Find the area of the shaded region in Fig. 11.9
Solution:
We have to find the area of the shaded region.
From the given figure,
We observe that the figure comprises an outer rectangle with inner rectangle which has semicircular ends.
PQRS is an outer rectangle
ABCD is inner rectangle with
ADM and BCN are two semicircular ends.
Considering rectangle PQRS
Length = 26 m
Breadth = 12 m
Area of rectangle = length × breadth
So, area of rectangle PQRS = 26 × 12
= 312 square m
Considering rectangle ABCD,
Breadth of rectangle = diameter of semicircle
Breadth = 12 - 4 - 4
= 12 - 8
Breadth = 4 m
Length = 26 - 2(radius of semicircle) - 3 - 3
= 26 - 2(4/2) - 6
= 26 - 4 - 6
= 26 - 10
= 16 m
Now, length of inner rectangle = 16 m
Breadth of inner rectangle = 4 m
Area of inner rectangle = 16(4)
= 64 m²
Considering semicircles AMB and DNC,
Diameter of semicircle = 4 m
Radius = 4/2 = 2 m
Area of semicircle = πr²/2
= (22/7)(2)²/2
= (22/7)(2)
= 44/7
= 6.29 m²
Area of semicircular ends = 2(6.29)
= 12.58 m²
Area of the shaded region = area of outer rectangle - area of inner rectangle - area of semicircular ends.
So, area of the shaded region = 312 - 64 - 12.58
= 312 - 76.58
= 235.42 m²
Therefore, the area of the shaded region is 235.42 m²
✦ Try This: 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 14m as shown in Fig. 12.31. Find the area of the remaining Part of the park.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 12
NCERT Exemplar Class 10 Maths Exercise 11.3 Problem 9
Find the area of the shaded region in Fig. 11.9
Summary:
A rectangle is a two-dimensional figure with four sides, four vertices, and four angles. The area of the shaded region is 235.42 m²
☛ Related Questions:
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