Math Ncert Exemplar 2019 Solutions for Class 9 Maths Chapter 12 - Heron's Formula

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Page No 113:

Question 1:

An isosceles right triangle has area 8 cm². The length of its hypotenuse is

(A) $\sqrt{32}$ cm

(B) $\sqrt{16}$ cm

(C) $\sqrt{48}$ cm

(D) $\sqrt{24}$ cm

Answer:

Let base and height be $x$ .
Area = 8 cm²
$\Rightarrow \frac{1}{2} \times Base \times height = 8 \Rightarrow \frac{1}{2} \times x \times x = 8 \Rightarrow x^{2} = 16 \Rightarrow x = 4 cm$
Length of hypotenuse = $\sqrt{4^{2} + 4^{2}}$
= $\sqrt{32} cm$

Hence, the correct answer is option A.

Page No 114:

Question 2:

The perimeter of an equilateral triangle is 60 m. The area is

(A) $10 \sqrt{3} m^{2}$

(B) $15 \sqrt{3} m^{2}$

(C) $20 \sqrt{3} m^{2}$

(D) $100 \sqrt{3} m^{2}$

Answer:

Let the length of side of an equilateral triangle be $x$ .
Perimeter of equilateral triangle = $3 \times x$
⇒ $60 = 3 \times x$
⇒ $x = 20 m$
$\begin{array}{rcl} Area of equilateral triangle & = & \frac{\sqrt{3}}{4} \times {(side)}^{2} \\ = & \frac{\sqrt{3}}{4} \times 20 \times 20 \\ = & 100 \sqrt{3} m^{2} \end{array}$

Hence, the correct answer is option D.

Page No 114:

Question 3:

The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then the area of the triangle is
(A) 1322 cm²
(B) 1311 cm²
(C) 1344 cm²
(D) 1392 cm²

Answer:

By heron's formula,
Area of triangle = $\sqrt{s (s - a) (s - b) (s - c)}$
Where s = $\frac{a + b + c}{2}$
We have, a = 56 cm, b = 60 cm and c = 52 cm.
Now,
$s = \frac{56 + 60 + 52}{2} = 84$
$\begin{array}{rcl} Area of triangle & = & \sqrt{84 (84 - 56) (84 - 60) (84 - 52)} \\ = & \sqrt{84 \times 28 \times 24 \times 32} \\ = & \sqrt{3 \times 7 \times 4 \times 4 \times 7 \times 4 \times 2 \times 3 \times 16 \times 2} \\ = & 3 \times 7 \times 4 \times 2 \times 2 \times 4 \\ = & 1344 {cm}^{2} \end{array}$

Hence, the correct answer is option C.

Page No 114:

Question 4:

The area of an equilateral triangle with side $2 \sqrt{3}$ cm is
(A) 5.196 cm²
(B) 0.866 cm²
(C) 3.496 cm²
(D) 1.732 cm²

Answer:

$\begin{array}{rcl} ∴ Area of equilateral triangle & = & \frac{\sqrt{3}}{4} \times {(side)}^{2} \\ = & \frac{\sqrt{3}}{4} \times 2 \sqrt{3} \times 2 \sqrt{3} \\ = & 3 \sqrt{3} {cm}^{2} \\ = & 5.196 {cm}^{2} \end{array}$

Hence, the correct answer is option A.

Page No 114:

Question 5:

The length of each side of an equilateral triangle having an area of $9 \sqrt{3}$ cm²is
(A) 8 cm
(B) 36 cm
(C) 4 cm
(D) 6 cm

Answer:

Area of equilateral triangle = $\frac{\sqrt{3}}{4} \times {(side)}^{2}$
$\Rightarrow 9 \sqrt{3} = \frac{\sqrt{3}}{4} \times {(side)}^{2} \Rightarrow {(side)}^{2} = 36 \Rightarrow side = 6 cm$
Hence, the correct answer is option D.

Page No 114:

Question 6:

If the area of an equilateral triangle is $16 \sqrt{3}$ cm², then the perimeter of the triangle is
(A) 48 cm
(B) 24 cm
(C) 12 cm
(D) 36 cm

Answer:

Area of equilateral triangle = $\frac{\sqrt{3}}{4} \times {(side)}^{2}$
$\Rightarrow 16 \sqrt{3} = \frac{\sqrt{3}}{4} \times {(side)}^{2}$
$\Rightarrow {(side)}^{2} = 64$
$\Rightarrow side = 8 cm$
$\begin{array}{rcl} Perimeter of equilateral triangle & = & 3 \times (side) \\ = & 3 \times 8 \\ = & 24 cm \end{array}$

Hence, the correct answer is option B.

