC² = AB² + BC² [Pythagoras theorem]
= 122 + 162
= 144 + 256
=400
∴ AC = \(\sqrt { 400 }\) [Taking square root of both sides]
= 20cm
∴ The diagonal of the rectangle is 20 cm.

Question 5.
Find the length of the side and perimeter of an equilateral triangle whose height is \(\sqrt { 3 }\) cm.
Solution:
Let ∆ABC be the given equilateral triangle.
∴ ∠B = 60° [Angle of an equilateral triangle]

AD ⊥ BC, B – D – C.
In ∆ABD, ∠B =60°, ∠ADB = 90°
∴ ∠BAD = 30° [Remaining angle of a triangle]
∴ ∆ABD is a 30° – 60° – 90° triangle.
∴ AD = \(\frac{\sqrt{3}}{2}\)AB [Side opposite to 600]
∴ \(\sqrt { 3 }\) = \(\frac{\sqrt{3}}{2}\)AB
∴ AB = \(\frac{2 \sqrt{3}}{\sqrt{3}}\)
∴ AB = 2cm
∴ Side of equilateral triangle = 2cm
Perimeter of ∆ABC = 3 × side
= 3 × AB
= 3 × 2
= 6cm
∴ The length of the side and perimeter of the equilateral triangle are 2 cm and 6 cm respectively.

Question 6.
In ∆ABC, seg AP is a median. If BC = 18, AB2 + AC2 = 260, find AP.
Solution:
PC = \(\frac { 1 }{ 2 } \) BC [P is the midpoint of side BC]
= \(\frac { 1 }{ 2 } \) × 18 = 9cm

in ∆ABC, seg AP is the median,
Now, AB² + AC² = 2 A² + 2 PC² [Apollonius theorem]
∴ 260 = 2 AP² + 2 (9)²
∴ 130 = AP² + 81 [Dividing both sides by 2]
∴ AP² = 130 – 81
∴ AP² = 49
∴ AP = \(\sqrt { 49 }\) [Taking square root of both sides]
∴ AP = 7 units

Question 7.
∆ABC is an equilateral triangle. Point P is on base BC such that PC = \(\frac { 1 }{ 3 } \) BC, if AB = 6 cm find AP.

Given: ∆ABC is an equilateral triangle.
PC = \(\frac { 1 }{ 3 } \) BC, AB = 6cm.
To find: AP
Consttuction: Draw seg AD ± seg BC, B – D – C.
Solution:
∆ABC is an equilateral triangle.
∴ AB = BC = AC = 6cm [Sides of an equilateral triangle]
pc = \(\frac { 1 }{ 3 } \) BC [Given]
= \(\frac { 1 }{ 3 } \) (6)
∴ PC = 2cm
In ∆ADC,
∠D = 90° [Construction]
∠C = 60° [Angle of an equilateral triangle]
∠DAC = 30° [Remaining angle of a triangle]
∴ ∆ ADC is a 30° – 60° – 90° triangle.
∴ AD = \(\frac{\sqrt{3}}{2}\) AC [Side opposite to 60°]
∴ AD = \(\frac{\sqrt{3}}{2}\) (6)
∴ AD = 3 \(\sqrt { 3 }\)cm
CD = \(\frac { 1 }{ 2 } \) AC [Side opposite to 30°]
∴ CD = \(\frac { 1 }{ 2 } \) (6)
∴ CD = 3cm
Now DP + PC = CD [D – P – C]
∴ DP + 2 = 3
∴ DP = 1cm
In ∆ADP,
∠ADP = 900
AP² = AD² + DP² [Pythagoras theorem]
∴ AP² = (3\(\sqrt { 3 }\))² + (1)²
∴ AP² = 9 × 3 + 1 = 27 + 1
∴ AP² = 28
∴ AP = \(\sqrt { 28 }\)
∴ AP = \(\sqrt{4 \times 7}\)
∴ AP = 2 \(\sqrt { 7 }\)cm

Question 8.
From the information given in the adjoining figure, prove that
PM = PN = \(\sqrt { 3 }\) × a

Solution:
Proof:
In ∆PMR,
QM = QR = a [Given]
∴ Q is the midpoint of side MR.
∴ seg PQ is the median.
∴ PM² + PR² = 2PQ² + 2QM² [Apollonius theorem]

∴ PM² + a² = 2a² + 2a²
∴ PM² + a² = 4a²
∴ PM² = 3a²
∴ PM,= \(\sqrt { 3 }\)a (i) [Taking square root of both sides]
SimlarIy, in ∆PNQ,
R is the midpoint of side QN.
∴ seg PR is the median.
∴ PN² + PQ² = 2 PR² + 2 RN² [Apollonius theorem]

∴ PN² + a² = 2a² + 2a²
∴ PN² + a² = 4a²
∴ PN² = 3a²
∴ PN = \(\sqrt { 3 }\)a (ii) [Taking square root of both sides]
∴ PM = PN = \(\sqrt { 3 }\) a [From (i) and (ii)]

Question 9.
Prove that the sum of the squares of the diagonals of a parallelogram ¡s equal to the sum of the squares of its sides.

Given: ꠸ABCD is a parallelogram, diagonals AC and BD intersect at point M.
To prove: AC² + BD² = AB² + BC² + CD² + AD²
Solution:
Proof:
꠸ABCD is a parallelogram.
∴ AB = CD and BC = AD (i) [Opposite sides of a parallelogram]
AM = \(\frac { 1 }{ 2 } \) AC and BM = \(\frac { 1 }{ 2 } \) BD (ii) [Diagonals of a parallelogram bisect each other]
∴ M is the midpoint of diagonals AC and BD. (iii)
In ∆ABC.
seg BM is the median. [From (iii)]
AB² + BC² = 2AM² + 2BM² (iv) [Apollonius theorem]
∴ AB² + BC² = 2(\(\frac { 1 }{ 2 } \) AC)² + 2(\(\frac { 1 }{ 2 } \) BD)² [From (ii) and (iv)]
∴ AB² + BC² = 2 × \(\frac{\mathrm{BD}^{2}}{4}+2 \times \frac{\mathrm{AC}^{2}}{4}\)
∴ AB² + BC² = \(\frac{B D^{2}}{2}+\frac{A C^{2}}{2}\)
∴ 2AB² + 2BC² = BD² + AC² [Multiplying both sides by 2]
∴ AB² + AB² + BC² + BC² = BD² + AC²
∴ AB² + CD² + BC² + AD² = BD² + AC² [From(i)]
i.e. AC² + BD² = AB² + BC² + CD² + AD²

