A door of length 1 m and breadth 0.5 m is on a wall. The length of the wall is 4.5 m and the breadth is 3.6 m as shown in figure. Find the cost of white washing the wall, if the rate of white washing the wall is Rs. 20 per m2.
We have | ||
L = length of the wall | = | 4.5 m |
b = Breadth of the wall | = | 3.6 m |
∴ Area of the wall | = | l × b |
= | 4.5 m × 3.6 m | |
= | 16.2 m2 | |
Area of the door | = | 1 m × 0.5 m |
= | 0.5 m2 | |
∴ Area to be white washed | = | 16.2 m2 – 0.5 m2 |
= | 15.7 m2 | |
Cost of white washing the wall | = | Rs. (15.7 × 20) |
= | Rs. 314 |
Let ABCD and EFGH be the two roads. | ||
Area of these two roads | = | (Area ABCD) + (Area EFGH) – (Area PQRS) |
= | {(60 × 5) + (40 × 5) – (5 × 5)} m2 | |
= | (300 + 200 – 25) m2 | |
= | 475 m2 | |
Cost of constructing the roads | = | Rs. (475 × 80) |
= | Rs. 38000 |
Calculate the area of the shaded region in the figure given below.
Area of the shaded region | = | (Area of rect. ABCDE) + (Area of rect. TPQRS) + (Area of rect. EDPT) |
= | {(13 × 2) + (16 × 1) + (9 × 2)} m2 | |
= | (26 + 16 + 18) m2 | |
= | 60 m2 | |
Hence, the area of the shaded region is 60 m2. |
A copper wire when bent in the form of a square encloses an area of 121 cm2. If the same wire is bent in the form of a circle. Find the area enclosed by it.
Area of the square | = | 121 cm2 |
⇒ Side of the square | = | √(121) cm |
= | 11 cm | |
⇒ Perimeter of the square | = | (11 × 4) cm |
= | 44 cm | |
⇒ Circumference of the circle | = | Perimeter of the square |
= | 44 cm | |
Let the radius of the circle be r cm. | ||
Then, its circumference | = | (2πr) cm |
∴ 2πr | = | 44 |
⇒ 2 × ![]() |
= | 44 |
r | = | 44 × ![]() |
= | 7 | |
∴ Area of the circle | = | πr2 |
= | (![]() |
|
= | 154 cm2 | |
Hence, the area enclosed by the circle is 154 cm2. |
Each side of a square park is 80 m. At each corner of the park, there is a flower bed in the form of a quadrant of a circle of radius 14 m, as shown in the figure. Find the area of the remaining part of the park.
Area of the park | = | (80 × 80) m2 |
= | 6400 m2 | |
Area of 4 flower beds | = | (4 × ![]() ![]() |
= | 616 m2 | |
Area of the remaining part | = | (6400 – 616) m2 |
= | 5784 m2 | |
Hence, the required area is 5784 m2. |
Four equal circles, each of radius 5 cm, touch each other as shown in figure. Find the area included between them.
Side of the square | = | 10 cm |
We know that the area of the square | = | Side × Side |
= | 10 × 10 | |
= | 100 cm2 | |
The radius of the quarter circle | = | 5 cm |
Area of quarter circle | = | ![]() |
Area of 4 quarter circle | = | 4 × ![]() |
= | πr2 | |
= | 3.14 × (5)2 | |
= | 78.5 cm2 | |
The area included in the quarter circle | = | Area of the square – Area of the four quarter circles |
Area included in them | = | (100 – 78.5) cm2 |
= | 21.5 cm2 |