# ICSE Class 8 Maths Selina Solutions for Chapter 20 Area of Trapezium and Polygon

### CHAPTER 20 – AREA OF TRAPEZIUM AND POLYGON

Question 1.

Find the area of a triangle, whose sides are:

(i) 10cm, 24cm and 26cm

Solution:-

Sides of Δ are

a=10cm

b=24cm

c=26cm

$\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}=\frac{10+24+26}{2}$ (Simplifying we get)

$=\frac{60}{2}=30$

Area of $\Delta=\sqrt{S(S-a)(S-b)(S-c)}$ (Formula)

$=\sqrt{30(30-10)(30-24)(30-26)}$ $=\sqrt{30 \times 20 \times 6 \times 4}$ $=\sqrt{10 \times 3 \times 10 \times 2 \times 2 \times 3 \times 2 \times 2}$ $=\sqrt{10 \times 10 \times 3 \times 3 \times 2 \times 2 \times 2 \times 2}$ $\mathrm{Ans}=10 \times 3 \times 2 \times 2=120 \mathrm{cm}^{2}$

(ii) 18mm, 24 mm and 30mm

Solution:-

Sides of Δ are

a=18mm

b=24mm

C=30mm

$\mathrm{S}=\frac{a+b+c}{2}=\frac{18+24+30}{2}$ $=\frac{72}{2}=36$

Area of $\Delta=\sqrt{S(S-a)(S-b)(S-c)}$ (Formula)

$=\sqrt{36(36-18)(36-24)(36-30)}$ $=\sqrt{36 \times 18 \times 12 \times 6}$ $=\sqrt{18 \times 2 \times 18 \times 2 \times 6 \times 6}$ $=\sqrt{18 \times 18 \times 2 \times 2 \times 6 \times 6}$ $\mathrm{Ans}=18 \times 2 \times 6=216 \mathrm{mm}^{2}$

(iii) 21 m, 28 m and 35 m

Solution:-

Sides of Δ are

a=21m

b=28m

c=35m

$S=\frac{a+b+c}{2}=\frac{21+28+35}{2}$ (Simplifying we get)

$=\frac{84}{2}=42$

Area of $\Delta=\sqrt{S(S-a)(S-b)(S-c)}$ (Formula)

$=\sqrt{42(42-21)(42-28)(42-35)}$ $=\sqrt{42 \times 21 \times 14 \times 7}$ $=\sqrt{7 \times 3 \times 2 \times 3 \times 7 \times 2 \times 7 \times 7}$ $=\sqrt{7 \times 7 \times 7 \times 7 \times 3 \times 3 \times 2 \times 2}$ $=7 \times 7 \times 3 \times 2$ $\mathrm{Ans}=294 \mathrm{m}^{2}$

Question 2.

Two sides of a triangle are 6 cm and 8 cm. If height of the triangle corresponding to 6 cm side is 4 cm; find.

(i) Area of the triangle

(ii) Height of the triangle corresponding to 8 cm side.

Solution:-

BC=6cm

Area of $\Delta=\frac{1}{2} \text { base } \times height$ $=\frac{1}{2} \times B C \times A D$ $=\frac{1}{2} \times 6 \times 4=12 \mathrm{cm}^{2}$

Area of $\Delta=\frac{1}{2} A C \times B E$ $12=\frac{1}{2} \times 8 \times B E$ $∴ \quad B E=\frac{12 \times 2}{8}$

BE=3cm

(i) $12 \mathrm{cm}^{2}$ (ii)3cm

Question 3.

The sides of a triangle are 16cm, 12cm and 20cm. Find:

(i) Area of the triangle;

(ii) Height of the triangle, corresponding to the largest side;

(iii) Height of the triangle, corresponding to the smallest side.

