Class 7 – Mensuration

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Topic Sub Topic Online Practice Test
Mensuration
  • Review of mensuration
  • Perimeter and area of quadrilaterals
  • Circumference and area of circles
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Mensuration
  • Area of paths
  • Surface area and volume of solids
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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:
Perimeter of triangle = (a + b + c) units.

Area of triangle

If sides are given, then
by Heron’s formula Area of triangle 

ii)   Equilateral Dle: Area of equilateral

Perimeter of equilateral

Height of equilateral

iii) Right angled Dle: Area of right angle

Perimeter
but hypotenuse

iv)  Isosceles right angled triangle: Area

Perimeter
∴ Hypotenuse

 

 

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.

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Area of an isosceles triangle with base b units and each equal side a units is

Perimeter = a + a + b = 2a + b units