Measuring Space Perimeter and Area Class 9 Extra Questions Maths Chapter 6

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Class 9 Measuring Space Perimeter and Area Extra Questions

Extra Questions on Measuring Space Perimeter and Area Class 9

Class 9 Ganita Manjari Chapter 6 Extra Questions

Measuring Space Perimeter and Area Class 9 Very Short Question Answer

Question 1.
Find the semi-perimeter of a triangle whose sides are 25 cm, 5 cm and 24 cm.
Solution:
Here, a = 25 cm, b = 5 cm, c = 24 cm
∴ s = \(\frac{a+b+c}{2}=\frac{25+5+24}{2}=\frac{54}{2}\)
= 27 cm
Hence, the semi-perimeter of the triangle is 27 cm.

Question 2.
The area of a triangle of base 35 cm is 420 cm². Find its altitude.
Solution:
Area of a triangle
= \(\frac{1}{2}\) × Base × Corresponding altitude
⇒ 420= \(\frac{1}{2}\) × 35 × Altitude
⇒ Altitude = \(\frac{420 \times 2}{35}\) = 24 cm

Question 3.
Find the area of a triangle whose sides are 6 cm, 8 cm and 10 cm respectively.
Solution:
∵ 6² + 8² = 10²
∴ Triangle is right angled.
∴ Area = \(\frac{a+b+c}{2}=\frac{6+8+10}{2}\) = 24 cm²
Aliter (Using Heron’s formula)
Here, a = 6 cm, b = 8 cm, c = 10 cm
s = \(\frac{a+b+c}{2}=\frac{6+8+10}{2}\)
= 12 cm
∴ Area = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{12(12-6)(12-8)(12-10)}\)
= \(\sqrt{12 \times 6 \times 4 \times 2}\)
= 24 cm²

Question 4.
The perimeter of an equilateral triangle is 60 cm. Find the area of the triangle.
Solution:
Let the side of the equilateral triangle be a cm.
Then, perimeter = 3 a cm
According to the question,
3a = 60 ⇒ a = 20
Area of the equilateral triangle
= \(\frac{\sqrt{3}}{4}\) a² = \(\frac{\sqrt{3}}{4}\) × 20 × 20 = 100√3 cm²

Question 5.
The area of an equilateral triangle is 16√3 sq. m. Find its perimeter.
Solution:
Let the side of the equilateral triangle be a m. Then,
area = \(\frac{\sqrt{3}}{4}\) a² m²
According to the question,
Area = 16√3 m²
\(\frac{\sqrt{3}}{4}\)a² = 16√3
⇒ a² = 64
⇒ a = 8
Perimeter = 3a
= 3 × 8 = 24 m

Question 6.
The height of an equilateral triangle measures 9√3 cm. Find its area.
Solution:
Let the side of the equilateral triangle ABC be a cm.

Then, AB = BC = CA = a cm …(1)
Let AD ⊥ BC
Now, AD = 9√3cm [Given]
∆ADB = ∆ADC [RHS Rule]
BD = CD [CPCT]
BD = \(\frac{1}{2}\)BC = \(\frac{a}{2}\) …(2)

In right triangle ADB,
AB² = AD² + BD² [By Pythagoras Theorem]
a = (9√3) + \(\left(\frac{a}{2}\right)^2\) [From (1) and (2)]
⇒ a² = 243 + \(\frac{1}{2}\)
⇒ \(\frac{3 a^2}{4}\) = 243
⇒ \(\frac{a^2}{4}\) = 81
⇒ a² = 81 × 4
⇒ a = 9 × 2 = 18 cm
Area of the equilateral triangle ABC
= \(\frac{1}{2}\)a2 = \(\frac{1}{2}\) × 18 × 18 = 8√3 cm²

Question 7.
Find the area of an equilateral triangle whose perimeter is 18 cm, using Heron’s formula.
(Use π = 1.73)
Solution:
Perimeter = 18 cm
∴ Side= \(\frac{18}{3}\) cm = 6 cm
Here, a = 6 cm, b = 6 cm, c = 6 cm
∴ S = \(\frac{a+b+c}{2}=\frac{6+6+6}{2}\) = 9 cm

Area = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{9(9-6)(9-6)(9-6)}\)
= 9√3 cm²
= 9 × 1.73 cm²
= 15.57 cm²

Question 8.
The base and the hypotenuse of a right angled triangle are 15 cm and 25 cm respectively. Find its area.
Solution:
In a right angled triangle,
(Hypotenuse)² = (Base)² + (Perpendicular)² [By Pythagoras Theorem]
⇒ (25)² = (15)² + (Perpendicular)²
⇒ 625 = 225 + (Perpendicular)²
⇒ (Perpendicular)² = 625 – 225
⇒ (Perpendicular)² = 400 = (20)²
⇒ Perpendicular = 20 m
∴ Area = \(\frac{\text { Base × Perpendicular }}{2}\)
= \(\frac{15 \times 20}{2}\)
= 150 cm²

