The document NCERT Solutions(Part- 3)- Mensuration Class 8 Notes | EduRev is a part of the Class 8 Course Class 8 Mathematics by VP Classes.

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**Exercise 11.2**

**Question 1. **The shape of the top surface of a table is a trapezium. Find its area it its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.

**Solution:** âˆµ Area of a trapezium = 1/2 * [Sum of parallel sides] * [Perpendicular distance between the parallel sides]

âˆ´

**Question 2.** The area of a trapezium is 34 cm^{2} and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.

**Solution: **Let the required other side = x cm

Area of the trapezium = 1/2 * [Sum of parallel sides ] * [Height]

or

or 10 + x = 17 â‡’ x = 17 â€“ 10 = 7 cm

Hence, the required other side = 7 cm.

**Question 3.** Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

**Solution: **Length of the fence = Perimeter

âˆ´ AB + BC + CD + DA = 120

or AB + 48 + 17 + 40 = 120

or AB + 105 = 120

or AB = 120 â€“ 105 = 15 m

âˆµ Area of a trapezium = 1/2 * [Sum of parallel sides] * Height

**Question 4.** The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.

**Solution:** Area of a quadrilateral

= 1/2 * Diagonal * [Sum of the perpendiculars on the diagonal from opposite vertices]

**Question 5.** The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

**Solution: **Area of the rhombus = 1/2 *Product of diagonals

**Question 6. **Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.

**Solution: **A rhombus is a parallelogram.

Area of a parallelogram = Base * Height

âˆ´ Area of the rhombus = Base * Height

= 6 * 4 cm2 = 24 cm^{2}

Let the other diagonal be d.

âˆ´ Area of the rhombus = 1/2 * 8 * d = 4d [âˆµ One of the diagonals = 8 cm]

or 4d = 24

or d = 24/4 = 6 cm

Thus, the required other diagonal is 6 cm.

**Question 7.** The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m^{2} is Rs 4.

Solution: Tiles are rhombus shaped, having d_{1} = 45 cm and d_{2} = 30 cm.

âˆ´ Area of a tile (rhombus) = 1/2 * d_{1} * d_{2}

= 1/2 * 45 * 30 cm^{2}

= 45 * 15 cm2 = 675 cm^{2}

Total number of tiles = 3000

âˆ´ Area of the floor = 675 * 3000 cm^{2}

= 2025000 cm^{2}

Cost of polishing the floor = Rs 4 * 2025/10

= Rs 2 * 405 = Rs 810

**Question 8.** Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along road. If the area of this field is 10500 m^{2} and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

**Solution: **Let the length of the side along the river = x m

âˆ´ Other side (parallel to the road) = 1/2 xm

âˆ´ Area of the trapezium shaped field

But area of the field = 10500 m^{2}

âˆ´ 75x m^{2} = 10500 m^{2}

or

Thus, length of the side along the river = 140 m

**Question 9.** Top surface of a raised platform is in the shape of regular octagon as shown in the figure. Find the area of the octagonal surface.

**Solution:** This regular octagon can be into two trapeziums (each having height 4 m and parallel sides as 11 m and 5 m) and a rectangular part with length 11 m and breadth 5 m.

Area of a trapezium

âˆ´ Area of both the trapeziums = 2 * 32 m^{2} = 64 m^{2}

Now, Area of the rectangular part = l * b

= 11 m * 5 m = 55 m^{2}

âˆ´ Area of the regular octagon = 64 m^{2} + 55 m^{2} = 119 m^{2}

**Question 10. **There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways.

Find the area of this park using both ways. Can you suggest some other way of finding its area?

**Solution:** Jyotiâ€™s diagram:

The given shape is splitted into two congruent trapeziums.

Area of one trapezium

âˆ´ Area of the pentagonal shape =

Kavitaâ€™s diagram:

The given shape is splitted into a square and a triangle.

Area of the square = 15 m * 15 m = 225 m^{2}

Area of the triangle = 1/2 * 15 * 15 =225/2 m^{2} = 112.5 m^{2}

âˆ´ Area of the pentagonal shape = 112.5 + 225 = 337.5 m^{2}

Another way of finding the area of the pentagonal shape:

By splitting the pentagonal shape into 3 triangles, we have:

Area of triangle I

Area of triangle II =

Area of triangle III

âˆ´ Area of the pentagonal shape =

**Question 11.** Diagram of the adjacent picture frame has outer dimensions = 24 cm Â¥ 28 cm and inner dimensions 16 cm Â¥ 20 cm. Find the area of each section of the frame, if the width of each section is same.

**Solution:** Area of trapezium I:

Parallel sides are 24 cm and 16 cm.

Area of trapezium II:

Parallel sides are 20 cm and 28 cm.

Area of trapezium III:

Area of trapezium III = Area of trapezium I = 80 cm^{2}

Area of trapezium IV:

Area of trapezium IV = Area of trapezium II = 96 cm^{2}

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