Question 10.
Pranali and Prasad started walking to the East and to the North respectively, from the same point and at the same speed. After 2 hours distance between them was 15\(\sqrt { 2 }\) km. Find their speed per hour.
Solution:
Suppose Pranali and Prasad started walking from point A, and reached points B and C respectively after 2 hours.
Distance between them = BC = 15\(\sqrt { 2 }\) km

Since, their speed is same, both travel the same distance in the given time.
∴ AB = AC
Let AB = AC = x km (i)
Now, in ∆ ABC, ∠A = 90°
∴ BC² = AB² + AC² [Pythagoras theorem]
∴ (15\(\sqrt { 2 }\))² = x² + x² [From (i)]
∴ 225 × 2 = 2 x²
∴ x² = 225
∴ x = \(\sqrt { 225 }\) [Taking square root of both sides]
∴ x = 15 km
∴ AB = AC = 15km
\(\text { Now, speed }=\frac{\text { distance }}{\text { time }}=\frac{15}{2}\)
= 7.5 km/hr
∴ The speed of Pranali and Prasad is 7.5 km/hour.

Question 11.
In ∆ABC, ∠BAC = 90°, seg BL and seg CM are medians of ∆ABC. Then prove that 4 (BL² + CM²) = 5 BC².

Given : ∠BAC = 90°
seg BL and seg CM are the medians.
To prove: 4(BL² + CM²) = 5BC²
Solution:
Proof:
In ∆BAL, ∠BAL 90° [Given]
∴ BL² = AB² + AL² (i) [Pythagoras theorem]
In ∆CAM, ∠CAM = 90° [Given]
∴ CM² = AC² + AM² (ii) [Pythagoras theorem]
∴ BL² + CM² = AB² + AC² + AL² + AM² (iii) [Adding (i) and (ii)]
Now, AL = \(\frac { 1 }{ 2 } \) AC and AM = \(\frac { 1 }{ 2 } \) AB (iv) [seg BL and seg CM are the medians]
∴ BL² + CM²
= AB² + AC² + (\(\frac { 1 }{ 2 } \) AC)² + (\(\frac { 1 }{ 2 } \) AB)² [From (iii) and (iv)]
\(=A B^{2}+A C^{2}+\frac{A C^{2}}{4}+\frac{A B^{2}}{4}\)
\(=A B^{2}+\frac{A B^{2}}{4}+A C^{2}+\frac{A C^{2}}{4}\)
\(=\frac{5 \mathrm{AB}^{2}}{4}+\frac{5 \mathrm{AC}^{2}}{4}\)
∴ BL² + CM² = \(\frac { 5 }{ 4 } \) (AB² + AC²)
∴ 4(BL² + CM²) = 5(AB² + AC²) (v)
In ∆BAC, ∠BAC = 90° [Given]
∴ BC² = AB² + AC² (vi) [Pythagoras theorem]
∴ 4(BL² + CM²) = 5BC² [From (v) and (vi)]

Question 12.
Sum of the squares of adjacent sides of a parallelogram is 130 cm and length of one of its diagonals is 14 cm. Find the length of the other diagonal.
Solution:
Let ꠸ABCD be the given
parallelogram and its diagonals AC and BD intersect at point M.

∴ AB² + AD² = 130cm, BD = 14cm
MD = \(\frac { 1 }{ 2 } \) BD (i) [Diagonals of a parallelogram bisect each other]
= \(\frac { 1 }{ 2 } \) × 14 = 7 cm
In ∆ABD, seg AM is the median. [From (i)]
∴ AB² + = 2AM² + 2MD² [Apollonius theorem]
∴ 130 = 2 AM² + 2(7)²
∴ 65 = AM² +49 [Dividing both sides by 2]
∴ AM² = 65 – 49
∴ AM² = 16 [Taking square root of both sides]
∴ AM = \(\sqrt { 16 }\)
= 4cm
Now, AC =2 AM [Diagonals of a parallelogram bisect each other]
2 × 4 = 8 cm
∴ The length of the other diagonal of the parallelogram is 8 cm.

Question 13.
In ∆ABC, seg AD ⊥ seg BC and DB = 3 CD. Prove that: 2 AB2 = 2 AC2 + BC2.
Given: seg AD ⊥ seg BC
DB = 3CD
To prove: 2AB² = 2AC² + BC²

Solution:
DB = 3CD (i) [Given]
In ∆ADB, ∠ADB = 90° [Given]
∴ AB² = AD² + DB² [Pythagoras theorem]
∴ AB² = AD² + (3CD)² [From (i)]
∴ AB² = AD² + 9CD² (ii)
In ∆ADC, ∠ADC = 90° [Given]
∴ AC² = AD² + CD² [Pythagoras theorem]
∴ AD² = AC² – CD² (iii)
AB² = AC² – CD² + 9CD² [From (ii) and(iii)]
∴ AB² = AC² + 8CD² (iv)
CD + DB = BC [C – D – B]
∴ CD + 3CD = BC [From (i)]
∴ 4CD = BC
∴ CD = \(\frac { BC }{ 4 } \) (v)
AB² = AC² + 8(\(\frac { BC }{ 4 } \))² [From (iv) and (v)]
∴ AB² = AC² + 8 × \(\frac{\mathrm{BC}^{2}}{16}\)
∴ AB² = AC² + \(\frac{\mathrm{BC}^{2}}{2}\)
∴ 2AB² = 2AC² + BC² [Multiplying both sides by 2]

Question 14.
In an isosceles triangle, length of the congruent sides is 13 em and its.base is 10 cm. Find the distance between the vertex opposite to the base and the centroid.
Given: ∆ABC is an isosceles triangle.
G is the centroid.
AB = AC = 13 cm, BC = 10 cm.
To find: AG
Construction: Extend AG to intersect side BC at D, B – D – C.

Solution:
Centroid G of ∆ABC lies on AD
∴ seg AD is the median. (i)
∴ D is the midpoint of side BC.
∴ DC = \(\frac { 1 }{ 2 } \) BC
= \(\frac { 1 }{ 2 } \) × 10 = 5
In ∆ABC, seg AD is the median. [From (i)]
∴ AB² + AC² = 2 AD² + 2 DC² [Apollonius theorem]
∴ 13² + 13² = 2 AD² + 2 (5)²
∴ 2 × 13² = 2 AD² + 2 × 25
∴ 169 = AD² + 25 [Dividing both sides by 2]
∴ AD² = 169 – 25
∴ AD² = 144
∴ AD = \(\sqrt { 144 }\) [Taking square root of both sides]
= 12 cm
We know that, the centroid divides the median in the ratio 2 : 1.
∴ \(\frac { AG }{ GD } \) = \(\frac { 2 }{ 1 } \)
∴ \(\frac { GD }{ AG } \) = \(\frac { 1 }{ 2 } \) [By invertendo]
∴ \(\frac { GD+AG }{ AG } \) = \(\frac { 1+2 }{ 2 } \) [By componendo]
∴ \(\frac { AD }{ AG } \) = \(\frac { 3 }{ 2 } \) [A – G – D]
∴ \(\frac { 12 }{ AG } \) = \(\frac { 3 }{ 2 } \)
∴ AG = \(\frac{12 \times 2}{3}\)
= 8cm
∴ The distance between the vertex opposite to the base and the centroid is 8 cm.

