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

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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  6. 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 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. Find the area of the shaded region. (Use π = 3.14 and 3 = 1.73205)

10 circle area exercise solution

Solution: Area of equilateral triangle

10 circle area exercise solution

Radius of circle = 100 cm (half of a)

Area of circle = πr2 = π x 1002 = 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.

10 circle area exercise solution

Solution: Area of 9 circles = 9 x πr2

= 9 x π x 72 = 1386 sq cm

Side of square = 6 x 7 = 42 cm

Area of square = Side2 = 422 = 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.

10 circle area exercise solution

Solution: Area of quadrant = ¼ x πr2

= ¼ x π x 3.52 = 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.

10 circle area exercise solution

Solution: Using Pythagoras Theorem;

BO2 = OA2 + OC2

= 202 + 202

Or, BO = 20√2 cm = Radius of circle

Area of quadrant = ¼ x πr2

= ¼ x π x (20√2)2 = 628 sq cm

Area of square = Side2 = 202 = 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.

10 circle area exercise solution

Solution: Area of shaded region

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

Area of sector of bigger circle

10 circle area exercise solution

Area of sector of smaller circle

10 circle area exercise solution

Area of shaded region

10 circle area exercise solution

 

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.

10 circle area exercise solution

Solution: Area of quadrant = ¼ x πr2

= ¼ x π x 142 = 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 πr2

= 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.

10 circle area exercise solution

Solution: Area of square = Side2 = 82 = 64 sq cm

Area of shaded portion

= Double area of segment formed by diagonal of the square

Area of two quadrants = ½ πr2

= ½ x π x 82 = 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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