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 $=\frac{90}{(50)^{3}} \mathrm{~m}^{3}$
$=\frac{90}{50 \times 50 \times 50} \mathrm{~m}^{3}$
$=\frac{90 \times 100 \times 100 \times 100}{50 \times 50 \times 50} \mathrm{~cm}^{3}$
$=\frac{90000}{125} \mathrm{~cm}^{2}$
$=720 \mathrm{~cm}^{3}$
No comments:
Post a Comment