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
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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, OR2 = PQ2 + PR2
= 242 + 72
= 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 πr2
= ½ x π x 12.52 = 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 = Side2 = 142 = 196 sq cm
Area of two semicircles = πr2
= π x 72 = 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 = Side2 = 42 = 16 sq cm
Area of cut portion = Area of two circles = 1 circle + 4quadrants
= 2 x π x 12 = 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 = 30o
Hence, area of equilateral triangle
Area of circle = πr2 = π x 322 = 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 = Side2 = 142 = 196 sq cm
Area of four quadrants = Area of 1 circle
= πr2 = π x 72 = 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(R2 – r2)
Where, R = radius of outer circle and r = radius of inner circle
Area = π(402 – 302) = 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 = πr2 = 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