SChand CLASS 9 Chapter 16 Area of plane figures Exercise 16(B)

  Exercise 16 B

Question-1

For any triangle, complete the following table

$\begin{array}{|l|l|l|l|l|l|l|l|}\hline & \text { (i) } & \text { (ii) } & \text { (iii) } & \text { (iv) } & \text { (v) } & \text { (vi) } & \text{ (vii) } \\\hline \text { Base } & 6 & 8 & 5 & x & 6 & ? & ? \\\hline \text { Height } & 10 & 14 & 20 & 2 x & ? & 5 & 11 \\\hline \text { area } & ? & ? & ? & ? & 30 & 50 & 110 \\\hline\end{array}$

Answer:

Question-2

The area of a triangle is $6 \mathrm{~cm}^{2}$ and its base is $4 \mathrm{~cm}$. Find its height.

Answer-2

Area $=6 \mathrm{~cm}^{2}$

Base $=4 \mathrm{~cm}$

$6=1 / 2 \times 4 \times$ height

Or, height $=3 \mathrm{~cm}$.

Question-3

Find the base of a triangle is its

(i) area is 25 ares and height $20 \mathrm{~m}$.

(ii) area is 16 hectares and height 40 decametres

1 are $=100 \mathrm{~m}^{2}$

1 hectare $=10000 \mathrm{~m}^{2}$

1 decameter $=10 \mathrm{~m}$

Answer-3 (i) $\frac{1}{2} \times$ base $\times 20=25 \times 100$

base = 250m

(ii)$\frac{1}{2} \times \operatorname{base} \times 400=$ $16\times 10000$

base =800

Question-4

Find the area of the triangle whose sides are

(i) $26 \mathrm{~cm}, 28 \mathrm{~cm}, 30 \mathrm{~cm}$

(ii) $48 \mathrm{~cm}, 73 \mathrm{~cm}, 55 \mathrm{~cm}$

(iii) $21 \mathrm{~cm}, 20 \mathrm{~cm}, 13 \mathrm{~cm}$

(iv) $7.5 \mathrm{~cm}, 18 \mathrm{~cm}, 19.5 \mathrm{~cm}$

Answer-4

(i) $a=26 \mathrm{~cm}, b=28 \mathrm{~cm}, c=30 \mathrm{~cm}$

$S=(a+b+c) / 2$ $=(26+28+30) / 2$ $=84 / 2$ $=42$

Area $=\sqrt{42}(42-36)(42-28)(42-30)$

$=\sqrt{42} \times 16 \times 14 \times 12$

$=\sqrt{1} 12896$

$=\sqrt{336}$

(ii) $a=48 \mathrm{~cm}, b=73 \mathrm{~cm}, c=55 \mathrm{~cm}$

$S=(a+b+c) / 2$

$=(48+73+55) / 2$

$=176 / 2$

$=88$

$\therefore$ Area $=\sqrt{8}(88-48)(88-73)(88-55)$

$=\sqrt{8} 8 \times 40 \times 15 \times 33 \mathrm{~cm}^{2}$

$=\sqrt{1} 742400$

$=1320 \mathrm{~cm}^{2}$

(iii) $a=21 \mathrm{~cm}, b=20 \mathrm{~cm}, c=13 \mathrm{~cm}$

$S=(a+b+c) / 2$

$=(21+20+13) / 2$

$=54 / 2$

$=27$

$\therefore$ Area $=-\sqrt{27}(27-21)(27-20)(27-13)$

$=\sqrt{2} 7 \times 6 \times 7 \times 14$

$=\sqrt{1} 5876$

$=126$

(iv)$a=7.5 \mathrm{~cm}, b=18 \mathrm{~cm}, c=19.5 \mathrm{~cm}$

$S=(a+b+c) / 2$

$=(7.5+1.8+19.5) / 2$

$=45 / 2$

$=22.5$

$\therefore$ Area $=-\sqrt{2} 2.5(22.5-7.5)(22.5-18)(22.5-19.5)$

$=\sqrt{2} .5(15)(4.5)(3)$

$=\sqrt{4} 556.25$

$=67.5 \mathrm{~cm}^{2}$

Question-5

The perimeter of a triangle is $540 \mathrm{~m}$ and its sides are in the ration $25: 17: 12$. Find the area of the triangle.

