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.

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

Question 1

The diameter of a circle is 28 cm.
Find its :
(i) Circumference
(ii) Area.

Sol:

Let r be the radius of the circle.

(i) 2r = 28 cm
    circumference = 2πr
                           = 28π cm
                            = 88 cm

(ii) area = πr² 

             = π ${\displaystyle {\left(\frac{28}{2}\right)}^2}$

             = 196π cm²

             = 616 cm²

Question 2

The circumference of a circular field is 308 m. Find is:
(i) Radius
(ii) Area.

Sol:

Let r be the radius of the circular field
(i) 2πr = 308

⇒ r = ${\displaystyle \frac{308}{2\pi }}$

⇒ r = ${\displaystyle \frac{308}{2}\times \frac{7}{22}}$

⇒ r = 49 m

(ii) area = πr² 

            = ${\displaystyle \frac{22}{7}\times {\left(49\right)}^2}$

            = 7546 m²  

Question 3

The sum of the circumference and diameter of a circle is 116 cm. Find its radius. 

Sol:

Let r be the radius of the circle.

2πr + 2r = 116

2r( π + 1) = 116

               r = ${\displaystyle \frac{116}{2(\pi +1)}}$

                  = 14 cm

Question 4

The radii of two circles are 25 cm and 18 cm. Find the radius of the circle which has a circumference equal to the sum of circumferences of these two circles. 

Sol:

Circumference of the first circle

S₁ = 2π x 25
     = 50π cm

Circumference of the second circle

S₂ = 2π x 18
     = 36π cm

Let r be the radius of the resulting circle.
2πr = 50π + 36π
2πr = 86π

r = ${\displaystyle \frac{86\pi }{2\pi }}$
  = 43 cm

Question 5

The radii of two circles are 48 cm and 13 cm. Find the area of the circle which has its circumference equal to the difference of the circumferences of the given two circles.

Sol:

Circumference of the first circle

s₁ = 2π x 48
    = 96πcm

Circumference of the first circle

s₂ = 2π x 13
    = 26πcm

Let r be the radius of the resulting circle.
2πr = 96π - 26π
2πr = 70π

    r = ${\displaystyle \frac{70\pi }{2\pi }}$

      = 35cm
Hence area of the circle

A = πr² 
   = π x 35²
   = 3850cm²  

Question 6

The diameters of two circles are 32 cm and 24 cm. Find the radius of the circle having its area equal to the sum of the areas of the two given circles.

Sol:

Let the area of the resulting circle be r.

π x (16)² + π x ( 12 )² = π x r²

             256 π + 144 π = π x r²

                          400π = π x r²

                               r² = 400

                                r = 20 cm

Hence the radius of the resulting circle is 20cm.

Question 7

The radius of a circle is 5 m. Find the circumference of the circle whose area is 49 times the area of the given circle. 

Sol:

Area of the circle having radius 85m is

A = π x ( 85 )² 
   = 7225πm²

Let r be the radius of the circle whose area is 49times of the given circle.

πr² = 49 x ( π x 5² )

  r² = ( 7 x 5 )²

   r = 35 

Hence circumference of the circle
S = 2πr

  = 2π x 35

 = 220 m

Question 8

A circle of the largest area is cut from a rectangular piece of cardboard with dimensions 55 cm and 42 cm. Find the ratio between the area of the circle cut and the area of the remaining card-board.

Sol:

Area of the rectangle is given by
A = 55 x 42
   = 2310 cm² 

For the largest circle, the radius of the circle will be half of the sorter side of the rectangle.

Hence r = 21cm.

Area of the circle = π x ( 21 )²
                            = 1384.74 cm²

Area remaining = 2310 - 1384.74
                          = 925.26

Hence
the volume of the circle: area remaining
                        = 1384. 74: 915. 26
                        = 3.2

Question 9

The following figure shows a square cardboard ABCD of side 28 cm. Four identical circles of the largest possible sizes are cut from this card as shown below.

