# Class 10th Question Answers on Area of Circle Exercise 12.3 (NCERT Book) Part 2

**Class 10th Question Answers on Area of Circle Exercise 12.3 (NCERT Book) Part 2**–Hello All we are weekly increasing our service for our users we shared Quantitative aptitude question answers,Previous exam papers,Free E-Books And last time we started basic level question answers hope you like it.Class 10th Question Answers on Area of Circle or * Area of Circle Exercise 12.3 (NCERT Book) Part 1 *…And This time We are back with

**Class 10th Question Answers on Area of Circle Exercise 12.3 (NCERT Book) Part 2**

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- Class 10th Question Answers on Area of Circle Chapter 12.3 (NCERT Book) Part 1

Question: 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:** Area of equilateral triangle

Radius of circle = 100 cm (half of a)

Area of circle = πr^{2} = π x 100^{2} = 31400 sq cm

Area of three sectors = ½ Area of circle (Because all angles sum up to 180)

= ½ x 31400 = 15700 sq cm

So, area of shaded portion = 17302.5 – 15700 = 1602.5 sq cm

Question: 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:** Area of 9 circles = 9 x πr^{2}

= 9 x π x 7^{2} = 1386 sq cm

Side of square = 6 x 7 = 42 cm

Area of square = Side^{2} = 42^{2} = 1764 sq cm

Area of remaining portion = 1764 – 1386 = 378 sq cm

Question: 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:** Area of quadrant = ¼ x πr^{2}

= ¼ x π x 3.5^{2} = 9.625 sq cm

Area of ∆BDO = ½ x BD x OD

= ½ x 3.5 x 2 = 3.5 sq cm

Hence, area of shaded portion = 9.625 – 3.5 = 6.125 sq cm

Question 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.

**Solution:** Using Pythagoras Theorem;

BO^{2} = OA^{2} + OC^{2}

= 20^{2} + 20^{2}

Or, BO = 20√2 cm = Radius of circle

Area of quadrant = ¼ x πr^{2}

= ¼ x π x (20√2)^{2} = 628 sq cm

Area of square = Side^{2} = 20^{2} = 400 sq cm

Area of shaded portion = 628 – 400 = 228 sq cm

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

**Solution:** Area of shaded region

= Area of sector of bigger circle – Area of sector of smaller circle

Area of sector of bigger circle

Area of sector of smaller circle

Area of shaded region

Question: 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:** Area of quadrant = ¼ x πr^{2}

= ¼ x π x 14^{2} = 154 sq cm

Area of triangle = ½ x b x h

= ½ x 14 x 14 = 98 sq cm

Area of segment made by hypotenuse of triangle

= 154 – 98 = 56 sq cm

Diameter of external semicircle = 14√2 cm (since other two sides of triangle are 14 cm)

So, area of semicircle = ½ x πr^{2}

= ½ x π x (14√2)^{2} = 154 sq cm

Area of shaded portion = 154 – 56 = 98 sq cm

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

**Solution:** Area of square = Side^{2} = 8^{2} = 64 sq cm

Area of shaded portion

= Double area of segment formed by diagonal of the square

Area of two quadrants = ½ πr^{2}

= ½ x π x 8^{2} = 100.48 sq cm

Area of square = Area of two triangles formed by radii and the cord

So, Area of shaded portion = 100.48 – 64 = 36.48 sq cm

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