Question

Solution :
Length of park ABCD, L = 50 m
Width of park, B = 40 m
Length of path PQRS, I. = AD
Width of path PQRS, b,
= PQ
total area covered
area of EFGH - area of LMNO
= 100 + 125 - 6.25 = 218.75"
cost of paving 1 m of area = 15
the cost of paving the paths = 15 x 216.75
= 3281.25
= 40m
= 2.5 m
Exercise 15.1
1. The length of a rectangular-shaped label is 1 centimetre more than twice the width. The perimeter
is 110 centimetres. Find the area of the rectangle.
2. Find the dimensions of the rectangular field whose length and breadth are in the ratio 6:4 and
whose area is 21600 m.
3. A wire, bent in the shape of a circle, encloses an area of 3850 cm. If the wire is now bent in the
shape of a square, find the area enclosed by the square
4. The area of a square field is 256 ni. Find its perimeter
5. A traffic signal is in the form of an equilateral triangle. The cost of colouring it at the rate of
*2 per cm is 7 498.24. The edges of the sign are to be taped with a red tape. Find the cost of the
tape used at the rate of 1.5 per centimetre (Use . 3 = 1.73 ]
6. A parallelogram has a base of 3 cm and a height of 7 cm. What is its area?
7. A rectangular piece of paper has a length of 16 cm and an area of 192cm. What is its perimeter
8. The perpendicular sides of a right triangle are in the ratio 7:24 and its hypotenuse is 50 cm lon
Find the perimeter and the area of the triangle. (Hint: Use Pythagoras theorem)
-2
9. There is a circular path of width 2 m along the boundary and inside a circu
nark of radius 16 m. Find the cost of paving the path with bricks at the​

Answer 1

Answer:

2,000

Explanation:

length×width

=40×50