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**CBSE Class 8, Mathematics - Mensuration
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**Chapter 11, NCERT Mathematics **

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__NCERT solutions
of Mensuration Exercise 11.2__

__NCERT solutions of Mensuration Exercise 11.2__

Question 1:

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

__Solution__:

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__:

It is given that, area of trapezium = 34 cm

^{2}and height = 4 cm
Let the lengths of parallel sides are b

_{1}and b_{2}. We know that,
Area of trapezium = 1/2(b

_{1}+ b_{2}) × h
[h = distance between the parallel sides]

So, 34 = 1/2 (10 + b

_{2}) x 4
Or, 34 = (10 + b

_{2}) x 2
Or, 17 = 10 + b

_{2}
Or, b

_{2 }= 7 cm
Thus, the length of the other parallel side is 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 of trapezium ABCD = AB + BC + CD + DA

120
m = AB + 48 m + 17 m + 40 m

AB
= 120 m − 105 m = 15 m

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__:

Given that the length of the diagonal,

*d*= 24 m
Length of the perpendiculars,

*h*_{1}and*h*_{2}, from the opposite vertices to the diagonal are*h*

_{1}= 8 m and

*h*

_{2}= 13 m

Question 5:

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

__Solution__:

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__:

Let the length of the other diagonal of the rhombus be

*x*.
Area of the given rhombus = Base × Height = 6 cm × 4 cm = 24 cm

^{2}
Also, area of rhombus =

**1/2(Product of its diagonals)**

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__:

Area of rhombus = 1/2(Product
of its diagonals)

Hence, area of each tile

= 675 cm

^{2}
Area of 3000 tiles = (675 × 3000) cm

^{2}= 2025000 cm^{2}= 202.5 m^{2}
Cost of polishing 202.5 m

^{2}area @ Rs 4 per m^{2}
= Rs (4 × 202.5) = Rs 810

Thus, the cost of polishing the floor is 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 the road. It 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 us assume the length of the field along the road is

*l*m. So according to the given condition, the length of the field along the river will be 2*l*m.
Area of trapezium = 1/2(Sum
of parallel sides) (Distance between the parallel sides)

Question 9:

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

__Solution__:

Since
the shape of the platform is a regular octagon (each side = 5 cm),

So,
area of trapezium ABCH = area of trapezium DEFG

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__:

__Jyoti’s diagram__:

__Kavita’s diagram__:

Question 11:

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

__Solution__:

Area
of outer frame in the picture = length x breadth

= 28 x
24 = 672 cm

^{2}
Area
of inner frame in the picture =

*l*x*b*
= 16 x
20 = 320 cm

^{2}
Hence,
area of remaining portion = 672 - 320 = 352 cm

^{2}
So,
area of each section = 352 ÷ 4 = 88 cm

^{2}
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