Advertisements
Advertisements
प्रश्न
Find the area enclosed by the following figure:
Advertisements
उत्तर
The given shape contains a semi-circle and a triangle.
Area of semi-circle
= `1/2 πr^2`
= `1/2 xx 22/7 xx 10 xx 10`
= `1100/7`cm2
Area of triangle
= `1/2` × Base × Height
= `1/2 xx 20 xx 7`
= 70 cm2
Hence, total area enclosed by the shape
= `1100/7 + 70`
= `(1100 + 490)/7`
= `1590/7`
= 227 cm2 (approx)
APPEARS IN
संबंधित प्रश्न
A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path.
A path 1 m wide is built along the border and inside a square garden of side 30 m. Find:
1) the area of the path
2) the cost of planting grass in the remaining portion of the garden at the rate of Rs 40 per m2
In the following figure, find the area of the shaded portions
Find the area of the quadrilateral ABCD.
Here, AC = 22 cm, BM = 3 cm,
DN = 3 cm, and
BM⊥AC, DN⊥AC
A rectangular park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park. Find the area of the path.
A path 5 m wide runs along inside a square park of side 100 m. Find the area of the path. Also, find the cost of cementing it at the rate of Rs. 250 per 10 m2.
Two crossroads, each of width 5 m, run at right angles through the center of a rectangular park of length 70 m and breadth 45 m and parallel to its sides. Find the area of the roads. Also, find the cost of constructing the roads at a rate of Rs. 105 per m2.
Find the area enclosed by the following figure:
A floor is 5 m long and 4 m wide. A square carpet with sides of 3 m is laid on the floor. What is the area of the floor that is not carpeted?
The perimeter of a square field is 112 m. What is the area of the square field?