Question 9.
In a right angled triangle, base and perpendicular are respectively 120 m and 22 m. Find the area of the triangle and length of the hypotenuse.
Solution:
∴ Area = \(\frac{\text { Base × Perpendicular }}{2}\)
= \(\frac{120 \times 22}{2}\) = 60 × 22 = 1320 m²

Length of the hypotenuse
= \(\sqrt{(\text { Base })^2+(\text { Perpendicular })^2}\) [By Pythagoras Theorem]
= \(\sqrt{(120)^2+(22)^2}=\sqrt{14400+484}\)
= \(\sqrt{14884}\) = 122 m

Question 10.
An isosceles right angled triangle has area 8 cm2. Find the length of its hypotenuse.
Solution:
Let each equal side be a cm. Then,
Area = 8 cm [Given]
⇒ a² = 16
⇒ a = 4 cm
Hypotenuse
= \(\sqrt{(\text { Base })^2+(\text { Perpendicular })^2}\) [By Pythagoras Theorem]
= \(\sqrt{(4)^2+(4)^2}\)
= 4√2 cm

Question 11.
In the figure, diagonals AC and BD of a trapezium ABCD with AB ∥CD intersect each other at O. Show that ar (∆AOD) = ar(∆BOC).

Solution:
Given: Diagonals AC and BD of a trapezium ABCD with AB ∥ CD intersect each other at O.
To Prove: ar(ΔAOD) = ar(ΔBOC)
Proof: ∵ ΔADB and ΔACB are on the same base AB and between the same parallels AB and DC
∴ ar(ΔADB) = ar(ΔACB)
∵ Two triangles on the same base and between the same parallels are equal in area
⇒ ar(ΔADB) = ar(ΔAOB)
= ar(ΔACB) – ar(ΔAOB)
⇒ ar(ΔAOD) = ar(ΔBOC)

Question 12.
ABCD is a quadrilateral and BD is one of its diagonals as shown in figure. Show that ABCD is a parallelogram and find its area.

Solution:
Given: ABCD is a quadrilateral and BD is one of its diagonals.
To Prove: ABCD is a parallelogram and to determine its area.
Proof: ∠ABD = ∠BDC (= 90°) [Given]
But these angles form a pair of equal alternate interior angles for lines AB, DC and a transversal BD
∴ AB ∥ DC
Also, AD = DC (= 3 cm) [Given]
Hence, quadrilateral ABCD is a parallelogram. A quadrilateral is a parallelogram if its one pair of opposite sides are parallel and equal Now,
ar(∥gm ABCD) = Base × corresponding altitude
= 3 × 4
= 12 cm²

Measuring Space Perimeter and Area Class 9 Short Question Answer

Question 1.
Length of a rectangular field is 15 m and its diagonal is of length 17 m. Find its area and the perimeter.
Solution:
Let ABCD be a rectangular field.

Then, AB = 15m, BD = 17m In right triangle BAD,
BD² = AB² + AD² [By Pythagoras Theorem]
⇒ (17)² = (15)² + AD²
⇒ 289 = 225 + AD²
⇒ AD² = 64
⇒ AD = 8 m
Area = AB × AD
= 15 × 8 = 120 m²
Perimeter = 2 (AB + AD)
= 2 (15 + 8)
= 46 m

Question 2.
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 126 cm, 120cm and 130cm and parallelogram stands on the base 120 cm, find the corresponding height of the parallelogram.
Solution:
For triangle
a = 126 cm, b = 120 cm, c = 130 cm
s = \(\frac{a+b+c}{2}=\frac{126+120+130}{2}\)
= \(\frac{376}{2}\)
= 188 cm
∴ Area = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{188(188-126)(188-120)(188-130)}\)
= \(\sqrt{188 \times 62 \times 68 \times 58}\)
= \(\sqrt{2 \times 2 \times 47 \times 2 \times 31 \times 2 \times 2 \times 17 \times 2 \times 29}\)
= 2 × 2 × 2\(\sqrt{47 \times 31 \times 17 \times 29}\)
= \(\sqrt{718301}\)
= 8 × 847.5
= 6780 cm² (approx.)

