The Baudhayana Pythagoras Theorem Class 8 Solutions Maths Ganita Prakash Part 2 Chapter 2

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Class 8 Maths Ganita Prakash Part 2 Chapter 2 Solutions

Class 8 Maths The Baudhayana Pythagoras Theorem Solutions

Class 8 Ganita Prakash Part 2 Chapter 2 Solutions The Baudhayana Pythagoras Theorem

Figure it Out (NCERT Page 34)

Question 1.
Moreover, all these small triangles are congruent to each other. Can you explain? Why?

Solution:
In this figure,
Each triangle has two equal sides (marked equal sign in the figure).
The included angle between those sides is the right angle.
Hence, by Side-Angle-Side (SAS) congruence criterion, all the small triangles are congruent.

NCERT (Page 35)

Question 1.
Cut out two identical squares of paper. Draw, label and cut as follows :

Now, place the pieces 5, 6, 7 and 8 around square 1 to get a square with double the area.
Solution:
Do yourself

NCERT (Page 36)

Question 1.
Why is the smaller inside square half the area of the larger square? Again, adding some East-West and North-South lines can explain it.

Solution:
The East-West and North-South lines divide the larger square into 4 equal small squares.
∴ Area of the larger square = 4 × Area of the smaller squares.
The inner square is made of 4 congruent right triangles, each triangle is equal to the half of the area of one small square.
So, total area of inner square
= 4 × \(\frac{1}{2}\) (area of small square)
= 2 × area of small squares
Thus, total area of inner square is half of the area of the larger square.

Question 2.
Cut out a square from a piece of paper. Now, make a square, whose area is half the area of the first square.
Solution:
Do yourself

Question 3.
Will the square having half the sidelength have half the area? Why not? How many such squares will fill the original square?
Solution:
No, a square with half the sidelength does not have half the area.
Reason Area of a square is proportional to the square of its side.
If the side is halved then the area of the new square
= \(\frac{1}{4}\) of the area of the original area
∴ A square with half the sidelength has one-fourth the area, not half. Hence, 4 such smaller squares are needed to fill the original square.

NCERT (Page 37)

Question 1.
Why is PQRS a square? Why is its area half that of the original paper? Explain by connecting QS and PR, finding the different angles formed and then using triangle congruence.

Solution:
Do yourself.

Figure it Out (NCERT Page 39 and 40)

Question 1.
Earlier, we saw a method to create a square with double the area of a given square paper. There is another method to do this in which two identical square papers are cut in the following way.

Can you arrange these pieces to create a square with double the area of either square?
Solution:
Do yourself.

Question 2.
The length of the two equal sides of an isosceles right triangle is given. Find the length of the hypotenuse. Find bounds on the length of the hypotenuse such that they have atleast one digit after the decimal point.
(i) 3
(ii) 4
(iii) 6
(iv) 8
(v) 9
Solution:
Let the length of two equal sides and the hypotenuse of the isosceles right triangle be a and c, respectively.
We know that c = a√2 and √2 = 1.414
(i) Here, a = 3
So, the length of the hypotenuse = 3√2 = 3 X 1.414
= 4.242 units
Since, the value ofc between 4.2 and 4.3.
Thus, the bounds are 4.2 < 3√2 < 4.3.

(ii) Here, a = 4
So, the length of hypotenuse = 4√2 = 4 × 1.414
= 5.656 units
Since, the value ofc is between 5.6 and 5.7.
Thus, the bounds are 5.6 < 4√2 < 5.7.

(iii) Here, a = 6
So, the length of the hypotenuse = 6√2 = 6 × 1.414
= 8.484 units
Since, the value ofc is between 8.4 and 8.5.
Thus, the bounds are 8.4 < 6√2 < 8.5

(iv) Here, a = 8
So, the length of the hypotenuse = 8√2 = 8 × 1.414
= 11.312 units
Since, the value of c is between 11.3 and 11.4.
Thus, the bounds are 11.3 < 8√2 < 11.4.

(v) Here,a = 9
So, the length of the hypotenuse = 9√2 = 9 × 1.414
= 12.726 units
Since, the value ofc is between 12.7 and 12.8.
Thus, the bounds are 12.7 < 9√2 < 12.8

Question 3.
The hypotenuse of an isosceles right-angled triangle is 10 units. What are its other two sidelengths?
Solution:
Let a be the length of the equal side and c be the length of the hypotenuse of an isosceles right-angled triangle. Given, c = 10
We know that c² = 2a²
So, (10)² = 2a²
⇒ 100 = 2a²
⇒ a² = 50
⇒ a = 5√2
Hence, the other two sides are 5√2 units each.