Answer-5

Let, the sides of the triangular $=25 x, 17 x, 12 x$

Perimeter $=(25 x+17 x+12 x)$ $=54 x$

The perimeter of a triangle is $540 \mathrm{~m}$ given

$\therefore 54 x=540$

$x=10$

$\therefore$ Sides are $=250,170,120$

$a=250 \mathrm{~cm}, b=170 \mathrm{~cm}, c=120 \mathrm{~cm}$

$S=(a+b+c) / 2$

$=(250+170+120) / 2$

$=540 / 2$

$=270$

$\therefore$ Area $=-\sqrt{270}(270-250)(270-170)(270-120)$

$=\sqrt{270}(20)(100)(150)$

$=\sqrt{81000000}$

$=9000 \mathrm{~m}^{2}$

Question-6

The given figure, $A B C D$ represents a square of side $6 \mathrm{~cm} . F$ is a point on $D C$ such that the area of the triangle ADF is one third of the area of the square. Find the length of FD.

Answer-6

Area of square $=6 \times 6=36 \mathrm{~cm}^{2}$

Area of $\triangle \mathrm{ADF}=1 / 3 \times 36=12 \mathrm{~cm}^{2}$

$\therefore 1 / 2 \times \mathrm{DF} \times 6=12$

$\mathrm{DF}=4 \mathrm{~cm}$

Question-7

Find the area of a triangle with base $5 \mathrm{~cm}$ and whose height is equal to that of a rectangle with base $5 \mathrm{~cm}$ and area $20 \mathrm{~cm}^{2}$

Answer-7

Let, I be the length of the rectangle,

$\begin{aligned}&\therefore \mid \times 5=20 \\&\mid=4\end{aligned}$

$\therefore$ Area of triangle $=1 / 2 \times 5 \times 4$

$=10 \mathrm{~cm}^{2}$

Question-8

Answer true or false:

(i) The area of a triangle with base $4 \mathrm{~cm}$ and perp. Height $6 \mathrm{~cm}$ is $24 \mathrm{~cm}^{2}$

(ii) The area of a triangle with sides measuring, $a, b, c$, is given by $\sqrt{s}(s-a)(s-b)(s-c)$ where $s$ is the perimeter of the triangle.

(iii) The area of a rectangle and a triangle are equal if they stand on the same base and are between the same parallels.

Answer-8

(i) base $=4 \mathrm{~cm}$, height $=6 \mathrm{~cm}$

Area $=1 / 2 \times 4 \times 6$

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$=12 \mathrm{~cm}^{2}$

False

(ii) False

$\mathrm{S}$ is semi-perimeter not perimeter.

(iii) False,

Area of triangle $=1 / 2 \times$ Area of rectangle.

Question- 9

ABC is triangle in which $A B=A C=4 \mathrm{~cm}$ and $\angle A=90^{\circ}$, calculate the area of $\triangle A B C .$

Answer- 9

$\therefore$ Area of $\triangle A B C=1 / 2($ base $\times$ height $)$

$\therefore$ Area of $\triangle A B C=1 / 2(4 \times 4)$

$=8 \mathrm{~cm}^{2}$

Question- 10

the side of triangle field are $320 \mathrm{~m} 200 \mathrm{~m}$ and $180 \mathrm{~m}$ long....... in $\mathrm{m}^{2}$

Answer- 10

on reducing $50: 1$

side will be $=320 / 50 \mathrm{~m}=6.4$

$200 / 50 \mathrm{~m}=4$

and $180 / 50 \mathrm{~m}=3.6$

area $=1 / 2($ base $\times$ height $)$

area $=1 / 2(320 \times 102.5)$

$=16400 \mathrm{~m}^{2}$

Question-11

The sides of a triangular field are $975 \mathrm{~m}, 1050 \mathrm{~m}$ and $1125 \mathrm{~m} .$ If this field is sold at the rate

of Rs. 1000 per hectare, find its selling price.

Answer-11

Area of triangular field= $\sqrt{s}^{\star}(s-a)^{\star}(s-b)^{\star}(s-c)$. [heron's formula] $a, b, c$ are the sides of triangle.

s=975+1050+1125/2=1575

Area of triangular field= $\sqrt{1575^{*}}(1575-975)^{\star}(1575-1050)^{\star}(1575-1125)$

area $=\sqrt{1} 575^{\star} 600^525^{\star} 450$

area $=\sqrt{945000^{\star}} 236250$

area $=\sqrt{223256250000}$

area $=472500 \mathrm{~m}^{2}$

1 hectare $=10000 \mathrm{~m}^{2}$

Therefore, $472500 \mathrm{~m}^{2}=47.25$ hectares

Selling price $=47.25^{\star} 1000$

Selling price $=$ Rs. 47250

Question-12

Find the area of the equilateral triangle whose each side is (i) $12 \mathrm{~cm}$ (ii) $5 \mathrm{~cm}$