Let the corresponding height of the parallelogram be h cm. Then,
Area = Base × Corresponding height
= 120 × h
= 120 h cm²
According to the question,
Area of the parallelogram = Area of the triangle
⇒ 120 h = 6780
⇒ h = \(\frac{6780}{120}\)
⇒ h = 56.5 cm (approx.)
Hence, the height of the parallelogram is 12 cm.

Question 3.
A triangle and a parallelogram have the same base and same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Solution:
For triangle
a = 26 cm, b = 28 cm, c = 30 cm
∴ s = \(\frac{a+b+c}{2}=\frac{26+28+30}{2}\) = 42 cm
∴ Area of the triangle
= \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{42(42-26)(42-28)(42-30)}\)
= \(\sqrt{42 \times 16 \times 14 \times 12}\)
= \(\sqrt{2 \times 3 \times 7 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7 \times 2 \times 2 \times 3}\)
= \(\sqrt{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7 \times 7}\)
= 2 × 2 × 2 × 2 × 3 × 7
= 336 cm²

Let the corresponding height of the parallelogram be h cm. Then,
Area of the parallelogram = Base × Corresponding height
= 28 h cm²
According to the question,
Area of the parallelogram = Area of the triangle
⇒ 28h = 336
⇒ h = \(\frac{336}{28}\) = 12 cm
Hence, the height of the parallelogram is 12 cm.

Question 4.
The unequal side of an isosceles triangle is 4 cm and its perimeter is 20 cm. Find its area.
Solution:
a = 4 cm …(1)
Perimeter = 20 cm
⇒ a + b + b = 20
⇒ 4 + 2b = 20
⇒ 2b = 16
⇒ b = 8 cm
∴ Area = \(\frac{a}{4} \sqrt{4 b^2-a^2}=\frac{4}{4} \sqrt{4(8)^2-(4)^2}\)
= \(\sqrt{256-16}=\sqrt{240}\)
= \(\sqrt{2 \times 2 \times 2 \times 2 \times 3 \times 5}\)
= 2 × 2√5
= 4√5 cm²

Question 5.
The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 :2. Find the area of the triangle.
Solution:
a + b + b = 32 cm
⇒ a + 2b = 32 cm
b : a = 3 : 2
⇒ \(\frac{b}{a}=\frac{3}{2}\)
⇒ \(\frac{b}{3}=\frac{a}{2}\)= K (say)
⇒ b = 3K, a = 2K …(2)

From (1) and (2),
2K + 2(3K) = 32
2K + 6K = 32
8K = 32
K = 4
∴ b = 3 × 4 = 12 cm
a = 2 × 4 = 8 cm

Question 6.
Find the perimeter of an isosceles right angled triangle having an area of 5000 m2.
(Use √2 = 1.41)
Solution:
By Pythagoras Theorem,
a² = b² + b²
⇒ a² = 2b² …..(1)

Area of the isosceles right triangle
= \(\frac{\text { Base × Perpendicular }}{2}=\frac{b \times b}{2}=\frac{b^2}{2}\)
According to the question,
Area = 5000 m²
⇒ \(\frac{b^2}{2}\) = 5000

b² = 10000
b = \(\sqrt{10000}\)
b = 100 m …(2)
From (1) and (2),
a² = 2 × (100)²
a = 100√2 m …(3)
.’. Perimeter of the isosceles right triangle
= a + 2b = 100√2 + 2 × 100
= 100√2 (1 + √2) m

Question 7.
Find the area of an isosceles triangle having unequal side as 12 cm and each of the equal sides as 24 cm. Also, find its altitude corresponding to the
unequal side.
Solution:
a = 12 cm, b = 24 cm
Area of the isosceles triangle

= 36√15 cm² …(1)
= \(\frac{\text {Again, area of the isosceles triangle unequal side} \times \text { altitude corresponding to the unequal side }}{2}\)
= \(\frac{12 \times \text { altitude corresponding to the unequal side }}{2}\)
= 6 × altitude corresponding to the unequal side ⇒ 36\(\sqrt{15}\)
= 6 × altitude corresponding to the unequal side [From (1)]
⇒ Altitude corresponding to the unequal side = \(=\frac{36 \sqrt{15}}{6}\) = 6\(\sqrt{15}\) cm
Hence, the altitude corresponding to the unequal side is 6\(\sqrt{15}\) cm.

Question 8.
Each side of an equilateral triangle is 2x cm. If x√3 = 48, then find its area.
Solution:
x√3 = 48 ⇒ x = 16√3 ⇒ 2x = 32√3
a = 32√3 cm
Area = \(\frac{\sqrt{3}}{4}\) a² = \(\frac{\sqrt{3}}{4}\)(32√3)²
= 768√3 cm²

Question 9.
If the length of a median of an equilateral triangle is x cm, find its area.
Solution:
Let ABC be an equilateral triangle of side a.
∵ AD is a median.