Page No 114:

Question 7:

The sides of a triangle are 35 cm, 54 cm and 61 cm, respectively. The length of its longest altitude
(A) $16 \sqrt{5}$ cm
(B) $10 \sqrt{5}$ cm
(C) $24 \sqrt{5}$ cm
(D) 28 cm

Answer:

Let ABC be a triangle with AB = 35 cm, BC = 54 cm, CA = 61 cm.

Now, semi perimeter (s)
$\begin{array}{rcl} ∴ s & = & \frac{a + b + c}{2} \\ = & \frac{54 + 61 + 35}{2} \\ = & 75 cm \end{array}$
$\begin{array}{rcl} ∴ Area of triangle & = & \sqrt{s (s - a) (s - b) (s - c)} \\ = & \sqrt{75 (75 - 61) (75 - 54) (75 - 35)} \\ = & \sqrt{75 \times 14 \times 21 \times 40} \\ = & \sqrt{25 \times 3 \times 2 \times 7 \times 3 \times 7 \times 2 \times 5 \times 4} \\ = & 5 \times 3 \times 2 \times 7 \times 2 \sqrt{5} \\ = & 420 \sqrt{5} {cm}^{2} \end{array}$
Also, Area of triangle = $\frac{1}{2} \times base \times altitude$
$\Rightarrow 420 \sqrt{5} = \frac{1}{2} \times 35 \times$ Altitude
⇒ Altitude = $24 \sqrt{5} cm$

Hence, the correct answer is option C.

Page No 114:

Question 8:

The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is

(A) $\sqrt{15} {cm}^{2}$

(B) $\sqrt{\frac{15}{2}} {cm}^{2}$

(C) $2 \sqrt{15} {cm}^{2}$

(D) $4 \sqrt{15} {cm}^{2}$

Answer:

Area of the triangle by heron's formula = $\sqrt{s (s - a) (s - b) (s - c)}$
Where, s = $\frac{a + b + c}{2}$
We have, a = 4 cm, b = 4 cm and c = 2 cm.
Now,
s = $\begin{array}{rcl} ∴ s & = & \frac{a + b + c}{2} \\ = & \frac{4 + 4 + 2}{2} \\ = & 5 cm \end{array}$
$\begin{array}{rcl} ∴ Area of triangle & = & \sqrt{s (s - a) (s - b) (s - c)} \\ = & \sqrt{5 (5 - 4) (5 - 4) (5 - 2)} \\ = & \sqrt{5 \times 1 \times 1 \times 3} \\ = & \sqrt{15} {cm}^{2} \end{array}$
Hence, the correct answer is option A.

Page No 114:

Question 9:

The edges of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it at the rate of 9 paise per cm²is
(A) Rs 2.00
(B) Rs 2.16
(C) Rs 2.48
(D) Rs 3.00

Answer:

Area of triangle = $\sqrt{s (s - a) (s - b) (s - c)}$
Where, s = $\frac{a + b + c}{2}$
Now,
s = $\frac{6 + 8 + 10}{2} = \frac{24}{2} = 12 cm .$
Here,
$\begin{array}{rcl} Area of triabgle & = & \sqrt{12 (12 - 6) (12 - 8) (12 - 10)} \\ = & \sqrt{12 \times 6 \times 4 \times 2} \\ = & 6 \times 2 \times 2 \\ = & 24 {cm}^{2} \end{array}$
$\begin{array}{rcl} Cost of painting & = & 24 \times (9 paise) = 216 paise \\ = & Rs . 2.16 \end{array}$
Hence, the correct answer is option B.

Page No 115:

Question 1:

Write True or False and justify your answer:
The area of a triangle with base 4 cm and height 6 cm is 24 cm².

Answer:

False,
$\begin{array}{rcl} Area of triangle & = & \frac{1}{2} \times base \times height \\ = & \frac{1}{2} \times 4 \times 6 \\ = & 12 {cm}^{2} \end{array}$
Hence, the area of triangle will be 12 cm².

Page No 115:

Question 2:

Write True or False and justify your answer:
The area of âˆ†ABC is 8 cm²in which AB = AC = 4 cm and ∠A = 90º.

Answer:

True

We have, ABC is a right-angled triangle at A, where AB = AC = 4 cm.

$\begin{array}{rcl} Area of triangle & = & \frac{1}{2} \times base \times altitude \\ = & \frac{1}{2} \times 4 \times 4 \\ = & 8 {cm}^{2} \end{array}$

Here, area of $∆ ABC$ is 8 cm².

