﻿ Find the area the quadrilateral LMNOP (as shown in the figure). from Mathematics Mensuration Class 8 Manipur Board

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Find the area the quadrilateral LMNOP (as shown in the figure). Diagonal LN = 8 cm
Perpendicular MX = 4.5 cm
Perpendicular OY = 3.5 cm
∵   Sum of the perpendiculars = MX + OY = 4.5 cm + 3.5 cm = 8 cm
∴   Area of the quadrilateral = (A diagonal ) x (Sum of the lengths of the perpendiculars on it from opposite vertices) 949 Views

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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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 cm2
= 56 cm2
Thus, the required total surface area of the cuboid = 56 cm2.

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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
Radius  = 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    137 Views

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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