ΔADB = ΔADC
∠ADB = ∠ADC
But ∠ADB + ∠ADC = 180° [Linear Pair Axiom]
∠ADB = ∠ADC = 90°
In right triangle ADB,
AB² = AD² + BD² [By Pythagoras Theorem]
⇒ a² = \(\left(\frac{a}{2}\right)^2\) + x²
⇒ \(\frac{3 a^2}{4}\) = x²
⇒ a² = \(\frac{4 x^2}{3}\)
⇒ a = \(\frac{2 x}{\sqrt{3}}\)

Area of the equilateral triangle ABC
= \(\frac{\sqrt{3}}{4} a^2=\frac{\sqrt{3}}{4}\left(\frac{2 x}{\sqrt{3}}\right)^2=\frac{\sqrt{3}}{4} \frac{4 x^2}{3}=\frac{x^2}{\sqrt{3}}\) cm²

Question 10.
A regular hexagon has a side 8 cm. Determine its perimeter and area.
Solution:
For regular hexagon
(i) Side = 8 cm
Perimeter = 6 × Side
= 6 × 8 = 48 cm

(ii) Area of equilateral triangle OAB
= \(\)(side)²
= \(\)(8)²
= 16√3 cm2 4 4

Area of the regular hexagon
= 6 × Area of equilateral triangle OAB
= 6 × 16√3
= 9673 cm²

Question 11.
The sides of a triangular ground are 5 m, 7 m and 8 m respectively. Find the cost of levelling the ground at the rate of ₹ 10 per m².
(Use √3 = 1.73).
Solution:
For triangular ground
a = 5 m, b = 7m, c = 8m
s = \(\frac{a+b+c}{2}=\frac{5+7+8}{2}\) = 10 cm
Area = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{10(10-5)(10-7)(10-8)}\)
= \(\sqrt{10 \times 5 \times 3 \times 2}\)
= 10√3
= 10 × 1.73
= 17.3 m²

∴ Cost of levelling
= 17.3 × 10 = ₹ 173

Question 12.
Find the cost of turfing a triangular field at the rate of ₹ 5 per m² having lengths of its sides as 40 m, 70 m and 90 m.
(Take \(\sqrt{20}\) = 4.47)
Solution:
For triangular field
a = 40 m, 6 = 70 m, c = 90 m
s = \(\frac{a+b+c}{2}=\frac{40+70+90}{2}\)
Area
= \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{100(100-40)(100-70)(100-90)}\)
= \(\sqrt{100 \times 60 \times 30 \times 10}\)
= 10x10x \(\frac{1}{2}\)
= 100\(\frac{1}{2}\)
= 100 × 3\(\sqrt{20}\)
= 300\(\sqrt{20}\)
= 300 × 4.47
= 1341 m²
∴ Cost of turfing = 1341 × 5 = ₹ 6705

Question 13.
The perimeter of a right triangle is 24 cm. If its hypotenuse is 10 cm, find its area.
Solution:
Let the sides AB and BC forming the right angle ABC be a cm and b cm respectively.
Then,
According to the question, a + b + 10 = 24 [∵ Perimeter = 24 cm]
⇒ a + b = 14 …(1)
Also, Hypotenuse = 10 cm
⇒ \(\sqrt{a^2+b^2}\) = 10 [By Pythagoras Theorem]
⇒ a² + b² = 100 …(2) [Squaring both sides]
We know that (a + b)² = a² + b² + 2ab
⇒ (14)² = 100 + 2ab [From (1) and (2)]
⇒ 196 = 100 + 2ab
⇒ 2ab = 96
⇒ ab = 48 …(3)
Also, we know that
(a – b)² = a² + b² – lab
= 100 – 2 (48) [From (2) and (3)]
= 100 – 96 = 4
⇒ a – b = 2 ……….(4) [If a > b (say)]
Solving (1) and (4), we get
a = 8 cm, b = 6 cm ……….(4)
Area of the right triangle ABC
= \(\frac{1}{2}\) × BC × AB = \(\frac{1}{2}\) × b × a = \(\frac{1}{2}\) ab
= \(\frac{1}{2}\)8.6
= 24 cm²

Question 14.
Prove that of all parallelograms of which the sides are given, the parallelogram which is a rectangle, has the greatest area.
Solution:
Given: A parallelogram ABCD whose sides are given.
To Prove: The area of the parallelogram ABCD is the greatest when it is a rectangle.
Construction: Draw DE ⊥ AB.
Proof: ar(∥ gm ABCD)
= Base × Corresponding altitude = AB × DE
When parallelogram ABCD is a rectangle, then its area = AB × DA
∵ Then ∠DAB = 90° as such DA will be the altitude
In right triangle DEA,
∠DEA > ∠DAE
∴ DA > DE
∵ Side opposite to greater angle of a triangle is longer
∴ ar(rectangle ABCD) > ar(∥ gm ABCD)
i. e., the area of the parallelogram ABCD is the greatest when it is a rectangle.