Answer-12

(i) $12 \mathrm{~cm}$

Sol: sides $=12 \mathrm{~cm}$

We know that

area of equilateral triangle

$\begin{aligned}&=\frac{\sqrt{3}}{4} a^{2} \\&=\frac{\sqrt{3}}{4} \times(12)^{2} \\&=\frac{\sqrt{3}}{4} \times 144 \\&=\sqrt{3} \times 36 \\&=36 \sqrt{3} \mathrm{~cm}^{2}\end{aligned}$

(ii) 5 cm

Sol: Side of equilateral triangle $=5 \mathrm{~cm}$

Area $=\sqrt{3} a^{2} / 4$

$=\sqrt{3}(5)^{2} / 4$

$=25 \times 1.73 / 4$

$=10.8 \mathrm{~cm}^{2}$

Question-13

The perimeter of an equilateral triangle is $24 \mathrm{~cm}$. Find its area.

Answer-13

Let ' $a$ ' be the side of the equilateral triangle,

$\begin{aligned}&3 a=24 \\&a=8\end{aligned}$

$\begin{aligned}&\text { Area }=\sqrt{3} / 4 \times 8^{2} \\&=\sqrt{3} / 4 \times 64 \\&=16 \sqrt{3} \\&=27.712 \mathrm{~cm}^{2}\end{aligned}$

Question-14

The perimeter of an equilateral triangle is $\sqrt{3}$ times its area. Find the length of each side.

Answer-14

Let, a be the length of each side,

Perimeter $=\sqrt{3} \times$ area

$3 a=\sqrt{3} \times \sqrt{3} / 4 a^{2}$

Or, $a=4$

$\therefore$ length $=4$ unit.

Question-15

The area of an equilateral triangle is $173.2 \mathrm{~m}^{2}$. Find its perimeter.

Answer-15

Let, a be the side

$\begin{aligned}&\sqrt{3} / 4 a^{2}=173.2 \\&\text { Or, } a^{2}=173.2 \times 4 / \sqrt{3} \\&a=20 \mathrm{~m}\end{aligned}$

Perimeter $=3 \times 20=60 \mathrm{~m}$

Question-16

Find the area of the isosceles triangle whose

(i) each of the equal sides is $8 \mathrm{~cm}$ and the base is $9 \mathrm{~cm}$

(ii) each of the equal sides is $10 \mathrm{~cm}$ and the base is $12 \mathrm{~cm}$

(iii) each of the equal sides is $7.4 \mathrm{~cm}$ and the base is $6.2 \mathrm{~cm}$;

(iv) Perimeter is $11 \mathrm{~cm}$ and base is $4 \mathrm{~cm}$.

Answer-16

(i) each of the equal sides is $8 \mathrm{~cm}$ and the base is $9 \mathrm{~cm}$

Semi-perimeter $=(a+b+c) / 2$

$\begin{aligned}&=(8+8+9) / 2 \\&=25 / 2 \\&=12.5\end{aligned}$

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

$\begin{aligned}&=\sqrt{12.5}(12.5-8)(12.5-8)(12.5-9) \\&=\sqrt{12.5} \times 4.5 \times 4.5 \times 3.5 \\&=\sqrt{885.9375} \\&=29.77\end{aligned}$

So, $29.77$ is the area of triangle.

(ii) each of the equal sides is $10 \mathrm{~cm}$ and the base is $12 \mathrm{~cm}$

Since we have given that

Measure of equal sides $=10 \mathrm{~cm}$

Measure of third side $=12 \mathrm{~cm}$

so, we need to find the area of an isosceles triangle.

Semi perimeter $s=\frac{a+b+c}{2}=\frac{10+10+12}{2}=\frac{32}{2}=16 \mathrm{~cm}$

So, Area of isosceles triangle would be

$\begin{aligned}\text { Area of triangle } &=\sqrt{s}(s-a)(s-b)(s-c) \\&=\sqrt{16}(16-10)(16-10)(16-12) \\&=\sqrt{16 \times 6 \times 6 \times 4} \\&=\sqrt{2304} \\&=48 \mathrm{~cm}^{2}\end{aligned}$

(iii) each of the equal sides is $7.4 \mathrm{~cm}$ and the base is $6.2 \mathrm{~cm}$;

$=$ Area of isosceles $=1 / 2 \times \mathrm{b} \times \mathrm{h} .$

$=1 / 2 \times 6.2 \times 7.4 .$

$=[7.4$ divide by 2$] .$

$=1 \times 6.2 \times 3.7 .$

$=22.94 .$

(iv) perimeter is $11 \mathrm{~cm}$ and base is $4 \mathrm{~cm}$.