Find the area of the remaining card-board.

Sol:

Area of the square is given by

A = 28²
   = 784 cm² 

Since there are four identical circles inside the square.
Hence the radius of each circle is one-fourth of the side of the square.

Area of one circle = π x 7²  
                             = 154 cm²

Area of four circle = 4 x 154cm²  
                             = 616 cm² 

Area remaining = 784 - 616
                          = 168 cm² 

The area remaining in the cardboard is = 168 cm².

Question 10

The radii of two circles are in the ratio 3: 8. If the difference between their areas is 2695π cm², find the area of the smaller circle. 

Sol:

Let the radius of the two circles be 3r and 8r respectively.

area of the circle having radius 3r = π(3r)²
                                                      = 9πr² 

area of the circle having radius 8r = π(8r)²
                                                      = 64πr²  

According to the question
64πr² - 9πr² = 2695π
              55r² = 2695
                  r² = 49
                  r = 7 cm

Hence the radius of the smaller circle is  3 x 7 = 21 cm
Area of the smaller circle is given by

A = πr² =${\displaystyle \frac{22}{7}\times {21}^2=1386{\text{cm}}^2}$

Question 11

The diameters of three circles are in the ratio 3: 5: 6. If the sum of the circumferences of these circles is 308 cm; find the difference between the areas of the largest and the smallest of these circles.

Sol:

Let the diameter of the three circles be 3d, 5d, and 6d respectively.

Now

π x 3d + π  x 5d + π  x 6d = 308
                                 14πd = 308
                                       d = 7

radius of the smallest circle = ${\displaystyle \frac{21}{2}}$ = 10.5

                                    Area = π  x (10 .5)²
                                            = 346.5

radius of the largest circle = ${\displaystyle \frac{42}{2}}$ = 21

                                  Area = π  x ( 21 )²
                                          = 1386
                         difference = 1386 - 346.5
                                          = 1039.5 cm²  

Question 12

Find the area of a ring-shaped region enclosed between two concentric circles of radii 20 cm and 15 cm.

Sol:

Area of the ring = π( 20 )² - π( 15 )²
                          = 400π - 225π
                          = 175π
                          = 549.7 cm²

Question 13

The circumference of a given circular park is 55 m. It is surrounded by a path of uniform width of 3.5 m. Find the area of the path.

Sol:

Let r be the radius of the circular park.
2πr = 55
r = ${\displaystyle \frac{55}{2}\pi }$
  = 8.75 m

area of the park = π x ( 8.75 )² = 240.625 m² 

Radius of the outer circular region including the path is given by
R = 8.75 + 3.5
   = 12.25 m

Area of that circular region is A = π x ( 12.25 )² = 471.625 m² 
Hence area of the path is given by 
Area of the path = 471.625 - 240.625 = 231 m².

Question 14

There are two circular gardens A and B. The circumference of garden A is 1.760 km and the area of garden B is 25 times the area of garden A. Find the circumference of garden B.

Sol:

Let r be the radius of the circular garden A

Since the circumference of the garden A is 1.760 km =  1760 m, we have,
2πr = 1760 m
⇒ r = ${\displaystyle \frac{1760\times 7}{2\times 22}}$ = 280 m

Area of garden A = πr² = ${\displaystyle \frac{22}{7}}$ x 280² m²

Let R be the radius of the circular garden B.

Since the area of the garden, B is 25 times the area of garden A, We have,
πR² = 25 x πr² 
⇒ πR² = 25 x π x 280²
⇒ R² = 1960000
⇒ R = 1400 m
Thus circumference of garden B = 2πR

= ${\displaystyle 2\times \frac{22}{7}\times 1400}$ = 8800 m = 8.8 km.

Question 15

A wheel has a diameter of 84 cm. Find how many completer revolutions must it make to cover 3.168 km.