Question 15.
The diagonals of a parallelogram ABCD intersect at a point O. Through O, a line is drawn to intersect AD at P and BC at Q. Show that PQ divides the parallelogram into two parts of equal area.
Solution:
Given: The diagonals of a parallelogram ABCD intersect at a point 0. Through 0, a line is drawn to intersect AD at P and BC at Q.

To Prove: ar(☐PDCQ) = ar(☐PQBA).
Proof: ∵ AC is a diagonal of∥ gm ABCD
ar(∆ABC) = ar(∆ACD)
= \(\frac{1}{2}\) ar(∥ gm ABCD)

In ∆AOP and ∆COQ,
AO = CO
[Diagonals of a parallelogram bisect each other]
∠AOP = ∠COQ
[Vertically opposite angles]
∠OAP = ∠OCQ
[Alternate interior angles]
∆AOP ≅ ∆COQ
[ By ASA Congruence Rule]
ar(∆AOP) = ar(∆COQ)
[∵ Congruent figures have equal areas]
⇒ ar(∆AOP) + ar(☐OPDC)
= ar(∆COQ) + ar(☐OPDC)
⇒ ar(∆ACD) = ar(☐PDCQ)
⇒ \(\frac{1}{2}\) ar(|| gm ∆BCD) = ar(☐PDCQ)
[From (1)]
⇒ ar(☐PQBA) = ar(☐PDCQ)
⇒ ar(☐PDCQ) = ar(☐PQBA).

Question 16.
In figure, AD is median of triangle ABC, E is the mid-point of AD and F is the mid-point of AE. Prove that ar (ABF) = \(\frac{1}{8}\) ar (ABC).

Solution:
Given: AD is median of triangle ABC. E is the mid-point of AD and F is the mid-point of AE.
To Prove: ar(ABF) = \(\frac{1}{8}\) ar(ABC)
Proof: ∵ AD is a median of ∆ABC
∴ ar(∆ABD) = ar(∆ACD) = \(\frac{1}{2}\) ar(∆ABC) …(1)
[∵ A median of a triangle divides it into two triangles of equal areas]
∵ E is the mid-point of AD
BE is a median of ∆ABD
∴ ar(∆BED) = ar(∆BEA) = \(\frac{1}{2}\) ar(∆ABD)
[∵ A median of a triangle divides it into two triangles of equal areas]
⇒ ar(∆BEA) = \(\frac{1}{2}\).\(\frac{1}{2}\) ar(AABC) [From (1)]
= \(\frac{1}{4}\)ar(∆ABC) …(2)
∵ F is the mid-point of AE
∴ BF is a median of AABE
∴ ar(∆ABF) = ar(∆BEF) = \(\frac{1}{2}\) ar(AABE)
∵ A median of a triangle divides it into two triangles of equal areas
⇒ ar(∆ABF) = \(\frac{1}{2}\)ar(∆ABE)
= \(\frac{1}{2}\).\(\frac{1}{4}\) ar(∆ABC) [From (2)]
= \(\frac{1}{8}\) ar(∆ABC)

Question 17.
Parallelograms on the same base and between same parallels are equal in area. Prove this.
Solution:
Given: Two parallelograms ABCD and EFCD, on the same base DC and between the same parallels AF and DC.
To Prove: ar(ABCD) = ar(EFCD)

Proof: In ∆ADE and ∆BCF,
∠DAE = ∠CBF …(1)
Corresponding angles
(∵ AD ∥ BC and a transversal AF intersects them)
∠AED = ∠BFC …(2)
Corresponding angles (Y ED || FC and a transversal AF intersects them)
∠ADE = ∠BCF …(3)
[Angle sum property of a triangle]
Also AD = BC …(4)
Opposite sides of the parallelogram ABCD
∆ADE = ∆BCF
[By ASA congruence rule, using (1), (3) and (4) ar(AADE) = ar(ABCF)
[∵ Congruent figures have equal areas]
⇒ ar(AADE) + ar(EDCB)
= ar(ABCF) + ar(EDCB)
[Adding ar(EDCB) to both sides]
⇒ ar(ABCD) = ar(EFCD)
So, parallelograms ABCD and EFCD are equal in area.