Figure it Out (NCERT Page 47)

Question 1.
(i) If a right-angled triangle has shorter sides of lengths 5 cm and 12 cm then what is the length of its hypotenuse?
(ii) Draw the right-angled triangle with these sidelengths and measure the hypotenuse then check your answer using Baudhayana’s theorem.
Solution:
(i) Do same as Example 7. Ans. 13 cm
(ii) Do yourself.

Question 2.
(i) If a right-angled triangle has a shorter side of length 8 cm and length of hypotenuse is 17 cm, what is the length of the third side?
(ii) Again, try drawing the triangle and measuring and then check your answer using Baudhayana’s theorem.
Solution:
(i) Do same as Example 8. Ans. 15 cm
(ii) Do yourself

Question 3.
Using the constructions you have now seen, how would you construct a square, whose area is triple the area of a given square? Five times the area of a given square? (Baudhayana’s Sulba-Sutra, Verse 1.10)
Solution:
Let the side of the given square be a.
Its area = a × a = a²
To construct a square with triple the area.
(i) Construct a right-angled triangle with perpendicular sides a and √2a (note that √2a the diagonal of the square)

(ii) By Baudhayana’s theorem,
(AD)² = a² + (√2a)²
= a² + 2 a² = 3a²
⇒ AD = √3a
(iii) Draw a square on the hypotenuse of this triangle.
Area of square ADGF = √3a × √3a = 3a²
Hence, the new square has three times the area of the given square.

To construct a square with five times the area
(i) Construct a right-angled triangle with perpendicular sides a and 2a.

(ii) By Baudhayana’s theorem,
a² + (2a)² = a² + 4a²
(QR)² = a² + (2a)²
= a² + 4a² = 5a²
⇒ QR = 5 a

(iii) Draw a square on the hypotenuse of this triangle.
Area of square QSTR = √5a × √5a = 5a²
Hence, the new square has five times the area of the given square.

Question 4.
Let a, b and c denote the length of the sides of a right triangle, with c being the length of the hypotenuse. Find the missing sidelength in each of the following cases.
(i) a = 5, b = 7
(ii) a = 8,b = 12
(iii) a = 9, c = 15
(iv) a = 7, b = 12
(v) a =1.5, b = 3.5
Solution:
(i) Given, a = 5 and b = 7
By Baudhayana’s theorem,
c² = a² + b²
⇒ c² = a² + b²
⇒ c² = 5² + 7²
⇒ c² = 25 + 49
⇒ c² = 74
⇒ c = \(\sqrt{74}\)
The missing sidelength is \(\sqrt{74}\).

(ii) Do same as part (i). Ans.c = 4\(\sqrt{13\)

(iii) Given, a = 9 and c = 15
By Baudhayana’s theorem,
c² = a² + b²
⇒ b² = c² – a²
⇒ b² = 15² – 9²
⇒ b² =225 – 81
⇒ b² = 144
⇒ b = 12
The missing sidelength is 12

(iv) Do same as part (i). Ans. c = \(\sqrt{193\)
(v) Do same as part (i). Ans. c = \(\sqrt{14.5\)

NCERT (Page 48 and 49)

Question 1.
List down all the Baudhayana triples with numbers less than or equal to 20.
Solution:
Baudhayana triples are sets of three positive integers a,b and c that satisfy the equation a² + b² = c²
The list of Baudhayana triples with numbers less than or equal to 20 are (3, 4, 5), (5, 12, 13) (8, 15, 17), (6, 8, 10) (9, 12, 15) and (12, 16, 20).

Question 2.
Is (30, 40, 50) a Baudhayana triple?
Solution:
Lat a = 30, b = 40 and c = 50
By Baudhayana-Pythagoras theorem, a² + b² = c²
⇒ (30)² +(40)² = 900 + 1600
= 2500 = (50)²
Therefore, (30,40, 50) is a Baudhayana triple.

Question 3.
Is (300, 400, 500) a Baudhayana triple?
Solution:
Let a = 300, b = 400 and c = 500
By Baudhaana-Pythagoras theorem, a² + b² = c²
⇒ (300)² + (400)² = 90000 + 160000
= 250000 = (500)²
Therefore, (300,400, 500) is a Baudhayana triple.

