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# Class 10 RD SHARMA Solutions Maths Chapter 13 - Areas Related to Circles

## Areas Related to Circles Exercise Ex. 13.1

### Solution 11

Area of a circle = πr2 = (22/7) × 28 × 28 = 2464 cm2

## Areas Related to Circles Exercise Ex. 13.2

### Solution 22

*Note: Answer given in the book is incorrect.

## Areas Related to Circles Exercise Ex. 13.4

### Solution 25

*Answer is not matching with textbook.

### Solution 28

Consider the following figure:

### Solution 29

(i)

According to the figure in the question, there are 6 triangles.

Area of one triangle is 9 cm2.

Area of hexagon = 6 × 9 = 54 cm2

(ii)

Area of the equilateral triangle = 9 cm2

Area of the circle in which the hexagon is inscribed

=

=

=

= 65.26 cm2

### Solution 48

Since the data given in the question seems incomplete and inconsistent with the figure, we make the following assumptions to solve it:

1. ABCD a symmetric trapezium with AD = BC

2. AD = BC = 14 cm (the distance between AB and CD is not 14 cm)

Draw perpendiculars to CD from A and B to divide the trapezium into one rectangle and two congruent right angled triangles.

The base of the right angled triangle=(CD - AB) ÷ 2

=(32 - 18) ÷ 2=7 cm

cosD = base ÷ hypotenuse = 7 ÷ 14 =1/2

mD = 60°

Hence, mA = 120°

### Solution 52

According to the question,

Side of a square is 28 cm.

Radius of a circle is 14 cm.

Required area = Area of the square + Area of the two circles - Area of two quadrants …(i)

Area of the square = 282 = 784 cm2

Area of the two circles = 2πr2

=

= 1232 cm2

=

= 308 cm2

Required area = 784 + 1232 - 308 = 1708 cm2

### Solution 53

According to the question,

For a cylindrical tank

d = 2 m, r = 1 m, h = 5 m

Volume of the tank = πr2h

=

=

After recycling, this water is used irrigate a park of a hospital with length 25 m and breadth 20 m.

If the tank is filled completely, then

Volume of cuboidal park = Volume of tank

h = 0.0314 m = 3.14 cm = p cm

### Solution 54

Join OB.

Here,  is a right triangle.

By Pythagoras theorem,

Therefore, radius of the circle (r)

Area of the square

Area of the quadrant of a circle

Area of the shaded region = Area of quadrant - Area of square

= 128.25 cm2

### Solution 55

Join AC.

Here,  is a right triangle.

By Pythagoras theorem,

Therefore, diameter of the circle = 4 cm

So, the radius of the circle (r) = 2 cm

Area of the square

Area of the circle

Area of the shaded region = Area of the circle - Area of square

= 4.56 cm2

## Areas Related to Circles Exercise 13.69

### Solution 1

Correct Option :- (D)

### Solution 2

According to the question,

Circumference of a circle =

=

= 44 cm

### Solution 3

Correct option (c)

### Solution 4

correct option - (c)

### Solution 5

correct option - (b)

### Solution 6

Correct Option: d

### Solution 7

Correct Option: (c)

### Solution 8

Correct option - (c)

### Solution 9

Correct option (c)

### Solution 10

Correct Option ( d )

### Solution 11

Correct option (a)

### Solution 12

Let r be the radius of the circle.

2pr = 88

Perimeter of a triangle = 30 cm

Semi-perimeter = 15 cm

Hence,

Area of a triangle = r × s …(r = incircle radius, s =semi perimeter)

= 14 × 15

= 210 cm2

### Solution 13

Correct option - (c)

## Areas Related to Circles Exercise 13.70

### Solution 24

Correct option: (d)

Diameter of circle = side of square

2r = 10

r = 5 cm

Area of circle = πr2 = 25 π cm2

## Areas Related to Circles Exercise 13.71

### Solution 26

Correct option: (b)

outer radius = r + h

area of shaded region = area of outer circle - area of inner circle

= π (r + h)2 - πr2

= π {(r + h)2 - r}

= π (r + h - r) (r + h + r)

= π (2r + h) h

### Solution 28

Correct option: (b)

area = circumference

πr2 = 2πr

r = 2 units

area = πr2

= 4π sq. units

** img pending

## Areas Related to Circles Exercise 13.74

### Solution 46

Correct option: (b)

radius of Circle = 5 cm

area = π (5)2

= 25 π

rdius of circle 2 = 12 cm

area = π (12)2

= 144 π

area of larger circle = 144 π + 25π

= 169 π

πr2 = 169 π

r2 = 169

r = 13

diameter = 2r

= 26