Page No 115:

Question 3:

Write True or False and justify your answer:
The area of the isosceles triangle is $\frac{5}{4} \sqrt{11}$ cm², if the perimeter is 11 cm and the base is 5 cm.

Answer:

True
Base = 5 cm, Perimeter = 11 cm.
Let other equal sides be $x$ .
$∴ x + x + 5 = 11 \Rightarrow 2 x = 6 \Rightarrow x = 3 cm$

By Heron's formula
Area of triangle = $\sqrt{s (s - a) (s - b) (s - c)}$
Here, s = $\frac{a + b + c}{2}$
Now,
s = $\frac{5 + 3 + 3}{2} = \frac{11}{2} cm$
$\begin{array}{rcl} Area of triangle & = & \sqrt{\frac{11}{2} (\frac{11}{2} - 5) (\frac{11}{2} - 3) (\frac{11}{2} - 3)} \\ = & \sqrt{(\frac{11}{2}) (\frac{1}{2}) (\frac{5}{2}) (\frac{5}{2})} \\ = & \frac{5 \sqrt{11}}{4} {cm}^{2} \end{array}$
Hence, area of triangle is $\frac{5 \sqrt{11}}{4}$ cm².

Page No 115:

Question 4:

Write True or False and justify your answer:
The area of the equilateral triangle is $20 \sqrt{3}$ cm²whose each side is 8 cm.

Answer:

False
$\begin{array}{rcl} Area of equilateral triangle & = & \frac{\sqrt{3}}{4} \times {(side)}^{2} \\ = & \frac{\sqrt{3}}{4} \times 8 \times 8 \\ = & 16 \sqrt{3} {cm}^{2} \end{array}$
Hence, the area of an equilateral triangle is $16 \sqrt{3} {cm}^{2}$ .

Page No 115:

Question 5:

Write True or False and justify your answer:
If the side of a rhombus is 10 cm and one diagonal is 16 cm, the area of the rhombus is 96 cm².

Answer:

Ture,
To find area of rhombus we divide it into two triangles.

By Heron's formula,
Area of triangle = $\sqrt{s (s - a) (s - b) (s - c)}$ , where, s = $\frac{a + b + c}{2}$
Now,
s = $\frac{10 + 10 + 16}{2} = 18 cm$
$\begin{array}{rcl} Area of triangle & = & \sqrt{18 (18 - 10) (18 - 10) (18 - 16)} \\ = & \sqrt{18 (8) (8) (2)} \\ = & 3 \times 8 \times 2 \\ = & 48 {cm}^{2} \end{array}$
$\begin{array}{rcl} ∴ Area of rhombas & = & 2 \times Area of triangle \\ = & 2 \times 48 \\ = & 96 {cm}^{2} \end{array}$
Hence, the area of the rhombus is 96 cm².

Page No 115:

Question 6:

Write True or False and justify your answer:
The base and the corresponding altitude of a parallelogram are 10 cm and 3.5 cm, respectively. The area of the parallelogram is 30 cm².

Answer:

False,

$\begin{array}{rcl} Area of parallelogram & = & Base \times Alititude \\ = & 10 \times 3.5 \\ = & 35 {cm}^{2} \end{array}$

Hence, the area of the parallelogram is 35 cm².

Page No 115:

Question 7:

Write True or False and justify your answer:
The area of a regular hexagon of side ‘a’ is the sum of the areas of the five equilateral triangles with side a.

Answer:

False,

We know that a regular hexagon is divided into six equilateral triangles.
Area of a regular hexagon is the sum of the areas of six equilateral triangles.

Page No 115:

Question 8:

Write True or False and justify your answer:
The cost of levelling the ground in the form of a triangle having the sides 51 m, 37 m and 20 m at the rate of Rs 3 per m²is Rs 918.

Answer:

True,

Let a = 51 m, b = 37 m, c = 20 m.

By Heron's formula
Area of triangle = $\sqrt{s (s - a) (s - b) (s - c)}$
Here, s = $\frac{a + b + c}{2}$
Now, s = $\frac{51 + 37 + 20}{2} = 54$ m

$\begin{array}{rcl} ∴ Area of triangle & = & \sqrt{54 (54 - 51) (54 - 37) (54 - 2)} \\ = & 306 m^{2} \end{array}$
$\begin{array}{rcl} Cost of Painting & = & Area \times Cost per m^{2} \\ = & 306 \times 3 \\ = & ₹ 918 \end{array}$

Hence, cost of the Painting is â‚¹918.