Question 15.
In a trapezium ABCD, seg AB || seg DC, seg BD ⊥ seg AD, seg AC ⊥ seg BC. If AD = 15, BC = 15 and AB = 25, find A (꠸ABCD).

Construction: Draw seg DE ⊥ seg AB, A – E – B
and seg CF ⊥ seg AB, A – F- B.
Solution:
In ∆ ACB, ∠ACB = 90° [Given]
∴ AB2 = AC2 + BC2 [Pythagoras theorem]
∴ 252 = AC2 + 152
∴ AC2 = 625 – 225
= 400

∴ AC = \(\sqrt { 400 }\) [Taking square root of both sides]
= 20 units
Now, A(∆ABC) = \(\frac { 1 }{ 2 } \) × BC × AC
Also, A(∆ABC) = \(\frac { 1 }{ 2 } \) × AB × CF
∴ \(\frac { 1 }{ 2 } \) × BC × AC = \(\frac { 1 }{ 2 } \) × AB × CF
∴ BC × AC = AB × CF
∴ 15 × 20 = 25 × CF
∴ CF = \(\frac{15 \times 20}{25}\) = 12 units
In ∆CFB, ∠CFB 90° [Construction]
∴ BC² = CF² + FB² [Pythagoras theorem]
∴ 15² = 12² + FB²
∴ FB² = 225 – 144
∴ FB² = 81
∴ FB = \(\sqrt { 81 }\) [Taking square root of both sides]
= 9 units
Similarly, we can show that, AE = 9 units
Now, AB = AE + EF + FB [A – E – F, E – F – B]
∴ 25 = 9 + EF + 9
∴ EF = 25 – 18 = 7 units
In ꠸CDEF,
seg EF || seg DC [Given, A – E – F, E – F – B]
seg ED || seg FC [Perpendiculars to same line are parallel]
∴ ꠸CDEF is a parallelogram.
∴ DC = EF 7 units [Opposite sides of a parallelogram]
A(꠸ABCD) = \(\frac { 1 }{ 2 } \) × CF × (AB + CD)
= \(\frac { 1 }{ 2 } \) × 12 × (25 + 7)
= \(\frac { 1 }{ 2 } \) × 12 × 32
∴ A(꠸ABCD) = 192 sq. units

Question 16.
In the adjoining figure, ∆PQR is an equilateral triangle. Point S is on seg QR such that QS = \(\frac { 1 }{ 3 } \) QR. Prove that: 9 PS² = 7 PQ².

Given: ∆PQR is an equilateral triangle.
QS = \(\frac { 1 }{ 3 } \) QR
To prove: 9PS² = 7PQ²
Solution:
Proof:
∆PQR is an equilateral triangle [Given]
∴ ∠P = ∠Q = ∠R = 60° (i) [Angles of an equilateral triangle]
PQ = QR = PR (ii) [Sides of an equilateral triangle]
In ∆PTS, ∠PTS = 90° [Given]
PS² = PT² + ST² (iii) [Pythagoras theorem]
In ∆PTQ,
∠PTQ = 90° [Given]
∠PQT = 60° [From (i)]
∴ ∠QPT = 30° [Remaining angle of a triangle]
∴ ∆PTQ is a 30° – 60° – 90° triangle
∴ PT = \(\frac{\sqrt{3}}{2}\) PQ (iv) [Side opposite to 60°]
QT = \(\frac { 1 }{ 2 } \) PQ (v) [Side opposite to 30°]
QS + ST = QT [Q – S – T]
∴ \(\frac { 1 }{ 3 } \) QR + ST = \(\frac { 1 }{ 2 } \) PQ [Given and from (v)]
∴ \(\frac { 1 }{ 3 } \) PQ + ST = \(\frac { 1 }{ 2 } \) PQ [From (ii)]
∴ ST = \(\frac { PQ }{ 2 } \) – \(\frac { PQ }{ 3 } \)
∴ ST = \(\frac { 3PQ-2PQ }{ 6 } \)
∴ ST = \(\frac { PQ }{ 6 } \) (vi)
\(\mathrm{PS}^{2}=\left(\frac{\sqrt{3}}{2} \mathrm{PQ}\right)^{2}+\left(\frac{\mathrm{PQ}}{6}\right)^{2}\) [From (iii), (iv) and (vi)]
∴ \(\mathrm{PS}^{2}=\frac{3 \mathrm{PQ}^{2}}{4}+\frac{\mathrm{PQ}^{2}}{36}\)
∴ \(\mathrm{PS}^{2}=\frac{27 \mathrm{PQ}^{2}}{36}+\frac{\mathrm{PQ}^{2}}{36}\)
∴\(\mathrm{PS}^{2}=\frac{28 \mathrm{PQ}^{2}}{36}\)
∴PS² = \(\frac { 7 }{ 3 } \) PQ²
∴ 9PS² = 7 PQ²

Question 17.
Seg PM is a median of APQR. If PQ = 40, PR = 42 and PM = 29, find QR.
Solution:
In ∆PQR, seg PM is the median. [Given]
∴ M is the midpoint of side QR.
∴ PQ² + PR² = 2 PM² + 2 MR² [Apollonius theorem]

∴ 40² + 42² = 2 (29)² + 2 MR²
∴ 1600 + 1764 = 2 (841) + 2 MR²
∴ 3364 = 2 (841) + 2 MR²
∴ 1682 = 841 +MR² [Dividing both sides by 2]
∴ MR² = 1682 – 841
∴ MR² = 841
∴ MR = \(\sqrt { 841 }\) [Taking square root of both sides]
= 29 units
Now, QR = 2 MR [M is the midpoint of QR]
= 2 × 29
∴ QR = 58 units

Question 18.
Seg AM is a median of ∆ABC. If AB = 22, AC = 34, BC = 24, find AM.
Solution:
In ∆ABC, seg AM is the median. [Given]
∴ M is the midpoint of side BC.