Solution:-

Sides of Δ are

a=20cm

b=12cm

c=16cm

$S=\frac{a+b+c}{2}$ $=\frac{20+12+16}{2}$ $=\frac{48}{2}=24$

Area of $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{24(24-20)(24-12)(24-16)}$ $=\sqrt{24 \times 4 \times 12 \times 8}$ $=\sqrt{12 \times 12 \times 4 \times 12 \times 2 \times 4}$ $=\sqrt{12 \times 12 \times 4 \times 4 \times 2 \times 2}$ $=12 \times 4 \times 2=96 \mathrm{cm}^{2}$

AD is height of Δ corresponding to largest side.

$∴ \quad \frac{1}{2} \times B C \times A D=96$ $\frac{1}{2} \times 20 \times \mathrm{AD}=96$ $\mathrm{AD}=\frac{96 \times 2}{20}$

BE is height of Δ corresponding to smallest side.

$∴ \frac{1}{2} A C \times B E=96$ $\frac{1}{2} \times 12 \times B E=96$ $B E=\frac{96 \times 2}{12}$

BE=16cm

(i) $96 \mathrm{cm}^{2}$ (ii)9.6cm(iii)16cm

Question 4.

Two sides of a triangle are 6.4 m and 4.8 m. If height of the triangle corresponding to 4.8m side is 6m; find:

(i) Area of the triangle;

(ii) height of the triangle corresponding to 6.4 m side.

Solution:-

ABC is the Δ in which BC=4.8m

$∴ \text { Area of } \Delta \mathrm{ABC}=\frac{1}{2} \mathrm{BC} \times \mathrm{AD}$ $=\frac{1}{2} \times 4.8 \times 6$ $=14.4 \mathrm{m}^{2}$

BE is height of Δ corresponding to 6.4 m

$\frac{1}{2} A C \times B E=14.4$ $\frac{1}{2} \times 6.4 \times B E=14.4$ $B E=\frac{14 \cdot 4 \times 2}{64}$ $B E=\frac{14-4}{3 \cdot 2}$ $=\frac{9}{2}=4.5 \mathrm{m}$

Therefore, (i) $.4 m^{2}$ (ii) 4.5 m

Question 5.

The base and the height of a triangle are in the ratio 4:5. If the area of the triangle is 40 $m^{2};$ find its base  and height.

Solution:-

Consider base of $\Delta=4 \times \mathrm{m} and height of \Delta=5 \times \mathrm{m}$

Area of $\Delta=40 \mathrm{m}^{2}$ $∵ \frac{1}{2} \text { Base } \times \text { height }=\text { area of } \Delta$ $\frac{1}{2} \times 44 x \times 5 x=40$ $10 x^{2}=40$ $x^{2}=4$ $x=\sqrt{4}$

x=2

$∴ \quad \text { Base }=4 x=4 \times 2=8 \mathrm{m}$

Height $=5 x=5 \times 2=10 \mathrm{m}$

Therefore, the base and height of the triangle is 8m; 10m.

Question 6.

The base and the height of a triangle are in the ratio 5:3. If the area of the triangle is $67 m^{2}$.find its base and height.

Solution:-

Consider base $=5 \times \mathrm{m}$

Height $=3 \times \mathrm{m}$

Area of $\Delta=\frac{1}{2} \text { base } \times height$ $∴ \quad \frac{1}{2} \times 5 x \times 3 x=67.5$ $x^{2}=\frac{67.5 \times 2}{15}$ $x^{2}=4.5 \times 2$ $x^{2}=9.0$ $x=\sqrt{9}$

x=3

Base $=5 x=5 \times 3=15 \mathrm{m}$

Height $=3 \mathrm{x}=3 \times 3=9 \mathrm{m}$

Question 7.

The area of an equilateral triangle is $144 \sqrt{3} \mathrm{cm}^{2};$ find its perimeter.