Question 4.
If (5, 12, 13) a primitive Baudhayana triple? What are the other primitive Baudhayana triples with numbers less than or equal to 20?
Solution:
First, check the Baudhayana-Pythagoras condition,
5² + 12² = 25 + 144 = 169 = 13²
So, (5, 12, 13) is a Baudhayana (Pythagorean) triple.
Now, check whether it is primitive,
HCFof (5, 12, 13) = 1
Since, the three numbers have no common factor other than 1, the triple is primitive.
Other primitive Baudhayana triples with numbers < 20 are (3, 4, 5), (5, 12, 13), (8, 15, 17).

Question 5.
Generate 5 scaled versions of each primitive Baudhayana triple. Are these scaled versions primitive?
Solution:
No, the scaled versions are not primitive.
Primitive triples (< 20) are (3, 4, 5), (5, 12, 13), (8, 15, 17).
To generate scaled versions of a primitive triple, one multiplies each element of the triple by common scaling factor k.
Scaled versions of (3, 4, 5)

Multiply by 2, 3, 4, 5, 6
(6, 8, 10), (9, 12, 15), (12, 16, 20), (15, 20, 25), (18, 24, 30) and so on.

Scaled versions of (5, 12, 13)
(10, 24, 26), (15, 36, 39), (20, 48, 52), (25, 60, 65), (30, 72, 78) and so on.

Scaled versions of (8, 15, 17)
(16, 30, 34), (24, 45, 51), (32, 60, 68), (40, 75, 85), (48, 90, 102) and so on.
Hence, all scaled versions have a common factor greater than 1, so they are non-primitive Baudhayana triples.

Question 6.
If (a, b, c) is non-primitive and the integers have f greater than 1 as a common factor, then is \(\left(\frac{a}{f}, \frac{b}{f}, \frac{c}{f}\right)\) a Baudhayana triple? Check this statement for (9, 12, 15). Justify statement.
Solution:
Given, Baudhayana triple = (9, 12, 15)
Here, the common factor (f) = 3
On dividing each term by 3, we get
\(\left(\frac{9}{3}, \frac{12}{3}, \frac{15}{3}\right)\) = (3, 4, 5)
Now, by Baudhayana-Pythagoras theorem,
3² + 4² = 9 + 16 = 25 = 5²
Hence, the resulting triple is a Baudhayana triple.

Question 7.
What is the sum of the first (n – 1) odd numbers?
Solution:
We know that sum of first k odd number is k² .
So, the sum of first (n – 1) is (n – 1)².

Figure it Out (NCERT Page 50)

Question 1.
Find 5 more Baudhayana triples using this idea.
Solution:
We know that
(n – 1)² + (2n – 1) = n² ,..(i)
It 2n – 1 is an odd square, we get the Baudhayana triples.
If 2n – 1 = 7² =49
⇒ 2n = 49 + 1 = 50
⇒ n = 25

So, from Eq. (i),
(25 – 1)² + 7² =25²
⇒ (24)² + 7² = (25)²
So, the Baudhayana triple is (7, 24, 25).
If 2n – 1 = 9² = 81
⇒ 2n = 81 + 1
⇒ 2n = 82
⇒ n = 41

So, from Eq. (i), (41 – 1)² + 9² = (41)²
⇒ (40)² +9² = (41)²
So, the Baudhayana triple is (9, 40, 41).
If 2n – 1 = (11)² =121
⇒ 2n = 121 + 1
⇒ 2n = 12²
⇒ n = 61

So, from Eq. (i), (61 – 1)² = 11² =(61)²
= (60)² + 11² = 61
So, the Baudhayana triple is (11, 60, 61).
If 2n – 1 = (13)² =169
⇒ 2n = 169 + 1
⇒ 2n = 170
⇒ n = 85

So, from Eq. (i),(85 – 1)² + (13)² = (85)²
⇒ (84)² + (13)² =(85)²
So, the Baudhayana triple is (13, 84, 85).
If 2n – 1 = (15)² = 225
⇒ 2n = 225 + 1
⇒ 2n = 226
⇒ n = 113

So, from Eq(i), (113 – 1)² + (15)² = (113)²
(112)² + (15)² = (113)²
So, the Baudhayana triple is (15, 112, 113).

Question 2.
Does this method yield non-primitive Baudhayana triples?
[Hint Observe that among the triples generated, one of the smaller sidelengths is one less than the hypotenuse.]
Solution:
No, this methqd does not yield non-primitive Baudhayana triples.
Using this method, the triples obtained are of the form (2 n – 1, n – 1, n), where one of the smaller sides is one less than the hypotenuse.

