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

 EXERCISE 25D

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

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Question 1

Find the area of the following trapezium whose dimensions are:

(i) Height=10cm

Parallel sides=6cm and 8cm

Sol :

Area of trapezium$=\frac{1}{2}\times (\text{sum of parallel sides})\times (\text{height})$

$=\frac{1}{2}\times (6+8)\times 10$

$=\frac{14}{2}\times 10$

=7×10

=70 cm²

(ii) Height=4cm
Parallel sides=6.4cm and 4.6cm

Sol :

$=\frac{1}{2}\times (6.4+4.6)\times 4$

$=\frac{11}{2}\times 4=\frac{44}{2}$

=22 cm²

(iii) Height=8.4cm
Parallel sides=15cm and 9.2cm

Sol :

$=\frac{1}{2}\times (15+9.2)\times 8.4$

$=\frac{24.2}{2}\times 8.4$

=12.1×8.4

=101.64 cm²

(iv) Height=7cm
Parallel sides=12cm and 16cm

Sol :

$=\frac{1}{2}\times (12+16)\times 7$

$=\frac{28}{2}\times 7$

=14×7

=98 cm²

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

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Question 2

Find the height of a trapezium. whose sum of the lengths of the bases is 31 cm and whose area is 124 cm²

Sol :

Area = 124 cm²

Sum of parallel sides = 31cm

Area of trapezium$=\frac{1}{2}\times (\text{sum of parallel sides})\times (\text{height})$

$124 =\frac{1}{2}\times 31 \times \text{height}$

Height$=\frac{124\times 2}{31}$

Height = 8 cm

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

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Question 3

The ratio of the length of the parallel sides of a trapezium is 3:2 . The shortest distance between them is 15cm . If the area of the trapezium is 450cm² , what is the sum of the lengths of the parallel sides ?

Sol :

Height is the shortest distance between the parallel sides. So, Height=15cm

Parallel sides =3x and 2x

Area of trapezium=450cm²

$\frac{1}{2}\times (3x+2x)\times 15=450$

$\frac{5x}{2}\times 15=450$

$\frac{5x}{2}=\frac{450}{15}$

5x=30×2

5x=60cm

Sum of the lengths of the parallel sides

=5x=60cm

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

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Question 4

The difference between two parallel sides of a trapezium is 8cm . The perpendicular distance between them is 24m . Find the lengths of the two parallel sides , if the area of the trapezium is 312m²

Sol :

Given: (x-y)=8m..(i)

Height=24m

Area of trapezium is 312m²

$\frac{1}{2}\times (x+y)\times 24=312$

$(x+y)=\frac{312}{12}$

(x+y)=26m..(ii)

On adding (i) and (ii), we get

(x+y)+(x-y)=(26+8)m

x+y+x-y=34

2x=34

$x=\frac{34}{2}$m

x=17m

putting x in (i), we get

(x-y)=8m

17-y=8

17-8=y

y=9m

Lengths of two parallel sides=17m and 9m

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

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Question 5

In the given diagram, CDE is an isosceles triangle with an area of 24cm². If AB=8cm and AD=12cm , calculate the area of the trapezium ABED

Sol :

Area of rectangle ABFD= 12×8=96cm²

In isosceles triangle the altitude divides the triangle into 2 equal parts so area of triangle DFE= 24÷2=12cm²

Area of ABED=Area of ABFE-Area of DFE

=96-12 

=84cm²

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

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Question 6

The following figure shows a parallelogram and a trapezium whose areas are equal. Find the value of 'a'.

Sol :

Area of parallelogram=b×h

=8.8×3cm²

=26.4cm²

Area of parallelogram=Area of trapezium

Area of trapezium=26.4cm²

$\frac{1}{2}\times (\text{sum of parallel sides})\times (\text{height})=26.4$

$\frac{1}{2}\times (a+8.4)\times 4=26.4$

(a+8.4)2=26.4

$(a+8.4)=\frac{26.4}{2}$

(a+8.4)=13.2

a=13.2-8.4

a=4.8cm

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

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Question 7

Two parallel sides of an isosceles trapezium are 16cm and 24cm respectively. If the lengths of each non-parallel side is 5cm, find the area of trapezium.

Sol :

In triangle AED

$AE=\frac{24-16}{2}$

AE=4cm

Using Pythagoras theorem

DA²=DE²+AE²

DE²=DA²-AE²

DE=√(5)²-(4)²

DE=√25-16

DE=√9

DE=3cm=Height

Area of trapezium$=\frac{1}{2}\times (\text{sum of parallel sides})\times (\text{height})$

$=\frac{1}{2}\times (16+24)\times 3$

$=\frac{40}{2}\times 3$

=20×3

=60cm²

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

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Question 8

A grassy area is in the shape of a trapezoid whose bases measures 150m and 100m and the height is 52m . A 3m walkway is constructed inside this grassy area which runs perpendicular from the two bases. Calculate the area of the grassy patches of land after the construction of this walkway.

Sol :

Area of Trapezium= 1/2(sum of parallel sides)× hieght

Area of Trapezium = 1/2(150+100)×52

= 1/2×250×52

=6500 m²

Area of rectangle= l×b

L=52

B=3

Area = 52×3

=156m²

Area of grassy patches = area of Trapezium- area of rectangle

= 6500-156

=6344 m²