S.chand books class 8 maths solution chapter 25 Areas of Rectilinear Figures exercise 25 C

 EXERCISE 25 C

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Q1 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper

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Question 1

Find the area of each of the following parallelogram whose base=340mm , height=17cm.

Sol :

Area of parallelogram=base×height

=340mm×17cm

[1cm=10mm]

=34cm×17cm

=578cm²

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Q2 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper

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Question 2

The adjacent sides of a parallelogram are 8m and 5m .The distance between the longer sides is 4m. What is the distance between the shorter sides ?

Sol :

Left side shows distance between the longer sides(base) is 4m , Right side shows distance between the shorter sides(base) is h

Area of parallelogram with longer side as base=8×4

=32m²

Area of parallelogram with shorter side as base=32m²

b×h=32

5×h=32

$h=\frac{32}{5}$

h=6.4m

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Q3 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper

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Question 3

The area of a parallelogram is 98cm² and its altitude is twice the corresponding base. Find the altitude of the parallelogram.

Sol :

H=2B

Area of parallelogram=98cm²

B×H=98cm²

B×2B=98cm²

2B²=98cm²

$B^2=\frac{98}{2}$cm

B=√49 cm

B=7 cm

H=2×7=14cm

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Q4 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper

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Question 4

A parallelogram has a length of 15cm and a height of 20cm. It is divided into two congruent triangles. What is the area of each triangle ?

Sol :

Area of triangle$=\frac{1}{2}\times b\times h$

$=\frac{1}{2}\times 15\times 20$

=15×10cm²

=150cm²

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Q5 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper

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Question 5

The two adjacent sides of a parallelogram are 60m and 40m and one of the diagonals is 80m long. What is the area of the parallelogram ?

Sol :

Suppose we divide parallelogram in to two equal triangle

Area of triangle

$s=\frac{a+b+c}{2}$

$=\frac{60+40+80}{2}$

s=90m

Area=√s(s-a)(s-b)(s-c)

=√90(90-60)(90-40)(90-80)

=√90(30)(50)(10)

=√9×3×5×1×10000

=100√3×3×3×5

=300√3×5

=300√15

Area of parallelogram equal to two equal triangle=2×300√15

=600√15 cm²

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Q6 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper

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Question 6

The area of a parallelogram and square are the same. The perimeter of the square is 84cm and the base of the parallelogram is 7cm, find the height of the parallelogram.

Sol :

ATQ,

Area of the parallelogram = area of the square

Given perimeter of the square = 84cm

We know that,

Side × 4 = perimeter of a square

➡ side × 4 = 84cm

➡ side = $\frac{84}{4}$

➡ side = 21cm

Therefore it's area = side²

= 21 × 21

= 441cm²

Since area of the parallelogram = area of the square.

i.e Area of the parallelogram = 441cm²

Given it's base = 7cm

We know that,

Base × Height = area of a parallelogram

➡ 7 × Height = 441cm²

➡ Height $=\frac{441}{7}$

➡ Height = 63cm

Hence, the height of the parallelogram is 63cm.

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Q7 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper

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Question 7

Find the area of the rhombus whose base (b)=2.5cm and height (h)=13.2cm

Sol :

Area of the rhombus=base×height

=2.5×13.2 cm²

=33 cm²

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Q8 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper

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Question 8

Find the area of rhombus ABCD in each case.
(i) AC=8cm , BD=10cm
(ii) AC=15cm , BD=36cm

Sol :

(i)

Area of rhombus $=\frac{1}{2}\times (\text{first diagonal}) \times (\text{second diagonal})$

$=\frac{1}{2}\times 8\times 10$

=4×10 cm²

=40 cm²

(ii)

Area of rhombus $=\frac{1}{2}\times (\text{first diagonal}) \times (\text{second diagonal})$

$=\frac{1}{2}\times 15 \times 36$

=15×18 cm²

=270 cm²

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Q9 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper

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Question 9

The area of a rhombus is 72cm² . If one of its diagonals is 18cm, what is the length of the other diagonal ?

Sol :

Area of rhombus=72cm²

$\frac{1}{2}\times (\text{first diagonal})\times (\text{second diagonal})=72$

$\frac{1}{2}\times 18\times (\text{second diagonal})=72$

$9 \times  (\text{second diagonal})=72$

Second diagonal$=\frac{72}{9}$

=8cm

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Q10 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper

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Question 10

The area of a rhombus is 2016 cm² and its sides are each equal to 63 cm. Find its height,

Sol :

Area of the rhombus=base×height

2016=63×height

$height=\frac{2016}{63}$

Height=32 cm

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Q11 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper

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Question 11

The perimeter of a rhombus is 40cm and the length of one of its diagonals is 16cm. Find its area.

Sol :

Perimeter of a rhombus =40cm

4×side=40cm

$\text{side}=\frac{40}{4}$cm

Side=10cm

Diagonals of rhombus bisect each other at right angles.

By Pythagoras theorem

OB=√AB²-OA²

=√(10)²-(8)²

=√100-64

=√36

=6

OB=OD=6cm

diagonal BD=6+6=12cm

Area of rhombus$=\frac{1}{2}\times (\text{first diagonal})\times (\text{second diagonal})$

$=\frac{1}{2}\times 16\times 12$

=8×12 

=96cm²

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Q12 | Ex-25C | Class 8 | S.Chand | Composite maths| Areas of Rectilinear Figures |Ch-25 | myhelper

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Question 12

Two fields in the shape of parallelogram and rhombus are equal in area. The diagonals of rhombus are 110m and 44m , respectively. If one of the sides of the parallelogram is 88m, find its altitude.

Sol :

ATQ, 

Area of rhombus = Area of parallelogram

Given diagonals of the rhombus are 110m and 44m.

Area of the rhombus $=\frac{1}{2}\times (\text{product of diagonals})$

$=\frac{1}{2}\times 110\times 44$

= 55 × 44

= 2420m²

Therefore area of parallelogram is also 2420m²

Base(side) of the parallelogram is given 88m.

Therefore b × h = 2420m²

⇒88 × h = 2420m²

⇒h $= \frac{2420}{88}$

⇒h = 27.5m

Hence, altitude of the parallelogram is 27.5m