Since, the hypotenuse and this smaller side differ by 1, they cannot have a common factor greater than 1. Hence, all three numbers cannot have a common factor. Therefore, the Baudhayana triples obtained by this method are primitive, not non-primitive.

Question 3.
Are there primitive triples that cannot be obtained through this method? If yes, give examples.
Solution:
Yes, there are primitive Baudhayana triples that cannot be obtained by this method.
This method always produces triples, where one of the smaller sides is exactly one less than the hypotenuse. However, some primitive triples do not have this property.
Here are some examples,
(8, 15, 17)
(12, 35, 37)
(20, 21, 29)
In these triples, neither smaller side is one less than the hypotenuse, so they cannot be generated by this method, even though they are primitive.

Figure it Out (NCERT Page 52-54)

Question 1.
Find the diagonal of a square with sidelength 5 cm.
Solution:
Let the side and diagonal of a square be a and c.
Since, a square can be divided into two congruent isosceles right-angled triangles. The diagonal of the square is the hypotenuse of one such triangle.
We know that
c = √2a
⇒ c = 5√2 [∵ a = 5]
The diagonal of the square is 5√2 cm.

Question 2.
Find the missing sidelengths in the following right triangles.

Solution:
Let the shorter sides of a right-angle triangle be a and b and the hypotenuse be c.
(i) Given, a = 7 and b = 9
By Baudhayana’s theorem,
a² + b² = c²
⇒ 7² + 9² = c²
⇒ 49 + 81 = c²
⇒ c² = 130
⇒ c = \(\sqrt{130}\)
The missing sidelength is V130.

(ii) Given, a = 10 and b = 4
By Baudhayana’s theorem,
⇒ a² + b² = c²
⇒ 10² + 4² = c²
⇒ 100 + 16 = c²
⇒ 116 = c²
⇒ c = \(\sqrt{116}\)
⇒ c = 2\(\sqrt{29}\)
The missing sidelength is 2\(\sqrt{29}\)

(iii) Given, a = 40 and c = 41
By Baudhayana’s theorem,
40² + b² = 41²
⇒ 1600 + b² = 1681
⇒ b² = 1681 – 1600
⇒ b² = 81
⇒ b = 9
The missing sidelength is 9

(iv) Given, a = 10 and c = \(\sqrt{200}\) By Baudhayana’s theorem,
a² + b² = c²
⇒ 10² + b² =(\(\sqrt{200}\))²
⇒ 100 + b² = 200
⇒ b² = 200 – 100
⇒ b² = 100
⇒ b = 10
The missing sidelength is 10

(v) Given, a = 10 and b V150 By Baudhayana’s theorem,
a² + b² = c²
⇒ 10² + (VI50)² = c²
⇒ 100 + 150 = c²
⇒ 250 = c²
⇒ c = \(\sqrt{250}\)
⇒ c = 5\(\sqrt{10}\)
The missing sidelength is 5\(\sqrt{10}\).

(vi) Given, a =27 and c = 45
By Baudhayana’s Theorem,
⇒ a² + b² = c²
⇒ 27² + b² = 45²
⇒ 729 + b² =2025
⇒ b² =2025 – 729
⇒ b² = 1296
⇒ b = 36
The missing sidelength is 36.

Question 3.
Find the sidelength of a rhombus, whose diagonals are of length 24 units and 70 units.
Solution:
We know that the diagonals of a rhombus bisect each other at right angles. This property forms four right-angled triangles within the rhombus, where the hypotenuse of each triangle is the sidelength of the rhombus and the legs are half the lengths of the diagonals.
Given, diagonals d₁ = 24 units and d₂ = 70 units
Half of the first diagonal,
a = \(\frac{d_1}{2}=\frac{24}{2}\) = 12 units

Half of the second diagonal,
b = \(\frac{d_2}{2}=\frac{70}{2}\) = 35 units

Let c be the hypotenuse of right-angle triangle.
Then, by Baudhayana’s theorem,
a² + b² = c²
⇒ 122 + 35² = c²
⇒ 144 + 1225 = c²
⇒ 1369 = c²
⇒ c = 37
Hence, the sidelength of the rhombus is 37 units.

Question 4.
Is the hypotenuse the longest side of a right triangle? Justify your answer.
Solution:
Yes, the hypotenuse is the longest side of a right triangle because it is the side opposite the right angle, which is the largest angle in the triangle.