Page No 115:

Question 9:

Write True or False and justify your answer:
In a triangle, the sides are given as 11 cm, 12 cm and 13 cm. The length of the altitude is 10.25 cm corresponding to the side having length 12 cm.

Answer:

True,

Let, a = 11 cm, b = 12 cm and c = 13 cm.
By Heron's formula
Area of triangle = $\sqrt{s (s - a) (s - b) (s - c)}$
Here, s = $\frac{a + b + c}{2}$
Now, s = $\frac{11 + 12 + 13}{2} = 18 cm$
$\begin{array}{rcl} Area & = & \sqrt{18 (18 - 11) (18 - 12) (18 - 13)} \\ = & 61.5 {cm}^{2} \end{array}$
$∴$ Area of triangle = $\frac{1}{2} \times$ base $\times$ height
⇒ 61.5 = $\frac{1}{2} \times 12 \times$ height
⇒ Height = $\frac{61.5 \times 2}{12}$
⇒ Height = 10.25 cm

Hence, length of altitude is 10.25 cm.

Page No 117:

Question 1:

Find the cost of laying grass in a triangular field of sides 50 m, 65 m and 65 m at the rate of Rs 7 per m².

Answer:

Given: sides of triangle, a = 50 m, b = 65 m, c = 65 m
By Heron's formula
Area of triangle = $\sqrt{s (s - a) (s - b) (s - c)}$
Here, s = $\frac{a + b + c}{2}$
Now,
s = $\frac{65 + 65 + 50}{2} = 90 cm$
$\begin{array}{rcl} Area of triangle & = & \sqrt{90 (90 - 50) (90 - 65) (90 - 65)} \\ = & \sqrt{90 \times 40 \times 25 \times 25} \\ = & 25 \times \sqrt{9 \times 10 \times 4 \times 10} \\ = & 25 \times 3 \times 2 \times 10 \\ = & 1500 m^{2} \end{array}$
$\begin{array}{rcl} Cost of laying grass & = & Area \times Cost per m^{2} \\ = & 1500 \times 7 \\ = & Rs 10500 \end{array}$

Hence, cost of laying grass is Rs. 10500.

Page No 117:

Question 2:

The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 13 m, 14 m and 15 m. The advertisements yield an earning of Rs 2000 per m²a year. A company hired one of its walls for 6 months. How much rent did it pay?

Answer:

Given: sides of triangle a = 13 m, b = 14 cm, c = 15 m
From Heron's formula, we have
Area = $\sqrt{s (s - a) (s - b) (s - c)}$
Where s = $\frac{a + b + c}{2}$
Now,
s = $\frac{13 + 14 + 15}{2} = 21 cm$
$\begin{array}{rcl} Area & = & \sqrt{21 (21 - 13) (21 - 14) (21 - 15)} \\ = & \sqrt{21 \times 8 \times 7 \times 6} \\ = & 84 {cm}^{2} \end{array}$

Advertisements yield is â‚¹2000 for 1 year
For 6 months = $\frac{2000}{2} = ₹ 1000$
$\begin{array}{rcl} Total cost for 6 months & = & ₹ 1000 \times 84 \\ = & ₹ 84000 \end{array}$

Hence, the company has to pay â‚¹84000.

Page No 117:

Question 3:

From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are 14 cm, 10 cm and 6 cm. Find the area of the triangle.

Answer:

Let the side of an equilateral triangle be a cm.

Here OP, OQ and OR are perpendicular or side AB, BC and CA of equilateral triangle.
$\begin{array}{rcl} Area of △ OAB & = & \frac{1}{2} \times 14 \times AB \\ = & \frac{1}{2} \times 14 \times a \\ = & 7 a \end{array}$
$\begin{array}{rcl} Area of △ OBC & = & \frac{1}{2} \times 10 \times a \\ = & 5 a \end{array}$
$\begin{array}{rcl} Area of △ OAC & = & \frac{1}{2} \times 6 \times a \\ = & 3 a \end{array}$

Area of equilateral triangle $△ ABC$ = Area of $△ OAB + Area of △ OBC + Area of △ OAC$
$\Rightarrow \frac{\sqrt{3}}{4} a^{2} = 7 a + 5 a + 3 a$
$\Rightarrow \frac{\sqrt{3}}{4} a^{2} = 15 a$
$\Rightarrow a = \frac{60}{\sqrt{3}} = 20 \sqrt{3} cm$

$\begin{array}{rcl} ∴ Area of equilateral ∆ ABC & = & \frac{\sqrt{3}}{4} a^{2} \\ = & \frac{\sqrt{3}}{4} \times {(20 \sqrt{3})}^{2} \\ = & 300 \sqrt{3} {cm}^{2} \end{array}$

Hence, the area of triangle is $300 \sqrt{3} {cm}^{2}$ .