∴ MC = \(\frac { 1 }{ 2 } \) BC
= \(\frac { 1 }{ 2 } \) × 24 = 12 units
Now, AB² + AC² = 2 AM² + 2 MC² [Apollonius theorem]
∴ 22² + 34² = 2 AM² + 2 (12)²
∴ 484 + 1156 = 2 AM² + 2 (144)
∴ 1640 = 2 AM² + 2 (144)
∴ 820 = AM² + 144 [Dividing both sides by 2]
∴ AM² = 820 – 144
∴ AM² = 676
∴ AM = \(\sqrt { 676 }\) [Taking square root of both sides]
∴ AM = 26 units

Maharashtra Board Class 10 Maths Solutions

Maharashtra Board Class 9 Maths Solutions Chapter 4 Ratio and Proportion Practice Set 4.3

Question 1.
If \(\frac { a }{ b }\) = \(\frac { 7 }{ 3 }\), then find the aIues of the following ratios.

Solution:

Question 2.
If \(\frac{15 a^{2}+4 b^{2}}{15 a^{2}-4 b^{2}}=\frac{47}{7}\), then find the value of the following ratios.

Solution:

Question 3.
If \(\frac{3 a+7 b}{3 a-7 b}=\frac{4}{3}\)then find the value of the ratio \(\frac{3 a^{2}-7 b^{2}}{3 a^{2}+7 b^{2}}\).
Solution:

Question 4.
Solve the following equations.

Solution:

This equation is true for x = 0
∴ x = 0 is one of the solutions.
If x ≠ 0, then x² ≠ 0
∴ \(\frac { 1 }{ 12x – 20 }\) = \(\frac { 1 }{ 8x + 12 }\) … [Dividing both sides by x²]
∴ 8x + 12 = 12x – 20
∴ 12 + 20 = 12x – 8x
∴ 32 = 4x
∴ x = 8
∴ x = 0 or x = 8 are the solutions of the given equation.

∴ 21(x – 5) = 4(2x + 3)
∴ 21x – 105 = 8x + 12
∴ 21x – 8x = 12 + 105
∴ 13x = 117
∴ x = 9
∴ x = 9 ¡s the solution of the given equation.

∴ 9(4x + 1) = 25(x + 3)
∴36x + 925x + 75
∴ 36x – 25 = 75 – 9
∴11x = 66
∴ x = 6
∴ x = 6 is the solution of the given equation.


∴ 4(3x – 4) = 5(x + 1)
∴ 12x – 16 = 5x + 5
∴ 12x – 5x = 5 + 16
∴ 7x = 21
∴ x = 3
∴ x = 3 ¡s the solution of the given equation.

Maharashtra Board Class 9 Maths Solutions

Maharashtra State Board Class 9 Maths Solutions Chapter 4 Ratio and Proportion Practice Set 4.1

Question 1.
From the following pairs of numbers, find the reduced form of ratio of first number to second number.
i. 72,60
ii. 38,57
iii. 52,78
Solution:
i. 72, 60
\(\text { Ratio }=\frac{72}{60}=\frac{12 \times 6}{12 \times 5}=\frac{6}{5}=6 : 5\)

ii. 38, 57
\(\text { Ratio }=\frac{38}{57}=\frac{19 \times 2}{19 \times 3}=\frac{2}{3}=2 : 3\)

iii. 52, 78
\(\text { Ratio }=\frac{52}{78}=\frac{26 \times 2}{26 \times 3}=\frac{2}{3}=2 : 3\)

Question 2.
Find the reduced form of the ratio of the first quantity to second quantity.
i. ₹ 700, ₹ 308
ii. ₹ 14, ₹ 12 and 40 paise
iii. 5 litres, 2500 ml
iv. 3 years 4 months, 5 years 8 months
v. 3.8 kg, 1900 gm
vi. 7 minutes 20 seconds, 5 minutes 6 seconds
Solution:
i. ₹ 700, ₹ 308
\( \text { Ratio }=\frac{700}{308}=\frac{28 \times 25}{28 \times 11}=\frac{25}{11}=25 : 11\)

ii. ₹ 14, ₹12 and 40 paise
₹ 14 = 14 x 100 paise = 1400 paise
₹ 12 and 40 paise = 12 x 100 paise + 40 paise
= (1200 + 40) paise
= 1240 paise

iii. 5 litres, 2500 ml
5 litres = 5 x 1000 ml = 5000ml

iv. 3 years 4 months, 5 years 8 months
3 years 4 months = 3×12 months + 4 months
= (36 + 4) months
= 40 months
5 years 8 months = 5 x 12 months + 8 months
= (60 + 8) months
= 68 months

v. 3.8 kg, 1900 gm
3.8 kg = 3.8 x 1000 gm = 3800 gm

vi. 7 minutes 20 seconds, 5 minutes 6 seconds
7 minutes 20 seconds = 7 x 60 seconds + 20 seconds
= (420 + 20) seconds
= 440 seconds
5 minutes 6 seconds = 5 x 60 seconds + 6 seconds
= (300 + 6) seconds
= 306 seconds

Question 3.
Express the following percentages as ratios
i. 75 : 100
ii. 44 : 100
iii. 6.25%
iv. 52: 100
v. 0.64%
Solution:

Question 4.
Three persons can build a small house in 8 days. To build the same house in 6 days, how many persons are required?
Solution:
Let the persons required to build a house in 6 days be x.
Days required to build a house and number of persons are in inverse proportion.
∴ 6 × x = 8 × 3
∴ 6 x = 24
∴ x = 4
∴ 4 persons are required to build the house in 6 days.

Question 5.
Convert the following ratios into percentages.
i. 15 : 25
ii. 47 : 50
iii. \(\frac { 7 }{ 10 }\)
iv. \(\frac { 546 }{ 600 }\)
v. \(\frac { 7 }{ 16 }\)
Solution:
Let 15 : 25 = x %

∴ 15 : 25 = 60 %

ii. Let 47 : 50 = x%

∴ 47 : 50 = 94 %

iii. Let \(\frac { 7 }{ 10 }\) = x %

iv. Let \(\frac { 546 }{ 600 }\) = x %

v. Let \(\frac { 7 }{ 16 }\) = x %

Question 6.
The ratio of ages of Abha and her mother is 2 : 5. At the time of Abha’s birth her mothers age was 27 years. Find the present ages of Abha and her mother.
Solution:
The ratio of ages of Abha and her mother is 2 : 5.
Let the common multiple be x.
∴ Present age of Abha = 2x years and
Present age of Abha’s mother = 5x years
According to the given condition, the age of Abha’s mother at the time of Abha’s birth = 27 years
∴ 5x – 2x = 27
∴ 3x = 27
∴ x = 9
∴ Present age of Abha = 2x = 2 x 9 = 18 years
∴ Present age of Abha’s mother = 5x =5 x 9 = 45 years
The present ages of Abha and her mother are 18 years and 45 years respectively.