Sol:

The diameter of the wheel = 84 cm
Thus, the radius of the wheel = 42 cm

Circumference of the wheel = 2 x ${\displaystyle \frac{22}{7}}$ x 42 = 264 cm
In 264 cm, the wheel is covering one revolution.

Thus, in 3.168 km = 3.168 x 100000 cm, number of revolutions
covered by the wheel = ${\displaystyle \frac{3.168}{264}\times 100000}$ = 1200.

Question 16

Each wheel of a car is of diameter 80 cm. How many completer revolutions does each wheel make in 10 minutes when the car is traveling at a speed of 66 km per hour?

Sol:

The car travels in 10 minutes
= ${\displaystyle \frac{66}{6}}$   = 11 km  = 1100000 cm

Circumference of the wheel = distance covered by the wheel in one revolution 

Thus, we have,
Circumference = 2 x ${\displaystyle \frac{22}{7}\times \frac{80}{2}}$ = 251.43 cm

Thus, the number of revolutions covered by the wheel in 1100000 cm = ${\displaystyle \frac{1100000}{251.43}}$ ≈ 4375.

Question 17

An express train is running between two stations with a uniform speed. If the diameter of each wheel of the train is 42 cm and each wheel makes 1200 revolutions per minute, find the speed of the train.

Sol:

radius of the wheel = ${\displaystyle \frac{42}{2}}$ = 21 cm

Circumference of the wheel = 2π x 21 = 132 cm

Distance travelled in one minute
= 132 x 1200 
= 158400 cm
= 1.584 km

Hence the speed of the train
= ${\displaystyle \frac{1.584\text{km}}{\frac{1}{60}\text{hr}}}$

= 95.04 km/hr.

Question 18

The minute hand of a clock is 8 cm long. Find the area swept by the minute hand between 8.30 a.m. and 9.05 a.m.

Sol:

The time interval is 9.05 - 8.30 = 35 minutes

Area covered in one 60 minutes = π x 8² = 201 cm²

Hence area swept in 35 minutes is given by
A = ${\displaystyle \frac{201}{60}\times 35=117\frac{1}{3}}$cm².

Question 19

The shaded portion of the figure, given alongside, shows two concentric circles. If the circumference of the two circles is 396 cm and 374 cm, find the area of the shaded portion.

Sol:

Let R and r be the radius of the big and small circles respectively.
Given that the circumference of the bigger circle is 396 cm

Thus, we have,
2πR = 396 cm

⇒ R = ${\displaystyle \frac{396\times 7}{2\times 22}}$

⇒ R = 63 cm

Thus, the area of the bigger circle = πR²
= ${\displaystyle \frac{22}{7}\times {63}^2}$

= 12474 cm²

Also, given that the circumference of the smaller circle is 374 cm
⇒ 2πr = 374

⇒ r = ${\displaystyle \frac{374\times 7}{2\times 22}}$
⇒ r = 59.5 cm

Thus, the area of the smaller circle = πr²
= ${\displaystyle \frac{22}{7}}$ x 59.5²

= 11126.5 cm²
Thus the area of the shaded portion = 12474 - 11126.5 = 1347.5 cm²

Question 20

In the given figure, the area of the shaded portion is 770 cm². If the circumference of the outer circle is 132 cm, find the width of the shaded portion.

Sol:

From the given data, we can calculate the area of the outer circle and then the area of the inner circle and hence the width of the shaded portion.

Given that the circumference of the outer circle is 132 cm
Thus, we have, 2πR = 132 cm

⇒ R = ${\displaystyle \frac{132\times 7}{2\times 22}}$

⇒ R = 21 cm

Area of the bigger circle = πR²
                                        = ${\displaystyle \frac{22}{7}}$ x 21²
                                        = 1386 cm²

Also given the area of the shaded portion.
Thus the area of the inner circle = Area of the outer circle - Area of the shaded portion
= 1386 - 770
= 616 cm²

⇒  πr² = 616

⇒ r² = ${\displaystyle \frac{616\times 7}{22}}$

⇒ r² = 196
⇒ r = 14 cm
Thus, the width of the shaded portion = 21 - 14 = 7 cm.

