Area word problem

Solution: Given: ABCD is a rectangle and BC is semi circle Example 1: A paper is in the form of a rectangle ABCD in which AB = 20 cm and BC = 14 cm. A Semicircle portion with BC as diameter is cut off. Find the area of the remaining part. Solution: Given: ABCD is a rectangle and BC is semi circle Length AB = 20 cm , Breadth BC= 14 cm BC = d =14 cm. 14 cm 20 cm 14 cm Area of Inner Shape Inner shape is semicircle d=14 ,  r = d/2 = 14/2 = 7 cm Area of semicircle = 11 x 7 Area of Semi circle BC = 77 cm Area of shaded region = Area of outer shape - Area of inner shape = 280 – 77 = 203 cm Area of remaining part = 203 cm Area of outer shape Outer shape is rectangle A = l x b A = 20 x 14 = 280 cm Area of rectangle ABCD = 280 cm 11

Inner shape is 4 quadrants Example 2 : A square park has each side of 100 m. At each corner of the park there is a flower bed in the form of a quadrant of radius 14 m as shown in the figure. Find the area of the remaining portion of the park. Solution: Given, Side of Square a = 100 m radius of quadrant r = 14 m 100 m Area of Inner Shape Inner shape is 4 quadrants Radius of quadrant r= 7 cm Area of 4 quadrants= Area of 4 quadrants A = 616 m2 Area of outer shape Outer shape is a square Area of square = a x a A= 100 x 100 = 10000 m Area of square A = 10000 m2 Area of shaded region A= Area of outer shape - Area of inner shape A= 10000 – 616 A= 9384 m2 Area of the remaining portion of the park = 9384 m2 11 2 2

Example 2: A 14m wide athletic track consists of two straight Sections each 120 m long joined by semi-circular ends with Inner radius is 35 m. Calculate the area of the track. Solution: Given : Breadth of the track(b) = 14 m, length of the track(l)=120m Radius of the inner semi circle(r) = 35 m Radius of the outer semi circle(R) = 35 + 14 = 49 m Radius of the outer semi circle R = 49 m Area of the track is the sum of the areas of 2 semicircular tracks and the areas of 2 rectangular tracks. Rectangular tracks ABCD and EFGH have same length and breadth Area of the rectangular tracks ABCD and EFGH = 2 x (l x b) = 2 x 120 x 14= 3360 m2 Area of the rectangular tracks ABCD and EFGH = 3360 m2

Cont….. Since, (a2 – b2 = (a +b) (a- b)) The two semicircular track have same outer radius and inner radius. Area of the 2 semicircular tracks = 2 x (Area of the outer semicircle – Area of the inner semicircle) Area of the semicircular tracks = 3696 m2 Area of the track = Area of the rectangular tracks ABCD and EFGH + Area of the 2 semicircular tracks = 3360 + 3696 = 7056 m2 Ans : Area of the track = 7056 m2 semicircular track semicircular track Since, (a2 – b2 = (a +b) (a- b))

Try these Question A square park has each side of 28 m. At each corner of the park there is a flower bed in the form of a quadrant of radius 14 m as shown in the figure. Find the area of the remaining portion of the park.