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 Fig. 12.23. Find the area of the remaining portion of the square
Solution:
We use the concepts of area of sectors of circles and area of rectangles to solve this problem.
The diameter of the circle which is cut out = 2 cm
∴ Radius of this circle (r) = 1 cm
Radius of all quadrants cut out (r) = 1 cm
Since all quadrants cut out are of the same radius thus,
Area of portions cut out of square = Area of the circle + 4 × (Area of each quadrant)
= πr2 + 4 (90°/360° × πr2)
= πr2 + 4 × πr2/4
= πr2 + πr2
= 2πr²
= 2π
= 2 × 22/7 cm2 =
= 44/7 cm2
Area of the remaining portion of the square = Area of square - Area of portion cut out
= (4 cm)2 - 44/7 cm2
= 16 cm2 - 44/7 cm2
= (112 - 44)/7 cm2
= 68/7 cm2
☛ Check: NCERT Solutions for Class 10 Maths Chapter 12
Video Solution:
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 Fig. 12.23. Find the area of the remaining portion of the square.
NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.3 Question 5
Summary:
If 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 then the area of the remaining portion of the square is 68/7 cm2.
☛ Related Questions:
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- In Fig. 12.25, 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.
- Fig. 12.26 depicts a racing track whose left and right ends are semicircular. Fig. 12.26 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.
- In Fig. 12.27, 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.