$=$ Area of isosceles $=1 / 2 \times \mathrm{b} \times \mathrm{h} .$

$=1 / 2 \times 4 \times 11 .$

$=2 \times 11 .$

$=22 .$

Question-17

(i) The base of an isosceles triangle is $24 \mathrm{~cm}$ and its area is $192 \mathrm{sq} . \mathrm{cm}$. Find its perimeter.

(ii) Find the base of an isosceles triangle whose area is $12 \mathrm{~cm}^{2}$ and one of the equal sides is $5 \mathrm{~cm}$.

Answer-17

(i)

In $\Delta \mathrm{ABC}$

$A B=A C$ since the given triangle is isosceles.

Base $=B C=24 \mathrm{~cm}$

Altitude = AD

Area $=192$ sq.cm.

Area of triangle $=1 / 2$ (base $x$ height)

So,

$A B^{2}=A D^{2}+B D^{2}$

$A B^{2}=16^{2}+12^{2}$

$A B^{2}=400$

$A B=20$

Thus $A B=A C=20 \mathrm{~cm}$

$B C=24 \mathrm{~cm}$

Perimeter of triangle = Sum of all sides

$=\mathrm{AB}+\mathrm{BC}+\mathrm{AC}$

$=20+24+20$

$=64 \mathrm{~cm} \text {. }$

Hence the perimeter of triangle is $64 \mathrm{~cm}$

(ii)Area: $12 \mathrm{~cm} \mathrm{sq}$

side $=5 \mathrm{~cm}^{2}$

Question-18

The perimeter of an isosceles triangle is $42 \mathrm{~cm}$. Its base is $2 / 3$ times the sum of equal sides. Find the length of each side and the area of the triangle.

Answer: let each of the equal sides be $x$

let the base by $y$

$y=(2 / 3)(2 x)=4 x / 3$

so $x+x+(4 / 3) x=42$

multiply by 3

$3 x+3 x+4 x=126$

$10 x=126$

$x=12.6$

so the base $=(4 / 3)(12.6)=16.8$

use Pythagoras to find the height, $h$

$h^2+8.4 \wedge 2=12.6 \wedge$

$h^2=88.2$

$h \sqrt{8} 8.2$

Area $=(1 / 2)$ base $x$ height

$=(1 / 2)(16.8) \sqrt{8} 8.2$

$=$ appr. $78.888$

Question-19

$\mathrm{PQR}$ is an isosceles triangle whose equal sides $\mathrm{PQ}$ and $\mathrm{PR}$ are $13 \mathrm{~cm}$ each, and the base $\mathrm{QR}$ measures $10 \mathrm{~cm}$. PS is the perpendicular from P to QR and $\mathrm{O}$ is a point on PS such that $\angle Q O R=90^{\circ}$. Find the area of shaded region

Sol:

 $\begin{aligned} \therefore \quad P S^{2} &=P R^{2}-S R^{2} \\ &=13^{2}=5^{2} \\ &=169-25 \\ &=144 \end{aligned}$

$P S=12 \mathrm{~cm}$

$\triangle QO S \cong \triangle R O S$ (SAS)

$O Q=O R=x$ (suppose)

$\therefore \triangle QOR= x^{2}+m^{2}=10^{2}$

$2 x^{2}=100$

$x^{2}=50$

$\triangle Q D R=\frac{1}{2} \times OY \times O R$

$=\frac{1}{2} \times x^{2}$

$=\frac{1}{2} \times 50=25$ $\mathrm{cm}^{2}$

Area of shaded portion = (100-25) =35${cm}^{2}$

Question-20

The perimeter of a right triangle is $50 \mathrm{~cm}$ and the hypotenuse is $18 \mathrm{~cm}$. Find its area.

Answer-20

Perimeter = $50 \mathrm{~cm}$

Hypotenuse $=18 \mathrm{~cm} .$

Let the sides be $a b$ and $c$ where $c$ is Hypotenuse

Now $a+b+c=50$

$a+b=50-c$

$A \wedge+b \wedge=c \wedge$

$(a+b) \wedge 2=(50-c) \wedge 2$

$A \wedge 2+b \wedge 2+2 a b=2500-100 c+c \wedge 2$

$c \wedge 2+2 a b=2500-100 \times 18+c \wedge 2$

$2 a b=2500-1800$

$2 a b=700$

$a b=350$

Area of triangle $=1 / 2 \times$ base $\times$ height

Area $=1 / 2 \times a \times b$

Area $=1 / 2 \times 350$

Area $=175 \mathrm{~cm} \wedge 2$