Measuring Space Perimeter and Area Class 9 Long Question Answer

Question 1.
A traffic signal board, including ‘SCHOOL AHEAD’ is an equilateral triangle with side a. Find the area of signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?
Solution:
‘a’ = a, ‘b’ = a, ‘c’ = a
s = \(\frac{{ }^{\prime} a^{\prime}+{ }^{\prime} b^{\prime} c \prime}{2}\)
= \(\frac{a+a+a}{2}=\frac{3 a}{2}\)
Area of the signal board using Heron’s formula
= \(\sqrt{s\left(s-{ }^{\prime} a^{\prime}\right)\left(s-{ }^{\prime} b^{\prime}\right)\left(s-{ }^{\prime} c^{\prime}\right)}\)
= \(\sqrt{\frac{3 a}{2}\left(\frac{3 a}{2}-a\right)\left(\frac{3 a}{2}-a\right)\left(\frac{3 a}{2}-a\right)}\)
= \(\sqrt{\frac{3 a}{2}\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)}=\frac{\sqrt{3}}{4}\)a²
Again, perimeter = 180 cm
⇒ 3a =180
⇒ a = 60 cm
Area of the signal board
= \(\frac{\sqrt{3}}{4}\)(60)2 = 900√3 cm²

Question 2.
A triangular park has sides 120 m, 80 m and 50 m. A gardener Ramu has to put a fence all around it and also plant some trees inside the garden to get clean air.
(i) Find the cost of fencing at the rate of ₹ 5 per metre.
(ii) Find the area where Ramu plants the trees.
Solution:
For triangular park
a = 120 m, b = 80 m, c = 50 m

(i) ∴Perimeter = a + b + c
= 120 + 80 + 50
= 250 m

∴ Cost of fencing = 250 × 5 = ₹ 1250

(ii) s = \(\frac{a+b+c}{2}=\frac{120+80+50}{2}\)
= \(\frac{250}{2}\) = 125 m
∴ Area where Ramu plants the trees

Question 3.
The sides of a triangular park are 8 m, 10 m and 6 m respectively. A small circular area of diameter 2 m is to be left out and the remaining area is to be used for growing roses. How much area is used for growing roses? (Use π = 3.14)
Solution:
For triangular park
∵ 6² + 8² = 10²
∴ Triangle is right angled.
[By converse of Pythagoras Theorem]
∴ Area of the triangular park
= \(\frac{1}{2}\) × Base × Altitude
= \(\frac{1}{2}\) × 6 × 8 = 24 cm²
∵ Diameter of circular area = 2 m
∴ Radius of circular area (r) = \(\frac{2}{2}\) = 1 m
∴ Circular area = πr² = π(1)² = π m² = 3.14m²
∴ Area used for growing roses
= Area of the triangular park – circular area
= 24 – 3.14
= 20.86 m²

Question 4.
In the following figure, calculate the area of the shaded portion:

Solution:
In right triangle PSQ,
PQ² = PS² + QS²
[By Pythagoras Theorem]
= (12)² + (16)²
= 144 + 256 = 400
⇒ PQ = \(\sqrt{400}\) = 20 cm
For ∆PQR
PQ = 20 cm, QR = 48 cm, RP = 52 cm
∵ PQ² + QR² = 20² + 48² = 400 + 2304
= 2704
= 52² = PR²
∴Triangle PQR is right angled with Z Q = 90°.
[By converse of Pythagoras Theorem]
∴Area of ∆PQR = \(\frac{1}{2}\) × Base × Altitude
= \(\frac{1}{2}\) × QR × PQ
= \(\frac{1}{2}\) × 48 × 20
= 480 cm²
Area of ∆PSQ = \(\frac{1}{2}\) × Base × Altitude
= \(\frac{1}{2}\) × QS × PS
= \(\frac{1}{2}\) × 16 × 12
= 96 cm²
Area of the shaded portion = Area of ∆PQR – Area of ∆PSQ
= 480 – 96
= 384 cm²

Question 5.
If a cloth seller sells triangular shaped clothes of sides 18 units, 24 units, 30 units and 16 units, 26 units, 22 units to the customer at the same price, which is better deal for the customer
Solution:
For First triangular shaped cloth
a = 18 units, b = 24 units, c = 30 units
s = \(\frac{1}{2}\)
= \(\frac{72}{2}\) =36 units
∴Area = \(\sqrt{s(s-a)(s-b)(s-c)}\)
= \(\sqrt{36(36-18)(36-24)(36-30)}\)
= \(\sqrt{36 \times 18 \times 12 \times 6}\)
= \(\sqrt{36 \times 36 \times 6 \times 6}\)
= 36 × 6
= 216 square units

For second triangular shaped cloth
a = 16 units, b = 26 units, c = 22 units a + b + c

Question 6.
Black and white coloured triangular sheets are used to make a toy as shown in the figure. Find the total area of black and white coloured sheets used for making the toy.

