S Chand Class 10 CHAPTER 12 Similar triangles Exercise 12 E

 Exercise 12 E

Question 1

Ans: side of a square $=3 \mathrm{~cm}$

when it is enlarged the scale factor $=2$ i.e. $k=2$

Area of the square $=$ side$^{2}=(3)^{2}=9 \mathrm{~cm}^{2}$

so area of the image of square $=k^{2}$

$=(2)^{2} \times 9=4 \times 9=36 \mathrm{~cm}^{2}$   

Question 2

Ans: Area of the riangle $=12 \mathrm{~cm}^{2}$

Area of enlarged triangle $=108 \mathrm{~cm} 2$

scale factor $=k$ and then scale factor of area $=k^{2}$

Then $k^{2} \times 12=108$

$\begin{aligned}&k^{2}=\frac{108}{12}=9 \\&\text { So } k=\sqrt{9}=3\end{aligned}$

Question 3

Ans: Sides of given $\triangle A B C, A B=10 \mathrm{~cm}, B C=12 \mathrm{~cm}$ and $A C=16 \mathrm{~cm}$

So longest side AC = 16cm

But longest side of Enlarged $\triangle A^{\prime} B^{\prime} C^{\prime}=24 \mathrm{~cm}$

i.e. $A^{\prime} \mathrm{c'}=24 \mathrm{~cm}$

Let k be the scale factor 

Now length of side A'B' = 10cm $times \frac{3}{2}=15 \mathrm{~cm}$

and length of side $B^{\prime} C^{\prime}=12 \mathrm{~cm} \times \frac{3}{2}=18 \mathrm{~cm}$

Question 4

Ans: In right angled $\triangle 1 Q R, \angle Q=90^{\circ}$

 $P Q=8 \mathrm{~cm}$ and $Q R=15 \mathrm{~cm}$

So $\begin{aligned} P R &=\sqrt{P Q^{2}+Q R^{2}}=\sqrt{(8)^{2}+1151^{2}} \\=& \sqrt{64+225}=\sqrt{289}=17 \mathrm{~cm} \end{aligned}$

$\triangle P Q R$ is enlarged into $\triangle P'Q'R$ in which image of P is P' of Q is Q' and of R is R' length of P'R'= 42.5cm

(i) $K$ be the scale factor of enlargement

So $42.5=k \times P R=k \times 17$

So $k=\frac{42.5}{17}=2.5=\frac{25}{10}=\frac{5}{2}$.

(ii) Now length of $P^{\prime} Q^{\prime}=\frac{5}{2} \times P Q=\frac{5}{2} \times 8=20 \mathrm{~cm}$ and length of $Q' K^{\prime}=\frac{5}{2} Q K=\frac{5}{2} \times 15=\frac{75}{2}=37.5 \mathrm{~cm}$

 

(iii) Now area of $\triangle P Q R=\frac{1}{2} \times P Q \times Q R$

$=\frac{1}{2} \times 8 \times 15 \mathrm{~cm}^{2}=60 \mathrm{~cm}^{2}$

So area of $\triangle P^{\prime} Q^{\prime} R^{\prime}=K^{2}$ \times area of $\triangle P Q R$

$=\left(\frac{5}{2}\right)^{2} \times 60 \mathrm{~cm}^{2}=\frac{25}{4} \times 60=375 \mathrm{~cm}^{2}$

Question 5

Ans: scale of map $=1 \mathrm{~cm}: 1 \mathrm{~km}$ or $1: 100000$ Now area of island on the map $=3.5 \mathrm{~cm}^{2}$ so Actual area $=k^{2} \times$ area as the map $=(100000)^{2} \times 3.5 \mathrm{~cm}^{2}$ $=100000 \times 100000 \times 3.5 \mathrm{~cm}^{2}$

$=\frac{100000 \times 100000 \times 3.5}{100000 \times 100000} \mathrm{~km}^{2}$

$=3.5 \mathrm{~km}^{2}$

Question 6

Ans: Scale of map $=5 \mathrm{~m}$ to $1 \mathrm{~cm}$

$\text { or } 500: 1 \Rightarrow k=\frac{500}{1}$

side of a square on the map $=3 \mathrm{~cm}$

So its area $=(\text { side })^{2}=(3 \mathrm{~cm})^{2}=9 \mathrm{~cm}^{2}$

(i) So Actual area of the square = $k^{2} \times$ area of

square on the map $=(500)^{2} \times 9 \mathrm{~cm}^{2}$

$=500 \times 500 \times 9 \mathrm{~cm}^{2}$

$=\frac{500 \times 500 \times 9}{100 \times 100}=225 \mathrm{~m}^{2}$

(ii) Actual area of a square $=50625 \mathrm{~m}^{2}$

So Area on the map $=\frac{1}{k^{2}} \times$ actual area

$\begin{aligned}&=\frac{1}{(500)^{2}} \times 50625 \\&=\frac{50625 \times 100 \times 100}{500 \times 500} \mathrm{~cm}^{2} \\&=2025 \mathrm{~cm}^{2} \\&\text { So side }=\sqrt{2025}=45 \mathrm{~cm}\end{aligned}$

