SELINA Solution Class 9 Chapter 20 Area and perimeter of plane figures Exercise 20D

Question 1

The perimeter of a triangle is 450 m and its side are in the ratio 12: 5: 13. Find the area of the triangle.

Sol:

Let the sides of the triangle be
a = 12x
b = 5x
c = 13x

Given that the perimeter = 450 m
⇒ 12x + 5x + 13x = 450
⇒ 30x = 450
⇒ x = 15

Hence, the sides of a triangle are
a = 12x = 12(15) = 180 m
b = 5x = 5(15) = 75 m
c = 13x = 13(15) = 195 m

Now,
semi-perimeter of a triangle,
s = ${\displaystyle \frac{a+b+c}{2}=\frac{180+75+195}{2}=\frac{450}{2}=225\text{m}}$

∴ Area of triangle = ${\displaystyle \sqrt{s(s-a)(s-b)(s-c)}}$

= ${\displaystyle \sqrt{225(225-180)(225-75)(225-195)}}$

= ${\displaystyle \sqrt{225\times 45\times 150\times 30}}$

= ${\displaystyle \sqrt{15\times 15\times 9\times 5\times 25\times 6\times 5\times 6}}$

= ${\displaystyle \sqrt{15\times 15\times 3\times 3\times 25\times 25\times 6\times 6}}$

= 15 x 3 x 25 x 6

= 6750 m² 

Question 2

A triangle and a parallelogram have the same base and the same area. If the side of the triangle is 26 cm, 28 cm, and 30 cm and the parallelogram stands on the base 28 cm, find the height of the parallelogram.

Sol:

Let the sides of the triangle be
a = 26 cm, b = 28 cm and c = 30 cm
Now,

semi-perimeter of a triangle,
s = ${\displaystyle \frac{a+b+c}{2}=\frac{26+28+30}{2}=\frac{84}{2}=42cm}$

∴ Area of triangle = ${\displaystyle \sqrt{s(s-a)(s-b)(s-c)}}$

= ${\displaystyle \sqrt{42(42-26)(42-28)(42-30)}}$

= ${\displaystyle \sqrt{42\times 16\times 14\times 12}}$

= ${\displaystyle \sqrt{7\times 6\times 4\times 4\times 7\times 2\times 6\times 2}}$

= ${\displaystyle \sqrt{7\times 7\times 4\times 4\times 6\times 6\times 2\times 2}}$

= 7 x 4 x 6 x 2

= 336 cm² 

Base of a parallelogram = 28 cm
Given ,
Area of parallelogram = Area of triangle
⇒ Base x Height = 336
⇒ 28 x Height = 336
⇒ Height = 12 cm

Question 3

Using the information in the following figure, find its area.

Sol:

Construction : Draw CM ⊥ AB

In right-angled triangle CMB,
BM² = BC² - CM² = (15)² - (9)² = 225 - 81 = 144
⇒ BM = 12 m
Now, AB = AM + BM = 23 + 12 = 35 m

∴ Area of trapezium ABCD
=${\displaystyle \frac{1}{2}}$ x ( sum of parallel sides ) x Height

=${\displaystyle \frac{1}{2}}$ x (AB + CD) x AD

= ${\displaystyle \frac{1}{2}}$ x ( 23 + 35 ) x 9

= ${\displaystyle \frac{1}{2}}$ x 58 x 9

 = 261 m²   

Question 4

Sum of the areas of two squares is 400 cm². If the difference of their perimeters is 16 cm, find the sides of the two squares. 