Question 5.
True or False-Every Baudhayana triple is either a primitive triple or a scaled version of a primitive triple.
Solution:
True, a Baudhayana triple is a set of three positive integers that satisfies the Pythagorean theorem, a² + b² = c². Any Baudhayana triple can be generated by scaling a primitive triple by a positive integer factor.
Thus, every Baudhayana triple is either a primitive triple itself or a scaled version of one.

Question 6.
Give 5 examples of rectangles, whose sidelengths and diagonals are all integers.
Solution:
We know that for a rectangle with integer sidelengths a and b, the diagonal c be an integer, the relationship between the sides and the diagonal is given by the
Pythagoras theorem,
a² + b² = c²
Finding such rectangles is equivalent to finding Pythagorean triples (a, b, c).
Hence, the five examples of rectangle sidelengths and diagonals are
sidelengths 3 and 4, diagonal 5;
sidelengths 5 and 12, diagonal 13;
sidelengths 8 and 15, diagonal 17;
sidelengths 7 and 24, diagonal 25 and
sidelengths 20 and 21, diagonal 29.

Question 7.
Construct a square, whose area is equal to the difference of the areas of squares of sidelengths 5 units and 7 units.
Solution:
Area of square with side 7 units = 7² = 49 sq units
and area of square with side 5 units = 5² = 25 sq units
Now, difference in both the areas = (49 – 25) sq units
= 24 sq units
We have, in a right-angled triangle,
hypotenuse (c) = 7 units and one side (a) = 5 units then the other side (b).

By Baudhayana-Pythagoras theorem,
c² = a² + b²
⇒ 7² = 5² + b²
⇒ 49 = 25 + b²
⇒ b² = 24
⇒ b = \(\sqrt{24}\) units = 4.89 units
Now, on constructing a square on the side \(\sqrt{24}\) units, we get

Hence, this square has area 24 sq units, equal to the difference of the given areas.

Question 8.
(i) Using the dots of a grid as the vertices can you create a square that has an area of (a) 2 sq units, (b) 3 sq units, (c) 4 sq units and (d) 5 sq unit?
(ii) Suppose, the grid extends indefinitely. What are the possible integer-valued areas of squares you can create in this manner?

Solution:
(i) Let the grid consist of unit squares.
Take two points on the grid such that the horizontal distance between them is a units and the vertical distance is b units. Then, the side of the square formed is the hypotenuse of a right-angled triangle with legs a and b.
By Baudhayana-Pythagoras theorem,
Side² = a² + b²
Hence, area of the square is a² + b².
(a) Area = 2 sq units
Choose a = 1 and b = 1
Area = 1² + 1² = 2
So, a square of area 2 sq units can be formed.

(b) Area = 3sq units
Try all possible sums of two squares, 1² + 1² = 2,
1² + 2² = 5
No pair of integers gives, a² + b² = 3
So, a square of area 3 sq units cannot be formed.

(c) Area = 4 sq units
Choose a = 2 and b = 2
Area = 2² = 4
So, a square of area 2 sq units can be formed.

(d) Area = 5 sq units
Choose a = 1 and b = 2.
Area = 1² + 2² = 5
So, a square of area 5 sq units can be formed.

(ii) If the grid extends indefinitely.
Since, the grid is infinite, we can choose any integers a and b.
From part (i), the area of a square formed on the grid is a2 +b2.
So, all integers which can be expressed as the sum of two perfect squares are possible areas of squares formed using grid points.
Examples of possible areas
1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25,…
Examples of impossible areas 3, 6, 7, 11, 14, 15,…

Question 9.
Find the area of an equilateral triangle with sidelength 6 units.
[Hint Show that an altitude bisects the opposite side. Use this to find the height.]
Solution:
Given, AD ⊥ BCandABCis an equilateral triangle.

In ΔADB and ΔADC,
AB = AC [∵ ABC is an equilateral triangle]
∠ADB = ∠ADC [each 90°]
and AD = AD [common]
ΔADB ≅ ΔADC [by RHS congruency rule]
⇒ BD = DC [By CPCT]
Hence, the altitude AD bisects the opposite side BC of ΔABC.
Given, the sidelength of an equilateral triangle = 6 units
Since, an altitude bisects the opposite side of triangle.
So, base (BD) = DC = \(\frac{1}{2}\) = 3 units
In right-angled ΔADB,
AB² = AD² + BD²
36 = AD² + 9 [∵ AB = 6 units]
AD² = 36 – 9 = 27
AD = 3√3 units
Now, area of the equilateral triangle = \(\frac{1}{2}\) × base × Height
= \(\frac{1}{2}\) × 6 × 3√3
= 9√3 sq units