Page No 117:

Question 4:

The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3 : 2. Find
.

Answer:

Given: Perimeter of an isosceles triangle = 32 cm
Ratio of equal side to base = 3 : 2
Let equal side be $3 x$ and base be $2 x$ .

$\Rightarrow 3 x + 3 x + 2 x = 32 \Rightarrow x = 4$

$∴$ Equal sides = 12 cm, Base = 8 cm.
Let a = 12 cm, b = 12 cm, c = 8 cm.
From Heron's formula.
Area of triangle = $\sqrt{s (s - a) (s - b) (s - c)}$
Here s = $\frac{a + b + c}{2}$
Now,
s = $\frac{12 + 12 + 8}{2} = 16 cm$
$\begin{array}{rcl} Area & = & \sqrt{16 (16 - 12) (16 - 12) (16 - 8)} \\ = & \sqrt{16 \times 4 \times 4 \times 8} \\ = & 32 \sqrt{2} {cm}^{2} \end{array}$

Hence, the area of the triangle is $32 \sqrt{2} {cm}^{2}$ .

Page No 117:

Question 5:

Find the area of a parallelogram given in the figure. Also find the length of the altitude from vertex A on the side DC.

Answer:

In $△ BCD$ , let a = 12 cm, b = 17 cm, c = 25 cm
From Heron's formula,
Area of triangle = $\sqrt{s (s - a) (s - b) (s - c)}$
Where s = $\frac{a + b + c}{2}$
Now,
s = $\frac{12 + 17 + 25}{2} = 27 cm$
$\begin{array}{rcl} Area of triangle ∆ BCD & = & \sqrt{27 (27 - 12) (27 - 17) (27 - 25)} \\ = & \sqrt{27 \times 15 \times 10 \times 2} \\ = & \sqrt{9 \times 3 \times 3 \times 5 \times 5 \times 2 \times 2} \\ = & 3 \times 3 \times 5 \times 2 \\ = & 90 {cm}^{2} \end{array}$

$\begin{array}{rcl} Area of parallelogram ABCD & = & 2 \times Area of △ BCD \\ = & 2 \times 90 \\ = & 180 {cm}^{2} \end{array}$

Let the altitude of parallelogram be h.
Also, area of parallelogram ABCD = Base $\times$ Altitude
⇒ 180 = 12 $\times$ h
⇒ h = 15 cm

Hence, area of parallelogram is 180 cm², and altitude is 15 cm.

Page No 117:

Question 6:

A field in the form of a parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. Find the area of the parallelogram.

Answer:

In $△ ACB$ , let a = 60 m, b = 40 m, c = 80 m.
From Heron's formula,
Area of triangle = $\sqrt{s (s - a) (s - b) (s - c)}$
Here, s = $\frac{a + b + c}{2}; s = \frac{60 + 40 + 80}{2} = 90 m$
Now,
$\begin{array}{rcl} Area of △ BCA & = & \sqrt{90 (90 - 60) (90 - 40) (90 - 80)} \\ = & \sqrt{90 \times 30 \times 50 \times 20} \\ = & 100 \times 3 \times \sqrt{15} \\ = & 300 \sqrt{15} {cm}^{2} \end{array}$
$\begin{array}{rcl} Area of paralleogram ABCD & = & 2 \times Area of △ ACB \\ = & 2 \times 300 \sqrt{15} \\ = & 600 \sqrt{15} {cm}^{2} \end{array}$

Hence, the area of parallelogram is $600 \sqrt{15} {cm}^{2}$ .

Page No 117:

Question 7:

The perimeter of a triangular field is 420 m and its sides are in the ratio 6 : 7 : 8. Find the area of the triangular field.

Answer:

Given: Perimeter of triangle = 420 m
Ratio of sides length = 6 : 7 : 8.
Let the side length be $6 x, 7 x, 8 x$
$∴ 6 x + 7 x + 8 x = 420 \Rightarrow 21 x = 420 \Rightarrow x = 20$
Hence, sides of triangle are 120 cm, 140 cm, 160 cm.
By Heron's formula
Area of triangle = $\sqrt{s (s - a) (s - b) (s - c)}$
Here, s = $\frac{a + b + c}{2}$
Now, s = $\frac{120 + 140 + 160}{2} = 210 m$
$\begin{array}{rcl} Area of triangle & = & \sqrt{210 (210 - 120) (210 - 140) (210 - 160)} \\ = & \sqrt{210 (90) (70) (50)} \\ = & 2100 \sqrt{15} m^{2} \end{array}$

Hence, the area of the triangular field is $2100 \sqrt{15} m^{2}$ .