Sol:

Let the sides of two squares by a and b respectively.
Then, area of one square, S₁ = a²
And, area of second square, S₂ = b²
Given, S₁ + S₂ = 400 cm²
⇒ a² + b² = 400 cm² …..(1)
Also, difference in perimeter = 16 cm
⇒ 4a - 4b = 16 cm
⇒ a - b = 4
⇒ a = (4 + b)

Substituting the value of 'a' in (1), we get
(4 + b)² + b² = 400
⇒ 16 + 8b + b² + b² = 400
⇒ 2b² + 8b - 384 = 0
⇒ b² + 4b - 192 = 0
⇒ b² + 16b - 12b - 192 = 0
⇒ b(b + 16) - 12(b + 16) = 0
⇒ (b +16)(b - 12) = 0
⇒ b + 16 = 0 or b - 12 = 0
⇒ b = - 16 or b = 12

Since, the side of a square cannot be negative, we reject - 16.
Thus, b = 12
⇒ a = 4 + b = 4 + 12 = 16

Hence, the sides of a square are 16 cm and 12 cm respectively.

Question 5

Find the area and the perimeter of a square with diagonal 24 cm. [Take √2 = 1.41 ] 

Sol:

Diagonal of a square = 24 cm

Now, diagonal of a square = side of a square x ${\displaystyle \sqrt{2}}$

⇒ 24 = side of a Square x ${\displaystyle \sqrt{2}}$

⇒ Side of a square = ${\displaystyle \frac{24}{\sqrt{2}}=\frac{12\times \sqrt{2}\times \sqrt{2}}{\sqrt{2}} =12\sqrt{2}}$

∴ Area of a square = ( Side )²  = ( 12√2)² = 288 cm²  

And, perimeter of a square =
= 4 x Side
= 4 x 12${\displaystyle \sqrt{2}}$
= 48 x 1. 41
= 67. 68 cm.

Question 6

A steel wire, when bent in the form of a square, encloses an area of 121 cm². The same wire is bent in the form of a circle. Find area the circle. 

Sol:

Area of a square = ( Side )²
⇒ 121 = ( Side )²
⇒ Side of a square = 11 cm

Now,
The perimeter of a square = Perimeter of a circle

⇒ 4 x Side = Perimeter of a circle

⇒ 4 x 11 = Perimeter of a circle

⇒ Perimeter of a circle = 44 cm

⇒ 2πr = 44              ....( r is radius of a circle )

⇒ r = ${\displaystyle \frac{44}{2\pi }=\frac{44}{2\times \frac{22}{7}}}$ = 7 cm 

∴ Area of a circle = πr² = ${\displaystyle \frac{22}{7}\times 7\times 7=154{\text{cm}}^2}$

Question 7

The perimeter of a semicircular plate is 108 cm. find its area. 

Sol:

The perimeter of a semi-circular plate = 108 cm
⇒ πr + 2r = 108                 ...(r is radius of a circle )
⇒ r ( π + 2 ) = 108

⇒ r = ${\displaystyle \frac{108}{\pi +2}=\frac{108}{\frac{22}{7}+2} =\frac{108\times 7}{36}=21\text{cm}}$

Now, area of a semi-circular plate = ${\displaystyle \frac{\pi r^2}{2} =\frac{\frac{22}{7}\times 21\times 21}{2} =693{\text{cm}}^2}$ 

Question 8

Two circles touch externally. The sum of their areas is 130π sq. cm and the distance between their centers is 14 cm. Find the radii of the circles.

Sol:

Let the radii of two circles be r₁ and r₂ respectively.
Sum of the areas of two circles = 130π sq. cm

⇒ πr₁² + πr₂² = 130π

⇒ r₁² + r₂² = 130 ….(i)

Also, distance between two radii = 14 cm

⇒ r₁ + r₂ = 14

⇒ r₁ = (14 - r₂)

Substituting the value of r₁ in (i), we get

(14 - r₂)² + r₂² = 130

⇒ 196 - 28r₂ + r₂² + r₂² = 130

⇒ 2r₂² - 28r₂ + 66 = 0

⇒ r₂² - 14r₂ + 33 = 0

⇒ r₂² - 11r₂ - 3r₂ + 33 = 0

⇒ r₂ (r₂ - 11) - 3 (r₂ - 11) = 0

⇒ (r₂ - 11) (r₂ - 3) = 0

⇒ r₂ = 11 or r₂ = 3

⇒ r₁ = 14 - 11 = 3 or r₁ = 14 - 3 = 11

Thus, the radii of two circles are 11 cm and 3 cm respectively.

