Class 7 – Mensuration
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Introduction: Already we learnt about perimeter and area of triangles, cube and cuboid in earlier classes. Now we just recall the concept. Perimeter and area of a triangle: i) Scalene triangle:
Area of triangle If sides are given, then ii) Equilateral Dle: Area of equilateral 
Perimeter of equilateral Height of equilateral iii) Right angled Dle: Area of right angle
Perimeter iv) Isosceles right angled triangle: Area Perimeter


v) Rectangle: Area= length × breadth = lb sq.units
Perimeter= 2 (length + breadth) = 2(l + b) units
Diagonal
vi) Square: Area
Perimeter= 4 × side = 4a (or) 4s units
Diagonal
1. Find the perimeter of a rectangle whose length and breadth are 25 m and 15 m respectively.
Solution: We have,
Length= 25 m and Breadth = 15 m
∴Perimeter = 2 (Length + Breadth)
=2 (25 + 15) m
= 2 × 40 m = 80 m.
2. Find the side of a square whose perimeter is 100 cm.
Solution: We have,
Perimeter= 100 cm
3. Pinky runs around a square field of side 75 m, Bobby runs around a rectangular field with length 160 m and breadth 105 m. Who covers more distance by how much?
Solution: We have,
Distance covered by Pinky in one round = Perimeter of the square field of side 75 m
= 4 × Length of a side
= 4 × 75 m = 300 m.
Distance covered by Bobby in one round = Perimeter of the rectangular field
= 2 × (Length + Breadth)
= 2 × (160 m + 105 m)
=2 × 265 m = 530 m
∴Difference in distance covered = 530 m – 300 m
= 230 m.
Hence, Bobby covers 230 m more distance than Pinky.
4. Find the perimeter of each of the following shapes:
i) An equilateral triangle of side 9 cm.
ii) An isosceles triangle with equal side 8 cm each and third side 6 cm.
Solution: i) We have,
Perimeter= 3 × (Length of a side)
= 3 × 9 cm
= 27 cm
ii)We have,
Perimeter = 8 + 8 + 6 = 22 cm.
5. Find the area of a rectangle of its length and breadth are 28 cm and 120 mm,respectively.
Solution: We have,
Length = 28 cm
Breadth = 120 mm = 12 cm
∴ Area = (Length × Breadth)
= (28 × 12) cm2 = 336 cm2.
6. Find the area of a square whose side is 5 cm.
Solution: We know that, area of a square = (Side)2
Here,Side = 5 cm
Area of the square = (5)2 cm2 = 25 cm2.
7. Find the perimeter of a rectangle whose area is 650 cm2 and its breadth is 13 cm.
Solution: We have,
Area of the rectangle = 650 cm2
Breadth of the rectangle = 13 cm
Length of the rectangle
Perimeter of the rectangle = 2 × (Length × Breadth)
= 2 × (50 + 13) cm
= 2 × 63 cm = 126 cm.
8. A marble tile measures 25 cm by 20 cm. How many tiles will be required to cover a wall of size 4 m by 3 m?
Solution: We have,
Area of the wall = 4 × 3 m2
= 12 m2
= 120000 cm2
Area of one tile = 25 × 20 cm2
= 500 cm2
9. The total cost of flooring a room at ` 8.50per square metre is ` 510. If the length of the room is 8 metres, find its breadth.
Solution: We have,
Total cost of flooring = ` 510
Rate of flooring = ` 8.50 per m2
Length of the room = 8 metres
10. What is the area of the square ABCD shown in the diagram?
Solution:
11. What is the area of a figure formed by a square of side 8 cm and an is osceles triangle with base as one side of the square and the perimeter as 18 cm?
Solution: Each of the equal sides of the isosceles
altitude EF bisects the base AB, AF = FB = 4 cm
∴ Area of the figure = Area of +Area of square ABCD
12. If the sides of an equilateral triangle are increased by 20 %, 30 % and 50 % respectively to form a new triangle, what is percentage increase in the perimeter of the equilateral triangle?
Solution: Let each side of the equilateral triangle be x cm. Then, after increase of the three sides are
i.e.,
i.e.,
∴Original perimeter = 3x,
Increased perimeter
% increase in perimeter
13. Find the area of a triangle in which base = 25 cm and height = 14 cm.
Solution: Here, base = 25 cm and height= 14 cm.
∴ area of the triangle
14. Find the height of a triangle whose base is 15 cm and area 120 cm2.
Solution: Let the required height be hcm. Then,
∴Height of the triangle is 16 cm.
15. Find the area of a right triangle in which base = 24 cm and hypotenuse is 25 cm. If find BD.
Solution:
Consider a in which
Now,
Area of
Let
16. Find the area of a triangular field whose sides are 78 m, 50 m and
112 m.
Solution: Let
Hence, Area of the given triangle is 1680 m2.
Area of an isosceles triangle with base b units and each equal side a units is
Perimeter = a + a + b = 2a + b units