Question 7.
Present ages of Vatsala and Sara are 14 years and 10 years respectively. After how many years the ratio of their ages will become 5 : 4?
Solution:
Present age of Vatsala = 14 years
Present age of Sara = 10 years
After x years,
Vatsala’s age = (14 + x) years
Sara’s age = (10 + x) years
According to the given condition,
After x years the ratio of their ages will become 5 : 4
∴ \(\frac { 14 + x }{ 10 + x }\) = \(\frac { 5 }{ 4 }\)
∴ 4(14 + x) = 5(10 + x)
∴ 56 + 4x = 50 + 5x
∴ 56 – 50 = 5x – 4x
∴ 6 = x
∴ x = 6
∴ After 6 years, the ratio of their ages will become 5 : 4.

Question 8.
The ratio of present ages of Rehana and her mother is 2 : 7. After 2 years, the ratio of their ages will be 1 : 3. What is Rehana’s present age ?
Solution:
The ratio of present ages of Rehana and her mother is 2 : 7
Let the common multiple be x.
∴ Present age of Rehana = 2x years and Present age of Rehana’s mother = 7x years
After 2 years,
Rehana’s age = (2x + 2) years
Age of Rehana’s mother = (7x + 2) years
According to the given condition,
After 2 years, the ratio of their ages will be 1 : 3
∴ \(\frac { 2x + 2 }{ 7x + 2 }\) = \(\frac { 1 }{ 3 }\)
∴ 3(2x + 2) = 1(7x + 2)
∴ 6x + 6 = 7x + 2
∴ 6 – 2 = 7x – 6x
∴ 4 = x
∴ x = 4
∴ Rehana’s present age = 2x = 2 x 4 = 8 years
∴ Rehana’s present age is 8 years.

Maharashtra Board Class 9 Maths Solutions

Maharashtra State Board Class 9 Maths Solutions Chapter 3 Polynomials Problem Set 3

Question 1.
Write the correct alternative answer for each of the following questions.

i. Which of the following is a polynomial?

Answer:
(D) √2x² + \(\frac { 1 }{ 2 }\)

ii. What is the degree of the polynomial √7 ?
(A) \(\frac { 1 }{ 2 }\)
(B) 5
(C) 2
(D) 0
Answer:
(D) 0

iii. What is the degree of the polynomial ?
(A) 0
(B) 1
(C) undefined
(D) any real number
Answer:
(C) undefined

iv. What is the degree of the polynomial 2x² + 5xsup>3 + 7?
(A) 3
(B) 2
(C) 5
(D) 7
Answer:
(A) 3

v. What is the coefficient form of x³ – 1 ?
(A) (1, -1)
(B) (3, -1)
(C) (1, 0, 0, -1)
(D) (1, 3, -1)
Answer:
(C) (1, 0, 0, -1)

vi. p(x) = x² – x + 3, then p (7√7) = ?
(A) 3
(B) 7√7
(C) 42√7+3
(D) 49√7
Answer:
(D) 49√7

vii. When x = – 1, what is the value of the polynomial 2x³ + 2x ?
(A) 4
(B) 2
(C) -2
(D) -4
Answer:
(A) 4

viii. If x – 1 is a factor of the polynomial 3x² + mx, then find the value of m.
(A) 2
(B) -2
(C) -3
(D) 3
Answer:
(C) -3

ix. Multiply (x² – 3) (2x – 7x³ + 4) and write the degree of the product.
(A) 5
(B) 3
(C) 2
(D) 0
Answer:
(A) 5

x. Which is the following is a linear polynomials?
(A)  x + 5
(B)  x² + 5
(C) x³ + 5
(D) x⁴ + 5
Answer:
(A)  x + 5

Hints:
v. x³ – 1 = x³ + 0x² + 0x – 1

vi. p(7√ 7) = (7√ 7)² (7√ 7) (7√ 7) + 3
= 3

vii. p(-1) = 2(-1)³ + 2(-1)
= -2 – 2 = -4

vii. p(1) = 0
∴ 3(1)² + m(1) = 0
∴ 3 + m =0
∴ m = -3

ix. Here, degree of first polynomial = 2 and
degree of second polynomial 3
∴ Degree of polynomial obtained by multiplication = 2 + 3 = 5

Question 2.
Write the degree of the polynomial for each of the following.
i. 5 + 3x⁴
ii. 7
iii. ax⁷ + bx⁹ (a, b are constants)
Answer:
i. 5 + 3x⁴
Here, the highest power of x is 4.
∴Degree of the polynomial = 4

ii. 7 = 7x°
∴ Degree of the polynomial = 0

iii. ax⁷ + bx⁹
Here, the highest power ofx is 9.
∴Degree of the polynomial = 9

Question 3.
Write the following polynomials in standard form. [1 Mark each]
i. 4x² + 7x⁴ – x³ – x + 9
ii. p + 2p³ + 10p² + 5p⁴ – 8
Answer:
i. 7x⁴ – x³ + 4x² – x + 9
ii. 5p⁴ + 2p³ + 10p² + p – 8

Question 4.
Write the following polynomial in coefficient form.
i. x⁴ + 16
ii. m⁵ + 2m² + 3m+15
Answer:
i. x⁴ + 16
Index form = x⁴ + 0x³ + 0x² + 0x + 16
∴ Coefficient form of the polynomial = (1,0,0,0,16)

ii. m⁵ + 2m² + 3m + 15
Index form = m⁵ + 0m⁴ + 0m³ + 2m² + 3m + 15
∴ Coefficient form of the polynomial = (1, 0, 0, 2, 3, 15)

Question 5.
Write the index form of the polynomial using variable x from its coefficient form.
i. (3, -2, 0, 7, 18)
ii. (6, 1, 0, 7)
iii. (4, 5, -3, 0)
Answer:
i. Number of coefficients = 5
∴ Degree = 5 – 1 = 4
∴Index form = 3x⁴ – 2x³ + 0x² + 7x + 18

ii. Number of coefficients = 4
∴Degree = 4 – 1 = 3
∴ Index form = 6x³ + x² + 0x + 7

iii. Number of coefficients = 4
∴ Degree = 4 – 1 = 3
∴ Index form = 4x³ + 5x² – 3x + 0

