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

**Class 10th Question Answers on Area of Circle Area of Circle Chapter 12.3 (NCERT Book) Part 1****—**Hello All we are weekly increasing our service for our users we shared Quantitative aptitude question answers,Previous exam papers,Free E-Books And this time we started basic level question answers jope you like it.Class 10th Question Answers on Area of Circle or *Area of Circle Exercise 12.3 (NCERT Book) Part 1*

### You may like this more

### {#AFCAT EXAM $} AFCAT 2 Exam Full Analysis,AFCAT 2 Exam Question Papers With Answer Key 2015

### Problems Based on Ages Short-cut Tricks Or Concepts and Tricks On Ages And Age Calculation

### Reasoning Question And Answer On Problem On Numbers For Upcoming Exams

Question: 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:** Here, OR is the hypotenuse of ∆PQR, because lines from two ends of diameter always make a right angle when they meet at circumference.

So, OR^{2} = PQ^{2} + PR^{2}

= 24^{2} + 7^{2}

= 576 + 49 = 625

Or, OR = 25 cm

Or radius = 12.5 cm

Area of triangle = ½ x base x height

= ½ x 24 x 7 = 84 sq cm

Area of semicircle = ½ x πr^{2}

= ½ x π x 12.5^{2} = 245.3125 sq cm

So, area of shaded region = 245.3125 – 84 = 161.3125 sq cm

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 angle AOC = 40°.

**Solution:** Area of shaded region

= Area of Bigger Sector – Area of Smaller Sector

If R and r are the two radii, then area of shaded region

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:** Area of Square = Side^{2} = 14^{2} = 196 sq cm

Area of two semicircles = πr^{2}

= π x 7^{2} = 154 sq cm

Area of shaded region = 196 – 154 = 42 sq cm

Question: 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:** Area of sector outside triangle

Area of triangle

Area of shaded region = 94.28 + 62.352 = 156.632 sq cm

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:** Area of square = Side^{2} = 4^{2} = 16 sq cm

Area of cut portion = Area of two circles = 1 circle + 4quadrants

= 2 x π x 1^{2} = 6.28 sq cm

So, area of remaining portion of square = 16 – 6.28 = 22.28 sq cm

Question: 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 (shaded region).

**Solution:** In ∆OCB; OB = 32 cm, ∠OBC = 30^{o}

Hence, area of equilateral triangle

Area of circle = πr^{2} = π x 32^{2} = 3215.36 sq cm

Area of shaded region = 3215.36 – 1330.176 = 1885.184 sq cm

Question: 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:** Area of square = Side^{2} = 14^{2} = 196 sq cm

Area of four quadrants = Area of 1 circle

= πr^{2} = π x 7^{2} = 154 sq cm

Area of shaded portion = 196 – 154 = 42 sq cm

Question: 8. The following 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:** Distance = Length of straight track + circumference of inner loop

Circumference = 2πr = 3.14 x 60 = 188.4

Hence, distance = 188.4 + 106 + 106 = 400.4 m

Area of straight portion = 2 x length x width

= 2 x 106 x 10 = 2120 sq m

Area of circular portion = π(R^{2} – r^{2})

Where, R = radius of outer circle and r = radius of inner circle

Area = π(40^{2} – 30^{2}) = 2200 sq m

So, Total area of track = 2200 + 2120 = 4320 sq m

Question: 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:** Area of smaller circle = πr^{2} = π x (3.5)^{2} = 38.5 sq cm

Area of bigger circle = 4 x 38.5 (because radius is double that of smaller circle)

Hence, area of bigger semicircle = 2 x 38.5 = 77 sq cm

Area of ∆ABC = ½ x AB x OC

= ½ x 14 x 7 = 49 sq cm

Area of shaded portion in semicircle = 77 – 49 = 28 sq cm

Total area of shaded portion = 28 + 38.5 = 66.5 sq cm

Class 10th Question Answers on Area of Circle,Area of Circle

## Leave a Reply