1. Area word problem

2. 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

3. 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 = m
Area of square A = m2
Area of shaded region
A= Area of outer shape
- Area of inner shape
A= – 616
A= 9384 m2
Area of the remaining portion of the park = 9384 m2 11 2 2

4. 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) = = 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

5. 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 =
= m2
Ans : Area of the track = 7056 m2
semicircular track
semicircular track
Since, (a2 – b2 = (a +b) (a- b))

6. 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.