Solution:

Consider each side of an equilateral triangle = x cm

Area $=\frac{\sqrt{3}}{4}(\text {side})^{2}$ $=\frac{\sqrt{3}}{4} x^{2}=144 \sqrt{3}$ $\Rightarrow x^{2}=144 \sqrt{3} \times \frac{4}{\sqrt{3}}$ $\Rightarrow x^{2}=144 \times 4 \Rightarrow x^{2}=576$ $\Rightarrow x=\sqrt{576}=24 \mathrm{cm}$

Each side =24cm

Therefore, perimeter =3(24)=72cm

Question 8.

The area of an equilateral triangle is numerically equal to its perimeter. Find its perimeter correct to 2 decimal places.

Solution:-

Consider each side of the equilateral triangle = x

Area $=\frac{\sqrt{3}}{4} x^{2}$

Area perimeter =3x

By the given condition $=\frac{\sqrt{3}}{4} x^{2}=3 x$ $x^{2}=3 x \times \frac{4}{\sqrt{3}}$ $x^{1}=\frac{3 x \times 4 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{3 x \times 4 \times \sqrt{3}}{3}=4 x \sqrt{3}$ $\Rightarrow x^{2}=\sqrt{3}(4 x) \Rightarrow x=4 \sqrt{3}[∵ x \neq 0]$

Perimeter $=12 \sqrt{3} units$

=12(1.732) =20.784=20.78 Units

Question 9.

A field is in the shape of a quadrilateral ABCD in which side AB = 18m, side AD = 24m, side BC = 40m, DC = 50m and angle A = 90°. Find the area of the field.

Solution:-

∠A=90°

By Pythagoras Theorem,

In ∆ABD,

$\mathrm{BD}=\sqrt{\mathrm{AB}^{2}+\mathrm{AD}^{2}}=\sqrt{18^{2}+24^{2}}$ $=\sqrt{324+576}=\sqrt{900}=30 \mathrm{m}$

Area of $\Delta \mathrm{ABC}=\frac{1}{2}(18)(24)$ $= (18)(12)=216 \mathrm{m}^{2}$

In ΔBCD; sides are 30, 40, 50

By Pythagoras Theorem$\angle C B D=90^{°e}$ $\left[∴, \mathrm{DC}^{2}=\mathrm{BD}^{2}+\mathrm{BC}^{2}, \text { since }(50)^{2}=(30)^{2}+(40)^{2}\right]$ $\text { Area of } \Delta \mathrm{BCD}=\frac{1}{2}(40)(30)=600 \mathrm{m}^{2}$

Therefore, area of quadrilateral ABCD= Area of  ∆ABD+ area of ΔBCD

=216+600=816 $\mathrm{m}^{2}$

Question 10.

The lengths of the sides of a triangle are in the ratio 4 : 5 : 3 and its perimeter is 96  cm . Find its area.

Solution:-

Consider the sides of the triangle ABC be 4 x, 5 x and 3x

AB=4 x, AC=5 x and BC=3 x

Perimeter = 4 x+5x+3x=96

12 x=96

$x=\frac{96}{12}$

x=8

Sides are

BC =3(8) =24 cm AB=4(8) =32 cm,

AC = 5(8) = 40 cm

$(A C)^{2}=(A B)^{2}+(B C)^{2}$ $\left[(5 x)^{2}=(3 x)^{2}+(4 x)^{2}\right]$

By Pythagoras Theorem, ∠B=90°

$\text { Area of } \Delta$ $\mathrm{ABC}=\frac{1}{2}(\mathrm{BC})(\mathrm{AB})=\frac{1}{2}(24)(32)$ $=12 \times 32=384 \mathrm{cm}^{2}$

Question 11.

One of the equal sides of an isosceles triangle is 13cm and its perimeter is 50cm. Find the area of the triangle.