Question 21

The cost of fencing a circular field at the rate of ₹ 240 per meter is ₹ 52,800. The field is to be ploughed at the rate of ₹ 12.50 per m². Find the cost of pouching the field.

Sol:

Let the radius of the field is r meter.

Therefore circumference of the field will be 2πr meter.

Now the cost of fencing the circular field is 52,800 at rate 240 per meter.

Therefore,
2πr. 240 = 52800

r = ${\displaystyle \frac{52800\times 7}{2\times 240\times 22}}$

r = 35
Thus the radius of the field is 35 meter.

Now the area of the field will be :
πr2 = ${\displaystyle \left(\frac{22}{7}\right)}$.35² = 3850 m²

Thus the cost of ploughing the field will be:
3850 x 12.5 = 48,125 rupees.

Question 22

Two circles touch each other externally. The sum of their areas is 58π cm² and the distance between their centers is 10 cm. Find the radii of the two circles.

Sol:

Let r and R be the radius of the two circles.
r + R = 10                ....(1)
πr² + πR² = 58π      .....(2)

Putting the value of r in (2)
r² + R² = 58
( 10 - R )² + R² = 58
100 - 20R + R² + R² = 58
2R² - 20R + 42 = 0
R² - 10R + 21 = 0
( R - 3 )( R - 7 ) = 0
R = 3, 7.
Hence the radius of the two circles is 3cm and 7cm respectively.

Question 23

The given figures show a rectangle ABCD inscribed in a circle as shown alongside.
If AB = 28 cm and BC = 21 cm, find the area of the shaded portion of the given figure.

Sol:

From the figure,
AB = 28 cm
BC = 21 cm

AC = ${\displaystyle \sqrt{{\text{AB}}^2+{\text{BC}}^2}}$

= ${\displaystyle \sqrt{{28}^2+{21}^2}}$ 

= 35.

Hence diameter of the circle is 35 cm and hence 
Area = π x ${\displaystyle {\left(\frac{35}{2}\right)}^2}$ = 962.5 cm²

Area of the rectangle = 28 x 21 = 588 cm²

Hence area of the shaded portion is given by
A = 962 - 588 = 374.5 cm²

Question 24

A square is inscribed in a circle of radius 7 cm. Find the area of the square.

Sol:

Since the diameter of the circle is the diagonal of the square inscribed in the circle.
Let a be the length of the sides of the square.
Hence,
${\displaystyle \sqrt{2}a}$ = 2 x 7
a = ${\displaystyle \sqrt{2}}$ x 7
a² = 98
Hence the area of the square is 98 sq.cm.

Question 25

A metal wire, when bent in the form of an equilateral triangle of largest area, encloses an area of 484 ${\displaystyle \sqrt{3}}$ cm². If the same wire is bent into the form of a circle of largest area, find the area of this circle.

Sol:

Let 'a' be the length of each side of an equilateral triangle formed.
Now, the area of an equilateral triangle formed = 484√3  cm²
⇒ ${\displaystyle \frac{\sqrt{3}}{4}}$a² = 484√3  
⇒ a² = 4 x 484 
⇒ a = 2 x 22 = 44 cm

Then, perimeter of equilateral triangle = 3a = 3 x 44 = 132 cm

Now, length of wire = perimeter of equilateral triangle = circumference of circle

⇒ circumference of circle = 132 cm
⇒ 2πr =  132                 .....( r is radius of circle )
⇒ r = ${\displaystyle \frac{132\times 7}{2\times 22}}$ = 21 cm

∴ Area of circle = πr² = ${\displaystyle \frac{22}{7}}$ x 21 x 21 = 1386cm².