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:-

Class 8 Maths Selina Solutions for Chapter 20 Area of Trapezium and Polygon-1

BC=6cm

Height AD=4cm

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.

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

AD=9.6cm

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:-

Class 8 Maths Selina Solutions for Chapter 20 Area of Trapezium and Polygon-3

ABC is the Δ in which BC=4.8m

AC=6.4m and AD=6m

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

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,

Class 8 Maths Selina Solutions for Chapter 20 Area of Trapezium and Polygon-4

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

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

\(x=\frac{96}{12}\)

Class 8 Maths Selina Solutions for Chapter 20 Area of Trapezium and Polygon-5

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

Class 8 Maths Selina Solutions for Chapter 20 Area of Trapezium and Polygon-6

BC=50-(13+13)cm

=50-26=24cm

AD⊥BC

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

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}

AD=5cm.

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

Class 8 Maths Selina Solutions for Chapter 20 Area of Trapezium and Polygon-7

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

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.

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

Solution:-

Sol. AB=24cm, BC=7cm

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

Class 8 Maths Selina Solutions for Chapter 20 Area of Trapezium and Polygon-9

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

(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 \(=120 \mathrm{cm}^{2}\)

Breadth, b=8cm

Area \(=l \times b\) \(l \times 8=120\) \(l=\frac{120}{8}=15 \mathrm{cm}\)

Perimeter \(=2(1+b)=2(15+8)=2 \times 23=46 \mathrm{cm}\)

Length =15cm

Perimeter =46cm

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