**Download NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.3 – Area related to Circles**

**1. Find the area of the shaded region in the given figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.**

**Solution:**

Using Pythagoras theorem in △PQR (Angle of semi-circle is right angle)

QR^{2} = PQ^{2} + PR^{2}QR^{2} = 24^{2} + 7^{2} = 576 + 49 = 625 = 25^{2}QR = 25 cm

Radius of the circle = ½ QR = cm

Area of shaded region = Area of semicircle – Area of △PQR

= x Π x – x 24 x 7

= x x – 12 x 7

= – 84

=

= cm^{2}

**2. Find the area of the shaded region in the given figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠ AOC = 40°.**

**Solution:**

Shaded area = Area of sector of larger circle – Area of sector of smaller circle

= x Π x 14² – x Π x 7²

= x x 196 – x x 49

= –

=

=

= cm^{2}

**3. Find the area of the shaded region in the given figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.**

**Solution:**

Side of square = 14 cm

Radius of circles = x 14 = 7 cm

Area of shaded region = Area of square – Area of two semicircles

= 14 x 14 – 2 x ( x Π x 7²)

= 196 – x 49

= 196 – 154

= 42 cm^{2}

**4. Find the area of the shaded region in the given figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.**

**Solution:**

∠ AOB = 60° (OAB is equilateral triangle)

Reflex ∠ AOB = 360° – 60° = 300°

Shaded area = Area of triangle + Area of major sector of the circle

= x (12)² x x Π x 6²

= x 144 + x x 36

= (36√3 + ) cm^{2}

**5. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the figure. Find the area of the remaining portion of the square.**

**Solution:**

Side of the square = 4 cm

Radius of quadrants = 1 cm

Radius of inner circle = x 2 = 1 cm

Remaining area = Area of square – (Area of circle + Area of 4 quadrants)

= 4 x 4 – (Π x (1)² + 4 x ( x Π x (1)²))

= 16 – (Π + Π)

= 16 – 2 x

=

= cm^{2}

**6. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the figure. Find the area of the design.**

**Solution:**

Let O be the centre of the circle and the side of the triangle be s

Draw an altitude AOD from A on BC

∠ ABD = 60°

In △ABD

sin 60° =

=

AD =

AO = Radius = 32 cm

OD = AD – AO = – 32

Since, centroid O divides the median into the ratio 2:1

Therefore,

=

32 = √3s- 64

96 = s√3

S = 32√3 cm

Area of design = Area of circle – Area of triangle

= Π x 32² – x (32√3)²

= x 1024 – x 3072

= ( – 768√3) cm^{2}

**7. In the given figure, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.**

**Solution:**

Let the radius of each circle be **r**r + r = Side of the square = 14 cm

2r = 14 cm

r = 7 cm

Shaded area = Area of square – Area of four quadrants

= 14 x 14 – 4 x ( x Π x 7²)

= 196 – x 49

= 196 – 154

= 42 cm^{2}

**8. The given figure depicts a racing track whose left and right ends are semi-circular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:**

**i. The distance around the track along its inner edge
ii. The area of the track**

According to the question, the lengths and radii of different parts of the track are as shown in the figure

**Solution:**

The distance around the track along its inner edge

= AB + arc BCD + DE + arc EFA

= 106 + x 2Π() + 106 + x 2Π()

= 212 + 2 x x 30

= 212 +

= m

Area of the track

= (ar(PQST) + Area of semicircles QRS and TUP) – (ar(ABDE) + Area of semicircles BCD and EFA)

= (106 × 80 + x Π x + x Π x ) – (106 × 60 + x Π x + x Π x )

= (106 × 80 + x 40²) – (106 × 60 + x 30²)

= (8480 + ) – (6360 + )

= 8480 – 6360 +

= 2120 +

= 2120 + 2200

= 4320 m^{2}

**9. In the given figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.**

**Solution:**

Radius of smaller circle = cm

Shaded area = Area of semicircle ACB + Area of smaller circle – Area of △ABC

=

= x x 49 + x – x 14 x 7

= 77 + – 49

= 28 +

= 28 + 38.5

= 66.5 cm^{2}

**10. The area of an equilateral triangle ABC is 17320.5 cm ^{2}. With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle. Find the area of the shaded region.(Use π = 3.14 and √3 = 1.73205)**

**Solution:**

Let the side of the triangle be s

Area = x s²

17320.5 = x s²

s^{2} = 4 x

s^{2} = 40000

s = 200 cm

Therefore, radius of each circle = = 100 cm

Area of sectors of circle = 60° (ABC is equilateral triangle)

Shaded area = Area of triangle – Area of 3 sectors

= 17320.5 – 3 x ( x Π x 100² )

= 17320.5 – 3 x x 3.14 x 10000

= 17320.5 – 15700

= 1620.5 cm^{2}

**11. On a square handkerchief, nine circular designs each of radius 7 cm are made. Find the area of the remaining portion of the handkerchief.**

**Solution:**

Side of the square = 3 × diameter = 6 × radius = 42 cm

Remaining area = Area of square – Area of nine circles

= 42 × 42 – 9 × π × 7^{2}= 1764 – 9 x x 49

= 1764 – 1386

= 378 cm^{2}

**12. In the given figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the**

**i. Quadrant OACB
ii. Shaded region**

**Solution:**

Radius of the quadrant of the circle = 3.5 cm

Area of quadrant = x Π x 3.5² = x Π x

= x x

= cm^{2}

Area of shaded region = Area of quadrant – Area of △BOD

= – x 3.5 x 2

= –

=

= cm^{2}

**13. In the given figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)**

**Solution:**

Using Pythagoras theorem

OB^{2} = OC^{2} + OA^{2}OB^{2} = 20^{2} + 20^{2}OB^{2} = 2 × 20^{2}OB = 20√2 cm

Radius of quadrant = OB = 20√2 cm

Shaded area = Area of quadrant – Area of square

= x Π x (20√2)² – 20 x 20

= x 3.14 x 800 – 400

= 628 – 400

= 228 cm^{2}

**14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O. If ∠AOB = 30°, find the area of the shaded region.**

**Solution:**

Radius of smaller sector = 7 cm

Radius of larger sector = 21 cm

Angle of sectors = 30°

Shaded area = Area of larger sector – Area of smaller sector

= x Π x 21² – x Π x 7²

= x x 441 – x x 49

= –

= –

=

= cm^{2}

**15. In the given figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.**

**Solution:**

Using Pythagoras theorem

BC^{2} = AB^{2} + AC^{2}BC^{2} = 14^{2} + 14^{2}BC^{2} = 2 × 14^{2}BC = 14√2 cm

Radius of the semicircle = = 7√2 cm

Shaded area = Area of semicircle – Area of segment of quadrant

= Area of semicircle – (Area of quadrant – Area of △ABC)

=

= x x 98 – ( x x 196 – )

=154-(154-98)

= 98 cm^{2}

**16. Calculate the area of the designed region in the given figure common between the two quadrants of circles of radius 8 cm each.**

**Solution:**

The design is a combination of two segments

Area of one segment = Area of quadrant – Area of triangle

=

= x x 64 –

= – 32

=

= cm^{2}

Area of design = Area of two segments = 2 ×

= cm^{2}

**Download NCERT Solutions For Class 10 Maths Chapter 12 Exercise 12.3 – Area related to Circles**

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