Page No 117:

Question 8:

The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order) respectively, and the angle between the first two sides is a right angle. Find its area.

Answer:

Given: ABCD is a quadrilateral, AB = 6 cm, BC = 8 cm, CD = 12 cm, DA = 14 cm.

Here, ABC is a right triangle at B.
By Pythagoras theorem,
${(AC)}^{2} = {(AB)}^{2} + {(BC)}^{2} \Rightarrow {(AC)}^{2} = {(6)}^{2} + {(8)}^{2} \Rightarrow {(AC)}^{2} = 36 + 64 = 10^{2} \Rightarrow AC = 10 cm$

Now,
Area of quadrilateral ABCD = Area of ( $△ ABC + △ ACD$ )
$\begin{array}{rcl} ∴ Area of △ ABC & = & \frac{1}{2} \times base \times height \\ = & \frac{1}{2} \times 8 \times 6 \\ = & 24 {cm}^{2} \end{array}$
By Heron's formula
Area of $△ ACD = \sqrt{s (s - a) (s - b) (s - c)}$
Here, a = 14 cm, b = 12 cm, c = 10 cm
s = $\frac{12 + 10 + 14}{2} = 18 cm$
$\begin{array}{rcl} Area of △ ACD & = & \sqrt{18 (18 - 10) (18 - 12) (18 - 14)} \\ = & \sqrt{18 \times 8 \times 6 \times 4} \\ = & \sqrt{{(3)}^{2} {(4)}^{2} \times 4 \times 2 \times 3} \\ = & 24 \sqrt{6} {cm}^{2} \end{array}$
$\begin{array}{rcl} Area of quadrilateral ABCD & = & 24 + 24 \sqrt{6} \\ = & 24 (1 + \sqrt{6}) {cm}^{2} \end{array}$

Hence, the area of the quadrilateral is $\begin{array}{rcl} 24 (1 + \sqrt{6}) {cm}^{2} \end{array}$ .

Page No 117:

Question 9:

A rhombus shaped sheet with perimeter 40 cm and one diagonal 12 cm, is painted on both sides at the rate of Rs 5 per m². Find the cost of painting.

Answer:

Let the side length of rhombus be $x$ .
Perimeter of rhombus = 4 $\times$ side
⇒ 40 = 4 $\times x$
$\Rightarrow x = 10 cm$

By Heron's formula,
Area of triangle = $\sqrt{s (s - a) (s - b) (s - c)}$
Here, s = $\frac{a + b + c}{2}$
Now,
s = $\frac{10 + 10 + 12}{2} = 16 cm$
$\begin{array}{rcl} Area of △ ACB & = & \sqrt{16 (16 - 10) (16 - 10) (16 - 12)} \\ = & \sqrt{16 \times 6 \times 6 \times 4} \\ = & 96 {cm}^{2} \end{array}$
$\begin{array}{rcl} Area of rhombus & = & 2 \times Area of △ ACB \\ = & 2 \times 96 \\ = & 192 {cm}^{2} \end{array}$
$\begin{array}{rcl} Cost of Painting sheet & = & Rs (192 \times 5 \times 2) \\ = & Rs 960 \end{array}$

Hence, the cost of painting is â‚¹960.

Page No 117:

Question 10:

Find the area of the trapezium PQRS with height PQ given in the figure

Answer:

Drawing a line parallel to PQ from R.

By Pythagoras theorem,
${(RS)}^{2} = {(ST)}^{2} + {(TR)}^{2} \Rightarrow {(13)}^{2} = {(5)}^{2} + {(TR)}^{2} \Rightarrow {(TR)}^{2} = 144 = 12^{2} \Rightarrow TR = 12 cm$

Now,
TR = PQ = 12 cm
$\begin{array}{rcl} Area of Trapezium PQRS & = & \frac{1}{2} (sum of parallel sides) \times height \\ = & \frac{1}{2} (12 + 7) \times 12 \\ = & \frac{1}{2} \times 19 \times 12 \\ = & 114 {cm}^{2} \end{array}$

Hence, the area of the trapezium PQRS is 114 cm².

Page No 118:

Question 1:

How much paper of each shade is needed to make a kite given in the figure in which ABCD is a square with diagonal 44 cm.

