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**Question 1: **Find the area of the following trapezium.**Solution:** Here, parallel sides are 25 cm and 15 cm.

Height (i.e. distance between the parallel sides) = 12 cm

∵ Area of a trapezium ABCD = 1/2 x (Sum of the parallel sides) * Height

∴ Area of the trapezium ABCD

= 1/2 (25 + 15) x 12 cm

= 1/2 x 40 x 12 cm^{2} = 240 cm^{2}

Thus, the required area of the trapezium ABCD = 240 cm^{2}**Question 2: **Find the length of the diagonal BD when, the area of the quadrilateral is 32 cm^{2}.**Solution:** Let the length of the diagonal BD = x cm

Area of the quadrilateral ABCD = 32 cm^{2}

∵ Area of a quadrilateral = (1/2)* (A diagonal) * (Sum of the length of the perpendiculars on the

diagonal from the opposite vertices)

∴ Area of the quadrilateral

ABCD = 1/2 x BD x (AP + CQ)

= 1/2 * x * (4.5 +3.5)

= 1/2 * x * 8 cm^{2}

Since area of the quadrilateral ABCD = 32 cm^{2}

Thus, the required length of the diagonal BD = 8 cm.**Question 3:** 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.**Solution: **Here, side of the square = 6 cm

∴ Area of the square = Side * Side

= 6 cm * 6 cm

= 36 cm^{2}

Since, [Area of the rhombus] = [Area of the square]

∴ Area of the rhombus = 36 cm^{2}

One of the diagonal of the rhombus = 4 cm

Let the other diagonal of the rhombus = d cm

∵ Area of a rhombus = (1/2)* Product of the diagonals

Thus, the required length of the rhombus = 18 cm.**Question 4: **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.**Solution: **The edge (side) of the given cube = 2 cm

Since, three such cubes are joined, then

Total length (l) = (2 + 2 + 2) cm = 6 cm

Breadth (b) = 2 cm

Height (h) = 2 cm

∴ Total surface area of the resultant cuboid

= 2[lb + bh + hl]

= 2[6 * 2 + 2 * 2 + 2 * 6] cm^{2}

= 2[12 + 4 + 12] cm^{2}

= 2[28] cm^{2} = 56 cm^{2}

Thus, the required total surface area of the cuboid = 56 cm^{2}.**Question 5:** A closed cylindrical tank of radius 7 cm and height 5 m is made from a sheet of metal.

If the breadth of the rectangular sheet is 10 m, then find the length of the sheet.

Solution: Here, Radius (r) = 7 cm

Height (h) = 5 cm

Since total surface area of the closed cylinder = 2πr (r + h)

∴ Total surface area of the cylindrical tank

Let, length of the sheet required = ‘l’ m

∴ Area of the rectangular sheet = l * 10 m^{2}

⇒ l * 10 = 528

Thus, the required length of the sheet = 52.8 m.**Question 6:** A road roller makes 250 complete revolutions to move once over to level a road. Find the area of the road levelled if the diameter of the road roller is 84 cm and the length is 1 m.**Solution: **A road roller is a cylinder, such that

Radius =84/2 cm = 42 cm

Length (h) = 1m = 100 cm

∵ Lateral surface area of a cylinder = 2πrh

∴ Lateral surface area of the road roller

= 2 x 22/7 x 42 x 100 cm^{2}

= 2 * 22 * 6 * 100 cm^{2}

∴ Area of road levelled in 1 revolution = 26400 cm^{2}

⇒ Area of road levelled in 250 revolutions

= 250 * 26400 cm^{2 }**Question 7: **A cuboid is of dimensions 50 cm * 45 cm * 30 cm. How many small cubes with sides 5 cm can be placed in the given cuboid?**Solution:** Volume of a cuboid = Length * Breadth * Height

Here, Length of the cuboid = 50 cm

Breadth of the cuboid = 45 cm

Height of the cuboid = 30 cm

∴ Volume of the cuboid = 50 * 45 * 30 cm^{3}

Volume of the small cube = Side * Side * Side

= 5 * 5 * 5 cm^{3}

Let the number of the required small cubes = x

∴ x * (5 * 5 * 5) = 50 * 45 * 30

= 10 * 9 * 6 = 540

Thus, 540 small cubes can be placed in the given cuboid.

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