These NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Exercise 12.3

Question 1.

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

Solution:

Question 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 = 400.

Solution:

∠AOC = 40° (given)

Radius of the sector AOC = 14 cm

Question 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:

ABCD is a square

Given: side of the square = 14 cm

∴ Area of the square = (side)² = (14)² = 196 cm²

Radius of the semicircle APD = \(\frac { 1 }{ 2 }\)(side of square) = \(\frac { 1 }{ 2 }\) x 14 = 7 cm

Area of the semicircle APD = \(\frac { 1 }{ 2 }\) πr² = \(\frac { 1 }{ 2 }\) × \(\frac { 22 }{ 7 }\) × 7 × 7 = 11 × 7 = 77cm²

Similarly, area of the semicircle BPC = 77 cm²

Total area of both the semicircles = 77 + 77 = 154 cm²

Area of the shaded region = Area of square – area of both semicircles

= 196 – 154 = 42 cm²

Question 4.

Find the area of the shaded region in the 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:

Area of the equilateral triangle OAB

Question 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:

In the figure, area of the shaded region = area of the square ABCD with side 4 cm – area of the circle of radius 1 cm, centrally placed – area of the four quarter circles of radii 1 cm each, placed | at each corner of the square ABCD.

∴ Area of the remaining portion of the square = \(\frac { 68 }{ 7 }\) cm²

Question 6.

In a circular table cover of the 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 (shaded region).

Solution:

Question 7.

In the 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:

Edge of the square ABCD = 14 cm

Area of square ABCD = (14)² = 196 cm²

Here radius of each circle is 7 cm.

Area of shaded region = Area of square – Area of 4 sectors

196 – 154 = 42cm².

Question 8.

The given figure depicts a racing track whose left and right ends are semicircular. 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.

Solution:

(i) ABCD and EFGH are two rectangles and corner has two semicircles. The distance around the track along its inner edge

= 2 x length of rectangle + 2 x circumference of semicirlce

(ii) Area of the track = 2 [Area of rectangles] + 2 [Area of outer semicircle] – [Area of inner semicircles]

Question 9.

In the 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:

Since AB ⊥ CD,

therefore, ∠COB = ∠COA

∠DOA = ∠DOB = 90°

In the figure, area of the shaded region, area of the small circle of diameter (OD = OA = 7cm) + (area of the segment BMC with central angle BOC = 90c and radius (OB = OC = 7cm) + area of the segment ANC with central angle AOC = 90° and radius (OA = OC) = 7cm

Question 10.

The area of an equilateral triangle ABC is 17320.5 cm². 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 (see figure). Find the area of the shaded region. (Use π = 3.14 and \(\sqrt{3}\) = 1.73205).

Solution:

∆ABC is an equilateral triangle. About the angular points A, B and C as centres three circles half the length of the side of the triangle are described.

Area of equilateral ∆ABC(given) = 17320.5 cm²

Since ABC is an equilateral triangle, therefore ∠A = ∠B = ∠C = 60°

= 17320.5 cm²

There are 3 equal sectors in the figure of central angles 60° and radii 100 cm [∵ \(\frac { 200 }{ 2 }\) = 10 cm]

Then, area of the shaded region = area of ∆ABC – 3 (area of one sector of central angle 60° and radius 100 cm)

= 17320.5 cm² – [3 x \(\frac { 60° }{ 360° }\) x π x (100)²]

= 17320.5 cm² – (\(\frac { 3 }{ 6 }\) x 3.14 x 1000) cm²

= 17320.5 cm² – (5000 x 3.14) cm²

= (17320.5 – 15700) cm² = 1620.5 cm²

∴ area of the shaded region is 1620.5 cm²

Question 11.

On a square handkerchief, nine circular designs each of the radius 7 cm are made (see figure). Find the area of the remaining portion of the handkerchief.

Solution:

ABCD is a square

Edge of the square ABCD = 3 x diameter of circle

= 3 x 4 = 42 cm

Area of square ABCD = (42)²

= 1764 cm²

Radius of one circle = 7 cm

Area of one circle = πr²

= \(\frac { 22 }{ 7 }\) x (7)²

= 154 cm²

Area of nine circles = 9 x Area of one circle

= 9 x 154 = 1386 cm²

Area of remaining portion = Area of square – Area of circle

= 1764 – 1386 = 378 cm²

Question 12.

In the 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:

(i) Let r be the radius of the circle and r = 35 cm \(\frac { 35 }{ 10 }\) = \(\frac { 7 }{ 2 }\) cm

Radius of the circle = Radius of quadrant of circle

= r = \(\frac { 7 }{ 2 }\) cm

Area of the quadrant of the circle = \(\frac { 1 }{ 4 }\)πr²

Since it is quadrant of circle, therfore the central angle of it θ = ∠AOB = 90°

(ii) Area of the shaded portion = Area of the sector for area of the quadrant of circle) – area of ∆AOD

Question 13.

In the 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:

OABC is a square and OA is 20 cm. Join OB. Now we have a triangle QAB.

By Pythagoras theorem

(OB)² = (OA)² + (AB)²

(OB)² = (20)² + (20)²

= 400 + 400

Area of square

OABC = (20)² = 400 cm².

Area of shaded region = Area of sector OPBQ – Area of square OABC

= 628 – 400

= 228 cm²

Question 14.

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

Solution:

Question 15.

In the 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:

Let us find BC By Pythagoras theorem BC² = AB² + AC²

BC² = 14² + 14²

BC = 14\(\sqrt{2}\) cm

Required Area = Area of semicircle BRC – [Area of quadrant – area of ∆ABC)

Required Area = Area BCQB – (Area BACQB – Area of ∆ABC)

Question 16.

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

Solution:

Area of square = (8)² = 64 cm²

Area of shaded region = Area of both sectors – Area of square

= \(\frac { 704 }{ 7 }\) – 64

= \(\frac { 704-448 }{ 7 }\) = \(\frac { 256 }{ 7 }\) cm²