Answer:

$\begin{array}{rcl} Area of square sheet ABCD & = & \frac{1}{2} \times {(diagonal)}^{2} \\ = & \frac{1}{2} \times {(44)}^{2} \\ = & 968 {cm}^{2} \end{array}$
Area of yellow sheet = (Area of region 1) + (Area of region 2)
$\begin{array}{rcl} Therefore, Area of yellow sheet & = & \frac{1}{2} \times 968 \\ = & 484 {cm}^{2} . . . . . (1) \end{array}$
$\begin{array}{rcl} Area of red sheet & = & Area of region 4 \\ = & \frac{1}{4} \times 968 \\ = & 242 {cm}^{2} . . . . . (2) \end{array}$

By Heron's formula,
Area of $△ CEF = \sqrt{s (s - a) (s - b) (s - c)}$
Here, s = $\frac{a + b + c}{2}$
Now,
s = $\frac{20 + 20 + 14}{2} = 27 cm$
$\begin{array}{rcl} ∴ Area of △ CEF & = & \sqrt{27 (27 - 20) (27 - 20) (27 - 14)} \\ = & \sqrt{27 (7) (7) (13)} \\ = & 21 \sqrt{39} \\ = & 131.15 {cm}^{2} \end{array}$

$\begin{array}{rcl} ∴ Area of green sheet & = & Area of region (3) + Area of region (5) \\ = & 131.15 + \frac{1}{4} \times 968 \\ = & 131.15 + 242 \\ = & 373.15 {cm}^{2} \end{array}$
Therefore, the area of yellow sheet is 484 cm², area of red sheet is 242 cm²and area of green sheet is 373.15 cm².

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Question 2:

The perimeter of a triangle is 50 cm. One side of a triangle is 4 cm longer than the smaller side and the third side is 6 cm less than twice the smaller side. Find the area of the triangle.

Answer:

Let the smaller side be $x$ cm.
Then, larger side = $(x + 4) cm$ and third side = $(2 x - 6) cm$
Perimeter = 50 cm (given)
$\Rightarrow x + x + 4 + 2 x - 6 = 50 \Rightarrow 4 x - 2 = 50 \Rightarrow 4 x = 52 \Rightarrow x = 13 cm$

Therefore, sides of the triangle are 13 cm, 17 cm, 20 cm.
Let a = 13, b = 17 cm, c = 20 cm

By Heron's formula, we have
Area of triangle = $\sqrt{s (s - a) (s - b) (s - c)}$
Where, s = $\frac{a + b + c}{2}, s = \frac{13 + 17 + 20}{2} = 25 cm$

$\begin{array}{rcl} Now, Area of triangle & = & \sqrt{25 (25 - 13) (25 - 17) (25 - 20)} \\ = & \sqrt{25 \times 12 \times 8 \times 5} \\ = & 20 \sqrt{30} {cm}^{2} \end{array}$

Hence, the area of the triangle is $20 \sqrt{30} {cm}^{2}$ .

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Question 3:

The area of a trapezium is 475 cm²and the height is 19 cm. Find the lengths of its two parallel sides if one side is 4 cm greater than the other.

Answer:

Let one side of trapezium be $x cm$ .
Then, other side = $(x + 4) cm$
Area of trapezium = 475 cm²
$\Rightarrow \frac{1}{2} (x + x + 4) \times 19 = 475 \Rightarrow 2 x + 4 = \frac{475 \times 2}{19} \Rightarrow 2 x + 4 = 50 \Rightarrow 2 x = 46 \Rightarrow x = 23 cm$

Hence, length of two parallel sides are 23 cm and 27 cm.

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Question 4:

A rectangular plot is given for constructing a house, having a measurement of 40 m long and 15 m in the front. According to the laws, a minimum of 3 m, wide space should be left in the front and back each and 2 m wide space on each of other sides. Find the largest area where house can be constructed.

Answer:

Here, length of inner rectangle, EF = $40 - 3 - 3 = 34 m$
Breadth of inner rectangle, FG = $15 - 2 - 2 = 11 m$
$∴$ Area of inner rectangle EFGH = EF × FG = 34 $\times$ 11 = 374 m²

Hence, largest area where house can be construct is 374 m².

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Question 5:

A field is in the shape of a trapezium having parallel sides 90 m and 30 m. These sides meet the third side at right angles. The length of the fourth side is 100 m. If it costs Rs 4 to plough 1m²of the field, find the total cost of ploughing the field.