Question 6.
Add the following polynomials.
i. 7x⁴ – 2x³ + x + 10;
3x⁴ + 15x³ + 9x² – 8x + 2
ii. 3p³q + 2p²q + 7;
2p²q + 4pq – 2p³q
Solution:
i. (7x⁴ – 2x³ + x + 10) + (3x⁴ + 15x³ + 9x² – 8x + 2)
= 7x⁴ – 2x³ + x + 10 + 3x⁴ + 15x³ + 9x² – 8x + 2
= 7x⁴ + 3x⁴ – 2x³ + 1 5x³ + 9x² + x – 8x + 10 + 2
= 10x⁴ + 13x³ + 9x² – 7x + 12

ii. (3p³q + 2p²q + 7) + (2p²q + 4pq – 2p³q)
= 3p³q + 2p²q + 7 + 2p²q + 4pq – 2p³q
= 3p³q – 2p³q + 2p²q + 2p²q + 4pq + 7
= p³q + 4p²q + 4pq + 7

Question 7.
Subtract the second polynomial from the first.
i. 5x² – 2y + 9 ; 3x² + 5y – 7
ii. 2x² + 3x + 5 ; x² – 2x + 3
Solution:
i. (5x² – 2y + 9) – (3x² + 5y – 7)
= 5x² – 2y+ 9 – 3x² – 5y + 1
= 5x² – 3x² – 2y – 5y + 9 + 7
= 2x² – 1y + 16

ii. (2x²+ 3x + 5) – (x² – 2x + 3)
= 2x² + 3x + 5 – x² + 2x – 3
= 2x² – x² + 3x + 2x + 5 – 3
= x² + 5x + 2

Question 8.
Multiply the following polynomials.
i. (m³ – 2m + 3) (m⁴ – 2m² + 3m + 2)
ii. (5m³ – 2) (m² – m + 3)
Solution:
i. (m³ – 2m + 3) (m⁴ – 2m² + 3m + 2)
= m³(m⁴ – 2m² + 3m + 2) – 2m(m⁴ – 2m² + 3m + 2) + 3(m⁴ – 2m² + 3m + 2)
= m⁷ – 2m⁵ + 3m⁴ + 2m³ – 2m⁵ + 4m³ – 6m² – 4m + 3m⁴ – 6m² + 9m + 6
= m⁷ – 2m⁵ – 2m⁵ + 3m⁴ + 3m⁴ + 2m³ + 4m³ – 6m² – 6m² – 4m + 9m + 6
= m⁷ – 4m⁵ + 6m⁴ + 6m³ – 12m² + 5m + 6

ii. (5m³ – 2) (m² – m + 3)
= 5m³(m² – m + 3) – 2(m² – m + 3)
= 5m⁵ – 5m⁴ + 15m³ – 2m² + 2m – 6

Question 9.
Divide polynomial 3x³ – 8x² + x + 7 by x – 3 using synthetic method and write the quotient and remainder.
Solution:
Dividend = 3x³ – 8x² + x + 7
∴ Coefficient form of dividend = (3, – 8, 1,7)
Divisor = x – 3
∴ Opposite of – 3 is 3

Coefficient form of quotient = (3, 1,4)
∴ Quotient = 3x² + x + 4 and
Remainder =19

Question 10.
For which value of m, x + 3 is the factor of the polynomial x³ – 2mx + 21?
Solution:
Here, p(x) = x³ – 2mx + 21
(x + 3) is a factor of x³ – 2mx + 21.
∴ By factor theorem,
Remainder = 0
∴ P(- 3) = 0
p(x) = x³ – 2mx + 21
∴ p(-3) = (-3)³ – 2(m)(-3) + 21
∴ 0 = – 27 + 6m + 21
∴ 6 + 6m = 0
∴ 6m = 6
∴ m = 1
∴ x + 3 is the factor of x³ – 2mx + 21 for m = 1.

Question 11.
At the end of the year 2016, the population of villages Kovad, Varud, Chikhali is 5x² – 3y², 7y² + 2xy and 9x² + 4xy respectively. At the beginning of the year 2017, x² + xy – y², 5xy and 3x² + xy persons from each of the three villages respectively went to another village for education, then what is the remaining total population of these three villages ?
Solution:
Total population of villages at the end of 2016 = (5x² – 3y²) + (7y² + 2xy) + (9x² + 4xy)
= 5x² + 9x² – 3y² + 7y² + 2xy + 4xy
= 14x² + 4y² + 6xy …….(i)
Total number of persons who went to other village at the beginning of 2017 = (x² + xy – y²) + (5xy) + (3x² + xy)
= x² + 3x² – y² + xy + 5xy + xy
= 4x² – y² + 7xy … (ii)
Remaining total population of villages = Total population at the end of 2016 – total number of persons who went to other village at the beginning of 2017
= 14x² + 4y² + 6xy – (4x² – y² + 7xy) … [From (i) and (ii)]
= 14x² + 4y² + 6xy – 4x² + y² – 7xy
= 14x² – 4x² + 4y² + y² + 6xy – 7xy = 1
= 10x² + 5y² – xy
∴ The remaining total population of the three villages is 10x² + 5y² – xy.

Question 12.
Polynomials bx² + x + 5 and bx³ – 2x + 5 are divided by polynomial x – 3 and the remainders are m and n respectively. If m – n = 0, then find the value of b.
Solution:
When polynomial bx² + x + 5 is divided by (x – 3), the remainder is m.
∴ By remainder theorem,
Remainder = p(3) = m
p(x) = bx² + x + 5
∴ p(3) = b(3)² + 3 + 5
∴m = b(9) + 8
m = 9b + 8 …(i)
When polynomial bx³ – 2x + 5 is divided by x – 3 the remainder is n
∴ remainder = p(3) = n
p(x) = bx³ – 2x + 5
∴ P(3)= b(3)³ – 2(3) + 5
∴ n = b(27) – 6 + 5
∴ n = 27b – 1 …(ii)
Now, m – n = 0 …[Given]
∴ m = n
∴ 9b + 8 = 27b – 1 …[From (i) and (ii)]
∴ 8 + 1 = 27b – 9b
∴ 9 = 18b
∴ b = \(\frac { 1 }{ 2 }\)

Question 13.
Simplify.
(8m² + 3m – 6) – (9m – 7) + (3m² – 2m + 4)
Solution:
(8m² + 3m – 6) – (9m – 7) + (3m² – 2m + 4)
= 8m² + 3m – 6 – 9m + 7 + 3m² – 2m + 4
= 8m² + 3m² + 3m – 9m – 2m – 6 + 7 + 4
= 11m² – 8m + 5

Question 14.
Which polynomial is to be subtracted from x² + 13x + 7 to get the polynomial 3x² + 5x – 4?
Solution:
Let the required polynomial be A.
∴ (x² + 13x + 7) – A = 3x² + 5x – 4
∴ A = (x² + 13x + 7) – (3x² + 5x – 4)
= x² + 13x + 7 – 3x² – 5x + 4
= x² – 3x² + 13x – 5x + 7+4
= -2x² + 8x + 11
∴ – 2x² + 8x + 11 must be subtracted from x² + 13x + 7 to get 3x² + 5x – 4.