Question 9

The diameters of the front and the rear wheels of a tractor are 63 cm and 1.54 m respectively. The rear wheel is rotating at ${\displaystyle 24\frac{6}{11}}$ revolutions per minute. Find:
(i) the revolutions per minute made by the front wheel.
(ii) the distance traveled bu the tractor in 40 minutes.

Sol:

Given the diameter of the front and rear wheels are 63 cm = 0.63 m and 1.54 m respectively,
The radius of the rear wheel = ${\displaystyle \frac{1.54}{2}}$ = 0.77 m.

and radius of the front wheel = ${\displaystyle \frac{0.63}{2}}$ = 0.315 m

Distance traveled by tractor in one revolution of the rear wheel = circumference of the rear wheel
= 2πr
= 2 x ${\displaystyle \frac{22}{7}}$ x 0.77
= 4.84 m

The rear wheel rotates at ${\displaystyle 24\frac{6}{11}}$ revolutions per minute
= ${\displaystyle \frac{270}{11}}$ revolutions per minute

Since in one revolution, the distance traveled by the rear wheel = 4.84 m

So, in ${\displaystyle \frac{270}{11}}$ revolutions, the tractor travels ${\displaystyle \frac{270}{11}}$ x 4.84 = 118.8 m
Let the number of revolutions made by the front wheel be x.

(i) Now, the number of revolutions made by the front wheel in one minute x circumference of the wheel

= the distance traveled by tractor in one minute

⇒ ${\displaystyle x\times 2\times \frac{22}{7}\times 0.315=118.8}$

⇒ ${\displaystyle \times =\frac{118.8\times 7}{2\times 22\times 0.315}}$ = 60

(ii) Distance traveled by tractor in 40 minutes
= number of revolutions made by the rear wheel in 40 minutes
x circumference of the rear wheel

= ${\displaystyle \frac{270}{11}\times 40\times 4.84=4752}$ m

Question 10

Two circles touch each other externally. The sum of their areas is 74π cm² and the distance between their centers is 12 cm. Find the diameters of the circle.

Sol:

Let the radius of the circles be r₁ and r₂.
So, r₁ + r₂ = 12 ⇒ r₂ = 12 - r₁

Sum of the areas of the circles = 74π
⇒ πr₁² + πr₁² = 74π
⇒ r₁² + r₁² = 74

⇒ r₁² + ( 12 - r₁ )² = 74
⇒ r₁² + 144 - 24r₁ + r₁² = 74
⇒ 2r₁² - 24r₁ + 70 = 0
⇒ r₁² - 12r₁ + 35 = 0
⇒ ( r₁ - 7 )( r₁ - 5 ) = 0
⇒ r₁ = 7  or      r₁  = 5
If r₁ = 7 cm, then r₂ = 5 cm
If r₁ = 5 cm, then r₂ = 7 cm
So, the diameters of the circles will 10 cm and 14 cm.

Question 11

If a square in inscribed in a circle, find the ratio of the areas of the circle and the square.

Sol:

If AB = x, AC = x√2

Diameter of the circle = diagonal of the square
⇒ 2r = x√2

⇒ r = ${\displaystyle \frac{x\sqrt{2}}{2}}$

Area of the circle = πr²
                            = π${\displaystyle {\left(\frac{x\sqrt{2}}{2}\right)}^2}$

                            = π${\displaystyle \left(\frac{x^{2}2}{4}\right)}$

                            = ${\displaystyle \frac{\pi x^2}{2}}$

Area of the square = x²

Required ratio = ${\displaystyle \frac{\frac{\pi x^2}{2}}{x^2}}$

                        = ${\displaystyle \frac{\pi }{2}}$

                        = ${\displaystyle \frac{22}{7}\times \frac{1}{2}}$

                        = ${\displaystyle \frac{11}{7}}$
Hence, the required ratio is 11 : 7.