Solution:
For one black coloured sheet
a = 4 cm, b = 6 cm
Area = \(\frac{a}{4} \sqrt{4 b^2-a^2}\)
[∵ Sheet is an isosceles triangle]
= \(\frac{4}{4} \sqrt{4(6)^2-(4)^2}=\sqrt{128}\) = 8√2 cm²
∴ Area of two black coloured sheets = 8√2 × 2 = 16>/2 cm²
Similarly, area of two white coloured sheets = 16>/2 cm²
∴ Total area of black and white coloured sheets = 16√2 + 16√2 = 32√2 cm²

Question 7.
Show that the area of a rhombus is half the product of the lengths of its diagonals.
Solution:
Let ABCD be a rhombus whose diagonals are AC and BD.
Then,
Area of rhombus ABCD = Area of ΔABD + Area of ΔCBD
= \(\frac{(\mathrm{BD})(\mathrm{AO})}{2}+\frac{(\mathrm{BD})(\mathrm{OC})}{2}\)
[∵ Diagonals of a rhombus are perpendiculars to each other]

= \(\frac{(\mathrm{BD})}{2}\)(AO + OC) = \(\frac{(\mathrm{BD})(\mathrm{AC})}{2}\)
= \(\frac{1}{2}\) Product of the lengths of its diagonals.

Question 8.
Given two points A and B and a positive real number k. Find the locus of a point P such that ar(∆PAB) = k.
Solution:
Given: Two points A and B and a positive real number k.
To find: The locus of a point P such that ar(ΔPAB) = k.
Construction: Draw PM ⊥ AB.
Determination: Let PM = h
ar(ΔPAB) = k [Given]
⇒ \(\frac{1}{2}\)(AB)(PM) = k
⇒ h = \(\frac{2k}{AB}\)

∵ Points A and B are given.
∴ AB is fixed.
∴ Also, k being a positive real number k is fixed. h is a fixed positive real number.
∴ The locus of P is a line parallel to the line
AB at a fixed distance \(\frac{2k}{AB}\) on either side of it.

Question 9.
Show that the line segment joining the mid-points of a pair of opposite sides of a parallelogram divides it into two equal parallelograms.
Solution:
Given: A parallelogram ABCD in which M and N are the mid-points of a pair of its opposite sides AB and DC respectively.

To Prove: ar(∥ gm AMND) = ar(∥ gm MBCN).
Proof: ∵ABCD is a ∥ gm
∴ AB = DC and AB ∥ DC
⇒ \(\frac{1}{2}\)AB = \(\frac{1}{2}\)DC and AB ∥DN
⇒ AM = DN and AM ⇒ DN
⇒ ☐AMND is a ∥gm.
Similarly, we can prove that ☐MBCN is a parallelogram.
∵ ∥ gm AMND and ∥ gm MBCN are on equal bases AM and MB (∵ M is the mid-point of AB) and between the same parallels AB and DC.
∴ ar(∥ gm AMND) = ar(∥ gm MBCN).

Question 10.
In an equilateral triangle, O is any point is the interior of the triangle and perpendiculars are drawn from O to the sides. Prove that the sum of these perpendicular line segments is constant.
Solution:
Given: ABC is an equilateral triangle. 0 is any point in the interior of the triangle. Perpendiculars OD, OE and OF are drawn from O on the sides BC, AB and AC respectively of ∆ABC.
To Prove: OD + OE + OF = constant
Construction: Join O to A, B and C. Draw AH ⊥ BC.
Proof: ∵ ∆ABC is an equilateral triangle AB = BC = CA
ar(∆AOB) = \(\frac{(\mathrm{AB})(\mathrm{OE})}{2}=\frac{(\mathrm{BC})(\mathrm{OE})}{2}\) ……(1) [∵ AB = BC]

ar (∆BOC) = \(\frac{(\mathrm{BC})(\mathrm{OD})}{2}\) …(2)
ar(∆COA) = \(\frac{(\mathrm{CA})(\mathrm{OF})}{2}=\frac{(\mathrm{BC})(\mathrm{OF})}{2}\) ….. (3) [∵ CA = BC]

Adding (1), (2) and (3), we get
ar (∆AOB) + ar(∆BOC) + ar(∆COA)
⇒ ar (∆AOB) = \(\frac{(\mathrm{BC})(\mathrm{OE})}{2}+\frac{(\mathrm{BC})(\mathrm{OD})}{2}\) + \(\frac{(\mathrm{BC})(\mathrm{OF})}{2}\)
⇒ \(\frac{1}{2}\)(BC)(AH) = \(\frac{1}{2}\) BC(OE + OD + OF)
⇒ AH = OD + OE + OF
⇒ OD + OE + OF = AH which is constant for a given triangle.