Answer:

In $△ BCE$ ,
Using Pythagoras theorem,
$\begin{array}{l} {(BC)}^{2} = {(EC)}^{2} + {(BE)}^{2} \\ \Rightarrow {(100)}^{2} = {(60)}^{2} + {(BE)}^{2} \\ \Rightarrow 10000 = 3600 + {(BE)}^{2} \\ {\Rightarrow (BE)}^{2} = 6400 = 80^{2} \\ \Rightarrow BE = 80 cm \end{array}$

$\begin{array}{rcl} Area of field & = & Area of trapezium \\ = & \frac{1}{2} (sum of Parallel sides) \times Distance between them \\ = & \frac{1}{2} \times (30 + 90) \times 80 \\ = & 4800 m^{2} \end{array}$
Now, cost of ploughing 1 m² = â‚¹5
$\begin{array}{rcl} Cost of ploughing 4800 m^{2} & = & ₹ (4800 \times 5) \\ = & ₹ 24000 \end{array}$

Hence, the total cost of ploughing the field is â‚¹24000.

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Question 6:

In the figure âˆ†ABC has sides AB = 7.5 cm, AC = 6.5 cm and BC = 7 cm. On base BC a parallelogram DBCE of same area as that of âˆ†ABC is constructed. Find the height DF of the parallelogram.

Answer:

Let us first calculate are of $△ ABC$ .
Here, AB = a =7.5 cm, BC = b =7 cm, CA = c = 6.5 cm
$\begin{array}{rcl} Semi perimeter of △ ABC, s & = & \frac{a + b + c}{2} \\ s & = & \frac{7.5 + 7 + 6.5}{2} \\ s & = & 10.5 cm \end{array}$

$\begin{array}{rcl} ∴ Area of △ ABC & = & \sqrt{s (s - a) (s - b) (s - c)} (∴ By Heron' s formula) \\ = & \sqrt{10.5 (10.5 - 2.5) (10.5 - 7) (10.5 - 6.5)} \\ = & \sqrt{(10.5) (3) (3.5) (4)} \\ = & 21 {cm}^{2} . . . . . (1) \end{array}$

Now, Area of parallelogram BCED = Base $\times$ height = 7 $\times$ DF .....(2)
Area of triangle = Area of parallelogram (given)
$\Rightarrow 21 = 7 \times DF \Rightarrow DF = 3 cm$

Hence, the height of the parallelogram is 3 cm.

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Question 7:

The dimensions of a rectangle ABCD are 51 cm × 25 cm. A trapezium PQCD with its parallel sides QC and PD in the ratio 9 : 8, is cut off from the rectangle as shown in the figure. If the area of the trapezium PQCD is $\frac{5}{6}$ th part of the area of the rectangle, find the lengths QC and PD.

Answer:

Given: Parallel sides of trapezium are in ratio 9 : 8
i.e., QC : PD = 9 : 8
Let QC be 9x and PD be 8x.

From question,
Area of trapezium = $\frac{5}{6} (Area of rectangle)$
$\Rightarrow \frac{1}{2} (9 x + 8 x) \times 25 = \frac{5}{6} \times (51 \times 25) \Rightarrow \frac{1}{2} \times 17 x \times 25 = \frac{5}{6} \times 51 \times 25 \Rightarrow x = \frac{5 \times 51 \times 25 \times 2}{6 \times 17 \times 25} \Rightarrow x = 5$

$∴ QC = 9 \times 5 = 45 cm And, PD = 8 \times 5 = 40 cm$

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Question 8:

A design is made on a rectangular tile of dimensions 50 cm × 70 cm as shown in the figure. The design shows 8 triangles, each of sides 26 cm, 17 cm and 25 cm. Find the total area of the design and the remaining area of the tile.

Answer:

Given: Dimension of rectangular tile is 50 cm $\times$ 70 cm
$∴$ Area of rectangular tile = 50 $\times$ 70 = 3500 cm²
Let sides of triangle in a design be a = 25 cm, b = 17 cm, c = 26 cm.

By Heron's formula, we have
Area of triangle = $\sqrt{s (s - a) (s - b) (s - c)}$
Here, s = $\frac{a + b + c}{2},$
Now, s = $\frac{25 + 17 + 26}{2} = 34$
$\begin{array}{rcl} Area of triangle & = & \sqrt{34 (34 - 25) (34 - 17) (34 - 26)} \\ = & \sqrt{17 \times 9 \times 17 \times 18} \\ = & 204 {cm}^{2} \end{array}$

$∴$ Total area of 8 triangles = 8 $\times$ 204 = 1632 cm²
Now,
Area of design = Area of 8 triangles = 1632 cm²

$\begin{array}{rcl} Remaining area & = & Area of rectangle - Area of design \\ = & 3500 - 1632 \\ = & 1868 {cm}^{2} \end{array}$

Hence, the total area of the design is 1632 cm² and the remaining area of the tile is 1868 cm².

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