Question 7

Ans: scale af a ground plan of a nouse $=1 \mathrm{~cm}: 15 \mathrm{~m}$

So scale factor (k)= 1cm : 1500cm 

$=1 :1500=\frac{1}{1500}$

Now length of a room = 18m and breadth = 12m

So length on plan = $18 \times \frac{1}{1500} m$

$=\frac{18 \times 100}{1500} \mathrm{~cm}=\frac{6}{5} \mathrm{~cm}=1.2 \mathrm{~cm}$

and bread th $=12 \times \frac{1}{1500} \mathrm{~m}=\frac{1200}{1500} \mathrm{~cm}$

$=\frac{4}{5} \mathrm{~cm}=0.8 \mathrm{~cm}$

Area on the plan $=1 \mathrm{~cm}^{2}$

Area on the ground $=1 \times(1500)^{2} \mathrm{~cm}^{2}$

$=\frac{1 \times 1500 \times 1500}{100 \times 100}=225 \mathrm{~m}^{2}$

Question 8

Ans: scale on the map $=1 \mathrm{~cm}$ to $4 \mathrm{~km}$

$\begin{aligned}&=1 \mathrm{~cm}: 400000 \mathrm{dm} \\&=1: 400000 \\&\Rightarrow 1<=\frac{1}{400000}\end{aligned}$

Area of an estate on the map $=9.37 \mathrm{~cm}^{2}$ 

So Actual area $=k^{2}$ (area on the map $=\left(4000001^{2} \times 9.37 \mathrm{~cm}^{2}\right.$

$=\frac{937 \times(400000)^{2}}{100 \times 100000 \times 100000}$

$=\frac{14992}{100}=149.92 \mathrm{~km}^{2}$

$=150 \mathrm{~km}^{2}$ (nearest $\left.\mathrm{km}^{2}\right)$

Question 9

Ans: scalc of a map $=1: 20000$

measure of hield on the map $=20 \mathrm{~cm} \times 15 \mathrm{~cm}=300 \mathrm{~cm}^{2}$

So Actual area $=K^{2}$ ( Pree of the field on the mapl

$=\left(200001^{2} \times 300 \mathrm{~cm}^{2}\right.$

$=20000 \times 20000 \times 300 \mathrm{~cm}^{2}$

$=\frac{20000 \times 20000 \times 300}{100000 \times 100000} \mathrm{~km}^{2}$

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

Question 10

Ans:  scale on the map $=1: 50000$

on the map in right angled $\triangle A B C, A B=2 \mathrm{~cm}$ 

$B C=3.5 \mathrm{~cm}$ and $\angle A B C=90^{\circ}$

So Area $=\frac{1}{2} \times A B \times B C=\frac{1}{2} \times 2 \times 3.5 \mathrm{~cm}^{2}=3.5 \mathrm{~cm} 2$

(i) Now Actual length of $B C=3.5 \mathrm{~cm} \times 50000$

$\begin{aligned}&=\frac{3.5 \times 50000}{100000} \mathrm{~km} \\&=\frac{35 \times 50000}{10 \times 100000}=\frac{35 \times 5}{100} \mathrm{~km}\end{aligned}$

$=\frac{175}{100}=1.75 \mathrm{~km}$

and actual area $=3.5 \mathrm{~cm}^{2} \times\left(500001^{2}\right)$

$=\frac{3.5 \times 50000 \times 50000}{100000 \times 100000} \mathrm{~km}^{2}$

$=\frac{3.5 \times 25}{100}=\frac{3.5}{4} \mathrm{~km}^{2}=0.875 \mathrm{~km}^{2}$

Question 11

Ans Scale factor of the model $=1: 50$ 

Dimensions of a model of a multistory 

building are $1 \mathrm{~m} \times 60 \mathrm{~cm} \times 1.20 \mathrm{~m}$ or $100 \mathrm{~cm} \times 60 \mathrm{~cm} \times 120 \mathrm{~cm}$

So Actual length $=100 \times 50 \mathrm{~cm}=\frac{100 \times 50 \mathrm{~m}}{100}=50 \mathrm{~m}$

$\text { Breadth } m=60 \times 50 \mathrm{~cm}$

$=\frac{60 \times 50}{100}=30 \mathrm{~m}$

and height $=120 \times 50 \mathrm{~cm}$

$=\frac{120 \times 50}{100}=60 \mathrm{~m}$

So Actual dimensions $=50 \mathrm{~m} \times 30 \mathrm{~m} \times 60 \mathrm{~m}$

(i) floor area of a room on the model

$=50 \mathrm{sq} \cdot \mathrm{cm}$

So Actual area $=50 \times k^{2}=50 \times 50 \times 50 \mathrm{~cm}^{2}$

$=\frac{50 \times 2500}{100 \times 100} \mathrm{~m}^{2}=12.5 \mathrm{~m}^{2}$

(ii) Actual volume of the room space $=90 \mathrm{~m}^{3}$

So volume on the model $=\frac{90}{(50)^{3}} \mathrm{~m}^{3}$

$=\frac{90}{50 \times 50 \times 50} \mathrm{~m}^{3}$

$=\frac{90 \times 100 \times 100 \times 100}{50 \times 50 \times 50} \mathrm{~cm}^{3}$

$=\frac{90000}{125} \mathrm{~cm}^{2}$

$=720 \mathrm{~cm}^{3}$