Measuring Space Perimeter and Area Extra Questions for Practice

Very Short Answer Type Questions

Question 1.
Find the area of the sector of a circle of radius 5 cm. if the corresponding arc length is 3.5 cm.
Solution:
8.75 cm²

Question 2.
The length of the minute hand of a clock is 6 cm. Find the distance covered by the tip of the minute hand during the time period 6: 05 am and 6: 40 am.
Solution:
22 cm

Question 3.
Find the area of minor sector of circle of radius 21 cm and central angle 120°.
Solution:
462 cm²

Question 4.
Find the perimeter of a square whose area is same as that of a rectangle with sidelengths 9 cm and 4 cm.
Solution:
24 cm

Question 5.
Find the area of major sector of circle of radius 14 cm and central angle 90°.
Solution:
462 cm

Question 6.
Find the area of the minor segment of a circle of radius 14 cm, when the angle of the corresponding sector is 60°.
Solution:
13.23 cm²

Short Answer Type Questions

Question 1.
Sides of a triangular field are 15 m, 16 m and 17 m. With the three corners of the field a cow, a buffalo and a horse are tied separately with ropes of length 7 m each to graze in the field. Find the area of the field which cannot be grazed by the three animals.
Solution:
77 m²

Question 2.
Area of a sector of central angle 200° of a circle is 770 cm². Find the length of the corresponding arc of this sector.
Solution:
21 cm

Question 3.
A calf is tied with a rope of length 6 m at the comer of a square grassy lawn of side 20 m. If the length of the rope is increased by 5.5m, find the increase is areas of the grassy lawn in which the calf can graze.

Solution:
75.625 m²

Question 4.
To warm ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 9 km. Find the area of the sea over which the ships are warmed. (Use n = 3.14)
Solution:
56.52 m²

Question 5.
In the given figure, ABCD is trapezium with AB || DC, AB = 18 cm, DC = 32 cm and distance between AB and DC = 14 cm. If arcs of equal radii 7 cm with centres A, B, C and D have been drawn, then find the area of the shaded region of the figure.

Solution:
196 cm²

Question 6.
An umbrella has 9 ribs that are equally spaced. Assuming umbrella to be a flat circle of radius 42 cm, find the area between the two consecutive ribs of the umbrella.
Solution:
616 cm²

Question 7.
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (i) length of the arc (ii) area of the sector formed by the arc.
Solution:
(i) 22 cm
(ii) 231 cm²

Question 8.
A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14)
Solution:
13.08 cm²

Long Answer Type Questions

Question 1.
The length of the minute hand of a clock is 3.5 cm. Find the area swept by the minute hand during the time period 10 : 04 am and 10 : 40 am.
Solution:
3.58 cm²

Question 2.
A horse is tethered to one comer of a field, which is in the shape of an equilateral triangle of side 20 m. If the length of the rope is 10.5 m, find the area of field the horse cannot graze.
Solution:
57.5 m²

Question 3.
A car has two wipers which do not overlap. Each wiper has a blade of length 21 cm sweeping through an angle of 1200. Find the total area cleaned at each sweep of the blades.
Solution:
924 cm²

Question 4.
In the given figure, AOH is a quarter circle. OCB and OCA are semicircies with OB and OA as diameters where, area of region OCO = a Units and area of region BCAB = b units. Then show a = b.

Solution:
Do it Yourself

Question 5.
A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors shown in given figure. Find:
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.
Solution:
(i) 258 mm
(ii) 96.25 mm²

Question 6.
A horse is tied to a peg at one comer of a square shaped grass field of side 15 m by means of a 5 m long rope. Find
(i) the area of that part of the field in which the horse can graze.
(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)
Solution:
(i) 19.325 m²
(ii) 58.875 m²

Case Study Based Questions

Mona bought a pendulum clock for her living room, the clock has a pendulum of length 45 cm the minute hand and hour hand of the clock are 9 cm and 6 cm long respectively.

Question 1.
Find the angle described by hour hand in 10 minutes.
Solution:
60°

Question 2.
Find the area swept by the minute hand in 14 minutes.
Solution:
59.4 cm

Question 3.
Find the distance covered by the tip of hour hand in 3.5 hours.
Solution:
132 cm²

Question 4.
If the tip of pendulum covers a distance of 66 cm in complete oscillation, then find the angle described by pendulum at the centre.
Solution:
84°