Question 15.
Which polynomial is to be added to 4m + 2n + 3 to get the polynomial 6m + 3n + 10?
Solution:
Let the required polynomial be A.
∴ (4m + 2n + 3) + A = 6m + 3n + 10
∴ A = 6m + 3n + 10 – (4m + 2n + 3)
= 6m + 3n + 10 – 4m – 2n – 3
= 6m – 4m + 3n – 2n + 10 – 3
= 2m + n + 7
∴ 2m + n + 7 must be added to 4m + 2n + 3 to get 6m + 3n + 10.

Question 1.
Read the following passage, write the appropriate amount in the boxes and discuss.
Govind, who is a dry land farmer from Shiralas has a 5 acre field. His family includes his wife, two children and his old mother. He borrowed one lakh twenty five thousand rupees from the bank for one year as agricultural loan at 10 p.c.p.a. He cultivated soyabean in x acres and cotton and tur in y acres. The expenditure he incurred was as follows :
He spent ₹10,000 on seeds. The expenses for fertilizers and pesticides for the soyabean crop was ₹ 2000x and ₹ 4000x² were spent on wages and cultivation of land. He spent ₹ 8000y on fertilizers and pesticides and ₹ 9000y2 for wages and cultivation of land for the cotton and tur crops.

Let us write the total expenditure on all the crops by using variables x and y.
₹ 10000 + 2000x + 4000×2 + 8000y + 9000y²
He harvested 5x² quintals soyabean and sold it at ₹ 2800 per quintal. The cotton crop yield was \(\frac { 5 }{ 3 }\) y² quintals which fetched ₹ 5000 per quintal.
The tur crop yield was 4y quintals and was sold at ₹ 4000 per quintal. Write the total income in rupees that was obtained by selling the entire farm produce, with the help of an expression using variables x and y. (Textbook pg. no. 44)
Answer:
Total income = income on soyabean crop + income on cotton crop + income on tur crop
= ₹ (5x² x 2800) + ₹(\(\frac { 5 }{ 3 }\) y² x 5000) + ₹ (4y x 4000)
= ₹ ( 14000x² + \(\frac { 25000 }{ 3 }\)y² + 16000y)

Question 2.
We have seen the example of expenditure and income (in terms of polynomials) of Govind who is a dry land farmer. He has borrowed rupees one lakh twenty-five thousand from the bank as an agriculture loan and repaid the said loan at 10 p.c.p.a. He had spent ₹ 10,000 on seeds. The expenses on soyabean crop was ₹ 2000x for fertilizers and pesticides and ₹ 4000x² was spent on wages and cultivation. He spent ₹ 8000y on fertilizers and pesticides and ₹9000y² on cultivation and wages for cotton and tur crop.
His total income was
₹ (14000x² + \(\frac { 25000 }{ 3 }\)y² + 16000y)
By taking x = 2, y = 3 write the income expenditure account of Govind’s farming. (Textbook pg. no. 52)
Solution:
–           Credit (Income)
₹ 1,25,000   Bank loan
₹ 56000      Income from soyabean
₹ 75000      Income from cotton
₹ 48000      Income from tur
₹ 304000     Total income

–                     Debit (Expenses)
₹ 1,37,000       loan paid with interest for seeds
₹ 10000          For seeds
₹ 4000            Fertilizers and pesticides for soyabean
₹ 16000         Wages and cultivation charges for soyabean
₹ 24000          Fertilizers and pesticides for cotton & tur
₹ 81000         Wages and cultivation charges for cotton & tur
₹ 272000       Total expenditure

Maharashtra Board Class 9 Maths Solutions

Maharashtra State Board Class 10 Maths Solutions Chapter 2 Pythagoras Theorem Practice Set 2.2

Question 1.
In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13, find QR.

Solution:
In ∆PQR, point S is the midpoint of side QR. [Given]
∴ seg PS is the median.
∴ PQ² + PR² = 2 PS² + 2 SR² [Apollonius theorem]
∴ 11² + 17² = 2 (13)² + 2 SR²
∴ 121 + 289 = 2 (169)+ 2 SR²
∴ 410 = 338+ 2 SR²
∴ 2 SR² = 410 – 338
∴ 2 SR² = 72
∴ SR² = \(\frac { 72 }{ 2 } \) = 36
∴ SR = \(\sqrt{36}\) [Taking square root of both sides]
= 6 units Now, QR = 2 SR [S is the midpoint of QR]
= 2 × 6
∴ QR = 12 units

Question 2.
In ∆ABC, AB = 10, AC = 7, BC = 9, then find the length of the median drawn from point C to side AB.
Solution:
Let CD be the median drawn from the vertex C to side AB.
BD = \(\frac { 1 }{ 2 } \) AB [D is the midpoint of AB]
= \(\frac { 1 }{ 2 } \) × 10 = 5 units

In ∆ABC, seg CD is the median. [Given]
∴ AC² + BC² = 2 CD² + 2 BD² [Apollonius theorem]
∴ 7² + 9² = 2 CD² + 2 (5)²
∴ 49 + 81 = 2 CD² + 2 (25)
∴ 130 = 2 CD² + 50
∴ 2 CD² = 130 – 50
∴ 2 CD² = 80
∴ CD² = \(\frac { 80 }{ 2 } \) = 40
∴ CD = \(\sqrt { 40 }\) [Taking square root of both sides]
= 2 \(\sqrt { 10 }\) units
∴ The length of the median drawn from point C to side AB is 2 \(\sqrt { 10 }\) units.

Question 3.
In the adjoining figure, seg PS is the median of APQR and PT ⊥ QR. Prove that,

i. PR² = PS² + QR × ST + (\(\frac { QR }{ 2 } \))²
ii. PQ² = PS² – QR × ST + (\(\frac { QR }{ 2 } \))²
Solution:
i. QS = SR = \(\frac { 1 }{ 2 } \) QR (i) [S is the midpoint of side QR]

∴ In ∆PSR, ∠PSR is an obtuse angle [Given]
and PT ⊥ SR [Given, Q-S-R]
∴ PR² = SR² +PS² + 2 SR × ST (ii) [Application of Pythagoras theorem]
∴ PR² = (\(\frac { 1 }{ 2 } \) QR)² + PS² + 2 (\(\frac { 1 }{ 2 } \) QR) × ST [From (i) and (ii)]
∴ PR² = (\(\frac { QR }{ 2 } \))² + PS² + QR × ST
∴ PR² = PS² + QR × ST + (\(\frac { QR }{ 2 } \))²

ii. In.∆PQS, ∠PSQ is an acute angle and [Given]

PT ⊥QS [Given, Q-S-R]
∴ PQ² = QS² + PS² – 2 QS × ST (iii) [Application of Pythagoras theorem]
∴ PR² = (\(\frac { 1 }{ 2 } \) QR)² + PS² – 2 (\(\frac { 1 }{ 2 } \) QR) × ST [From (i) and (iii)]
∴ PR² = (\(\frac { QR }{ 2 } \))² + PS² – QR × ST
∴ PR² = PS² – QR × ST + (\(\frac { QR }{ 2 } \))²

Question 4.
In ∆ABC, point M is the midpoint of side BC. If AB² + AC² = 290 cm, AM = 8 cm, find BC.

