﻿ Find the length of the diagonal BD when, the area of the quadrilateral is 32 cm2. from Mathematics Mensuration Class 8 Manipur Board

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The area of a rhombus and that of a square are equal. The side of the square is 6 cm. If one of the diagonal of the rhombus is 4 cm, then find the length of its other diagonal.

Here, side of the square = 6 cm
∴                   Area of the square =  Side x Side
= 6 cm x 6 cm
= 36 cm2
Since,         [Area of the rhombus]  = [Area of the square]
∴                  Area of the rhombus =

One of the diagonal of the rhombus = 4 cm
Let the other diagonal of the rhombus  = d cm
∵                  Area of the rhombus  =

∴                  Area of the rhombus  =

Thus, the required length of the rhombus = 18 cm.
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The edge of a cube is 2 cm. Find the total surface area of the cuboid formed by three such cubes joined edge to edge.

The edge (side) of the given cube = 2 cm
Since, there such cubes are joined, then
Total length (l) = (2+2+2) cm
= 6 cm
Height (h) = 2 cm

∴   Total surface area of the resultant cuboid
= 2[lb + bh + hl]
= 2[6 x 2 + 2 x 2 + 2 x 6] cm2
= 2[28] cm2
= 56 cm2
Thus, the required total surface area of the cuboid = 56 cm2.

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# Find the length of the diagonal BD when, the area of the quadrilateral is 32 cm2.

Let the length of the diagonal BD = x cm
Area of the quadrilateral ABCD = 32 cm2
∵    Area of the quadrilateral  =  x (A diagonal) x (Sum of the length of the perpendicular on the diagonal from the opposite vertices)
∴    Area of the quadrilateral ABCD =
=

=
Since area of the quadrilateral ABCD = 32 cm2
∴

= 8 cm
Thus, the required length of the diagonal BD = 8 cm.
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A road roller makes 250 complete revolution to move once over to once level a road. Find the area of the road levelled if the diameter of the road roller is 84 cm and length is 1 m.

A road roller is a cylinder, such that
Length (h) = 1m = 100 cm
∵       Lateral surface area of a cylinder =
∴       Lateral surface area of the road roller

∴        Area of road levelled in 1 revolution = 26400 cm2
Area of road levelled in 250 revolutions

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A rectangle and a square have equal perimeter. If the breadth of the rectangle is 10 cm and the side the square is 15 cm, then find the difference between their areas.

∵                     Side of the square = 15 cm
∴             Perimeter of the square = 4 x 15 cm = 60 cm
Since, [Perimeter of the rectangle] = [Perimeter of the square]
∴     [Perimeter of the rectangle] = 60 cm
Let length of the rectangle = x cm
∴                  2[x + 10] = 60     [∵ Breadth of the rectangle = 10 cm]
2x + 20 = 60
2x = 60 - 20 = 40

Length of the rectangle = 20 cm
Now, Area of the rectangle  = Length x Breadth
= 20 x 10 cm2
= 200 cm2
Area of the square = Side x Side
= 15 x 15 cm2
= 225 cm2
∴ Difference in areas = 225 cm2 - 200 cm2 = 25 cm2

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