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