Solution:-

In Isosceles ΔABC

AB=AC=13cm But perimeter =50cm

BC=50-(13+13)cm

=50-26=24cm

$\mathrm{AD}=\mathrm{DC}=\frac{24}{2}=12 \mathrm{cm}$

In right ΔABD,

$\mathrm{AB}^{2}=\mathrm{AD}^{2}+\mathrm{BD}^{2}$ (Pythagoras Theorem)

$(13)^{2}=\mathrm{AD}^{2}+(12)^{2}$ $\Rightarrow 169=A D^{2}+144$ $\Rightarrow A D^{2}=169-144$

=25=(5)^{2}

Area of $\Delta A B C=\frac{1}{2} \text { Base } \times Altitude$ $=\frac{1}{2} \times B C \times A D$ $=\frac{1}{2} \times 24 \times 5=60 \mathrm{cm}^{2}$

Question 12.

The altitude and the base of a triangular field are in the ratio 6:5. If its cost is Rs.49, 57,200 at the rate of Rs.36, 720 per hectare and 1 hectare =10,000 sq. m, find (in meter) dimensions of the field,

Solution:-

Total cost =349,57,200

Rate = 736,720 per hectare

Total area of the triangular field

$=\frac{4957200}{36720} \times 10000 \mathrm{m}^{2}=1350000 \mathrm{m}^{2}$

Ratio in altitude and base of the field =6:5

Consider altitude =6x and base =5x

$\text { Area }=\frac{1}{2} \text { Base } \times Altitude$ $\Rightarrow 1350000=\frac{1}{2} \times 5 \mathrm{x} \times 6 \mathrm{x}$ $\Rightarrow 15 \mathrm{x}^{2}=1350000 \Rightarrow \mathrm{x}^{2}=\frac{1350000}{15}$ $\Rightarrow x^{2}=90000=(300)^{2}$

x=300

Base $=5 \mathrm{x}=5 \times 300=1500 \mathrm{m}$

Altitude $=6 \mathrm{x}=6 \times 300=1800 \mathrm{m}$

Question 13.

Find the area of the right-angled triangle with hypotenuse 40cm and one of the other two sides 24cm.

Solution:-

In right angled triangle ABC Hypotenuse AC =40cm

One side AB=24cm

$\mathrm{BC}=\sqrt{\mathrm{AC}^{2}-\mathrm{AB}^{2}}$ $=\sqrt{40^{2}-24^{2}}=\sqrt{1600-576}$ $=\sqrt{1024}=32 \mathrm{cm}$

Area

$=\frac{1}{2} \mathrm{AB} \times \mathrm{BC}=\frac{1}{2} \times 24 \times 32$ $\mathrm{cm}^{2}=384 \mathrm{cm}^{2}$

Question 14.

Use the information given in the adjoining figure to find:

(i) The length of AC.

(ii) The area of a ΔABC

(iii) The length of BD, correct to one decimal place.

Solution:-

Sol. AB=24cm, BC=7cm

(i) $\mathrm{AC}=\sqrt{\mathrm{AB}^{2}+\mathrm{BC}^{2}}$

$=\sqrt{24^{2}+7^{2}}$ $=\sqrt{576+49}$ $=\sqrt{625}=25 \mathrm{cm}$

(ii) Area of $\triangle \mathrm{ABC}=\frac{1}{2} \mathrm{AB} \times \mathrm{BC}$ $=\frac{1}{2} \times 24 \times 7=84 \mathrm{cm}^{2}$

(iii) BD⊥AC

Area $\Delta \mathrm{ABC}=\frac{1}{2} \mathrm{AC} \times \mathrm{BD}$ $84=\frac{1}{2} \times 25 \times \mathrm{BD}$ $\Rightarrow \mathrm{BD}=\frac{84 \times 2}{25}=\frac{168}{25}=6.72 \mathrm{cm}$

=6.7cm

Question 15.

Find the length and perimeter of a rectangle, whose area $=120 \mathrm{cm}^{2}$ and breadth =8cm

Solution:-

Area of rectangle $=120 \mathrm{cm}^{2}$

Area $=l \times b$ $l \times 8=120$ $l=\frac{120}{8}=15 \mathrm{cm}$
Perimeter $=2(1+b)=2(15+8)=2 \times 23=46 \mathrm{cm}$