Solution:
In ∆ABC, point M is the midpoint of side BC. [Given]
∴ seg AM is the median.
∴ AB² + AC² = 2 AM² + 2 MC² [Apollonius theorem]
∴ 290 = 2 (8)² + 2 MC²
∴ 145 = 64 + MC² [Dividing both sides by 2]
∴ MC² = 145 – 64
∴ MC² = 81
∴ MC = \(\sqrt{81}\) [Taking square root of both sides]
MC = 9 cm
Now, BC = 2 MC [M is the midpoint of BC]
= 2 × 9
∴ BC = 18 cm

Question 5.
In the adjoining figure, point T is in the interior of rectangle PQRS. Prove that, TS² + TQ² = TP² + TR². (As shown in the figure, draw seg AB || side SR and A – T – B)

Given: ꠸PQRS is a rectangle.
Point T is in the interior of ꠸PQRS.
To prove: TS² + TQ² = TP² + TR²
Construction: Draw seg AB || side SR such that A – T – B.
Solution:
Proof:
꠸PQRS is a rectangle. [Given]
∴ PS = QR (i) [Opposite sides of a rectangle]
In ꠸ASRB,
∠S = ∠R = 90° (ii) [Angles of rectangle PQRS]
side AB || side SR [Construction]
Also ∠A = ∠S = 90° [Interior angle theorem, from (ii)]
∠B = ∠R = 90°
∴ ∠A = ∠B = ∠S = ∠R = 90° (iii)
∴ ꠸ASRB is a rectangle.
∴ AS = BR (iv) [Opposite sides of a rectanglel

In ∆PTS, ∠PST is an acute angle
and seg AT ⊥ side PS [From (iii)]
∴ TP² = PS² + TS² – 2 PS.AS (v) [Application of Pythagoras theorem]
In ∆TQR., ∠TRQ is an acute angle
and seg BT ⊥ side QR [From (iii)]
∴ TQ² = RQ² + TR² – 2 RQ.BR (vi) [Application of pythagoras theorem]

TP² – TQ² = PS² + TS² – 2PS.AS
-RQ² – TR² + 2RQ.BR [Subtracting (vi) from (v)]
∴ TP² – TQ² = TS² – TR² + PS²
– RQ² -2 PS.AS +2 RQ.BR
∴ TP² – TQ² = TS² – TR² + PS²
– PS² – 2 PS.BR + 2PS.BR [From (i) and (iv)]
∴ TP² – TQ² = TS² – TR²
∴ TS² + TQ² = TP² + TR²

Question 1.
In ∆ABC, ∠C is an acute angle, seg AD Iseg BC. Prove that: AB² = BC² + A² – 2 BC × DC. (Textbook pg. no. 44)
Given: ∠C is an acute angle, seg AD ⊥ seg BC.
To prove: AB² = BC² + AC² – 2BC × DC
Solution:
Proof:
∴ LetAB = c, AC = b, AD = p,

∴ BC = a, DC = x
BD + DC = BC [B – D – C]
∴ BD = BC – DC
∴ BD = a – x
In ∆ABD, ∠D = 90° [Given]
AB² = BD² + AD² [Pythagoras theorem]
∴ c² = (a – x)² + [P²] (i)
∴ c² = a² – 2ax + x² + [P²]
In ∆ADC, ∠D = 90° [Given]
AC² = AD² + CD² [Pythagoras theorem]
∴ b² = p² + [X²]
∴ p² = b² – [X²] (ii)
∴ c² = a² – 2ax + x² + b² – x² [Substituting (ii) in (i)]
∴ c² = a² + b² – 2ax
∴ AB² = BC² + AC² – 2 BC × DC

Question 2.
In ∆ABC, ∠ACB is an obtuse angle, seg AD ⊥ seg BC. Prove that: AB² = BC² + AC² + 2 BC × CD. (Textbook pg. no. 40 and 4.1)
Given: ∠ACB is an obtuse angle, seg AD ⊥ seg BC.
To prove: AB² = BC² + AC² + 2BC × CD
Solution:
Proof:

Let AD = p, AC = b, AB = c,
BC = a, DC = x
BD = BC + DC [B – C – D]
∴ BD = a + x
In ∆ADB, ∠D = 90° [Given]
AB² = BD² + AD² [Pythagoras theorem]
∴ c² = (a + x)² + p² (i)
∴ c² = a² + 2ax + x² + p²
Also, in ∆ADC, ∠D = 90° [Given]
AC² = CD² + AD² [Pythagoras theorem]
∴ b² = x² + p²
∴ p² = b² – x² (ii)
∴ c² = a² + 2ax + x² + b² – x² [Substituting (ii) in (i)]
∴ c² = a² + b² + 2ax
∴ AB² = BC² + AC² + 2 BC × CD

Question 3.
In ∆ABC, if M is the midpoint of side BC and seg AM ⊥seg BC, then prove that
AB² + AC² = 2 AM² + 2 BM². (Textbook pg, no. 41)
Given: In ∆ABC, M is the midpoint of side BC and seg AM ⊥ seg BC.
To prove: AB² + AC² = 2 AM² + 2 BM²
Solution:
Proof:

In ∆AMB, ∠M = 90° [segAM ⊥ segBC]
∴ AB2 = AM2 + BM2 (i) [Pythagoras theorem]
Also, in ∆AMC, ∠M = 90° [seg AM ⊥ seg BC]
∴ AC2 = AM2 + MC2 (ii) [Pythagoras theorem]
∴ AB² + AC² = AM² + BM² + AM² + MC² [Adding (i) and (ii)]
∴ AB² + AC² = 2 AM² + BM² + BM² [∵ BM = MC (M is the midpoint of BC)]
∴ AB² + AC² = 2 AM² + 2 BM²

Maharashtra Board Class 10 Maths Solutions