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

 EXERCISE 25 B

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Q1 | Ex-25B | 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 triangles:

Sol :

(i)

$=\frac{1}{2} \times b \times h$

$=\frac{1}{2} \times 5 \times 12$

=5×6

=30 cm ²

(ii)

$=\frac{1}{2} \times b \times h$

$=\frac{1}{2} \times 13 \times 5$

=32.5 cm ²

(iii)

$=\frac{1}{2} \times b \times h$

$=\frac{1}{2} \times 8 \times 5$

=5×4

=20 cm ²

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

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Question 2

The base of a triangle measures 32 cm and its altitude is 45 cm. Find the area of the triangle in square metres.

Sol :

Area of triangle$=\frac{1}{2} \times 32 \times 45$

=720 cm ²

1 m²=10000 cm²

$\begin{aligned} \frac{1}{10000} \mathrm{~m}^{2} &=1 \mathrm{~cm}^{2} \\ \frac{720}{10000} m^2 &=720 \mathrm{cm}^{2} \end{aligned}$

=0.0720 m²

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

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Question 3

Find the height of a triangle, if its area is 24 cm² and its base is 15 cm.

Sol :

Area of triangle=24 cm²

$\frac{1}{2} \times b \times h=24$

$\frac{1}{2} \times 15 \times h=24$

$\begin{aligned} n &=\frac{24 \times 2}{15} \\ &=\frac{48}{18} \end{aligned}$

h=3.2 cm

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

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Question 4

The area of a right-angled triangle is 60 cm² and its smallest side is 8cm. What is its hypotenuse ?

Sol :

Area of triangle=60 cm²

$\frac{1}{2} \times 8 \times h=60$

$h=\frac{60}{4}$

h=15 cm

$x^{2}=15^{2}+8^{2}$

$x=\sqrt{225+64}$

$=\sqrt{289}$

x=17 cm

Hypotenuse=17 cm

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

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Question 5

Calculate the shaded area of each of the following

Sol :

(i)

Area of rectangle=l×b

=48×30=1440 cm²

Area of white triangle$=\frac{1}{2} \times 20 \times(48-16)$

=10×32=320 cm²

and Also smaller triangle$=\frac{1}{2} \times 16 \times 10$

=8×10=80 cm²

Area of shaded region=Area of rectangle-(Area of unshaded region)

=1440-(320+80)

=1440-400=1040 cm²

(ii)

Area of rectangle=l×b

=14×11=154 cm²

^{Area of unshaded region=}

$=\frac{1}{2} \times 11 \times 5+\frac{1}{2} \times 11 \times 4$

$=\frac{55}{2}+11 \times 2$

$=\frac{55}{2}+22$

$=\frac{55+44}{2}=\frac{99}{2}$

=49.5 cm²

Area of shaded region=154-49.5

=104.5 cm²

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

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Question 6

Find the area of an isosceles right triangle, if one of its right sides is 18 cm long.

Sol :

Area of isosceles Δ$=\frac{1}{2} \times 18 \times 18$ (isosceles triangle have two sides of equal length)

=18×9

=162 cm²

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

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Question 7

The area of an isosceles triangle is 300cm². The height of the triangle is 15cm. Find the perimeter of the triangle.

Sol :

Area of isosceles triangle=300cm²

$\frac{1}{2}\times b\times h=300$

$b\times 15=300\times 2$

$b=\frac{300\times 2}{15}$

b=40cm

Using Pythagoras theorem,

H²=15²+20²

H²=225+400

H²=625

H=25cm

Perimeter of triangle=25+25+40

=90cm

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

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Question 8

The area of a triangle is 1176cm². The ratio of the base to the corresponding altitude is 3 : 4. Find the altitude of the triangle.

Sol :

Base=3x , Height=4x where x is common factor

Area of triangle=1176

$\frac{1}{2}\times b\times h=1176$

$\frac{1}{2}\times 3x\times 4x=1176$

$\frac{12x^2}{2}=1176$

6x²=1176

$x=\frac{1176}{6}$

x=√196cm

x=14cm

Height=4x=4×14=56cm

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

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Question 9

Find the area of quadrilateral ABCD, where BD = 22 cm, CE = 7 cm and AF = 6 cm.

Sol :

Area of quadrilateral =Area of ΔDBC+Area of ΔDBA

$=\frac{1}{2}\times 22\times 7+\frac{1}{2}\times 22\times 6$

=11×7+11×6

=77+66

=143cm²

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

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Question 10

Find the area of the triangle in each of the following where the three sides are given as
(i) 26 cm, 28 cm, 30 cm (ii) 48 cm, 73 cm, 55 cm

Sol :

(i)

$s=\frac{26+28+30}{2}=\frac{84}{2}$

s=42

Area of triangle$=\sqrt{s(s-a)(s-b)(s-c)}$

 =√42(42-26)(42-28)(42-30)

=√42×16×14×12

 =√112896

=336cm²

(ii)

$s=\frac{48+73+55}{2}=\frac{176}{2}$

s=88

Area of triangle$=\sqrt{s(s-a)(s-b)(s-c)}$

 =√88(88-48)(88-73)(88-55)

=√88×40×15×33

 =√1742400

=1320cm²

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

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Question 11

Find the area of an equilateral triangle whose each side is 12 cm.

Sol :

Area of equilateral triangle$=\frac{\sqrt{3}}{4}a^2$

$=\frac{\sqrt{3}}{4}12^2$

$=\frac{\sqrt{3}}{4}12\times 12$

$=\sqrt{3}\times 3\times 12$

=36√3 cm²