Exercise 12 A
Question 1
Ans: (TO BE ADDED)
Plot the points A(3,4) ,B(9.4) C(9,6) D(3,6) on the graph
Join them to form rectangle ABCD
The ratio scale of transformation is $\left(x, y\right) \rightarrow\left(2 x, 2 y\right)$ ie $1: 2$
Now plot the points $A\left(6,81 \mathrm{~B}^{\prime}(18,8) C^{\prime}(18,12)\right.$ and $D^{\prime}(6,12)$ on the same graph and Join them to from another rectangle $A'B' C: D$
(i) Length of the side of the image of rectangle ABCD is 1:2 i.e double of the original rectangle
(ii) Yes the rectangle are similar and Ratio is 1:2
(iii) Ratio in area = $\frac{Area of image A,B,C,D}{Area of image ABCE}$ $=\frac{12 \times 4}{6 \times 2}=\frac{48}{12}$
i.e 4:1
Question 2
Ans: (i) If (x, y) = (3x, 3y)
So the image A,B ,C ,D is three times in length of the figure ABCD
(ii) If The area of the image is 16 times the area of the original rectangle
so $(x, y) \rightarrow(4 x, 4 y)$
(iii) In the given figure
The image is half the length of the corresponding sides of the original figure
So $(x, y) \rightarrow\left(\frac{1}{2} x, \frac{1}{2} y\right)$
Question 3
Ans: O is the center of dilatation and the dilatation factor 3
i.e (x,y) $\rightarrow(3 x, 3 y)
$\begin{array}{|l|l|}\hline \text { Perimage } & \text { image } \\\hline \text { 1. } A B=3 \text { unit } & A^{\prime} B^{\prime}=3 \times 5=15 \text { units } \\\text { 2. } B C=\frac{1}{3} \times 18 & B^{\prime} C^{\prime}=18 \text { units } \\& =6 \text { units } \\\text { 3. } A C=2 \text { units } & A^{\prime} C^{\prime}=3 \times 2=6 \text { units } \\\text { 4. } O A=7 \text { units } & O A^{\prime}=3 \times 7=21 \text { units } \\\text { 5. } O C=\frac{1}{3} \times 39=13 \text { units } & O C^{\prime}=39 \text { units } \\\text { 6. } O B=m \text { units } m \in R & O B^{\prime}=m \times 3=3 \text { m units } Mε R\end{array}$
(ii) if $(x, y) \rightarrow(3 x+3 y)$
(a) $\frac{O A'}{O A}=\frac{3}{1}$
(b) $\frac{O B' }{O B}=\frac{3}{1}$
(c)$\frac{O C'}{O C}=\frac{3}{1}$
(d) $\frac{A' B'}{A B}=\frac{3}{1}$
(e) $\frac{B^{\prime} C^{\prime}}{B C}=\frac{3}{1}$
(f) $ \frac{C'A'}{C A}=\frac{3}{1}$
Question 4
Ans: The coordinates of the vertices of a triangle are $A(1,1) B(1,3)$ and $C(3,1)$ respectively Enlargement of scale factor is 2 i.e (x, y) $\rightarrow(2 x, 2 y)$
(IMAGE TO BE ADDED)
So coordinates of enlargement triangle will be
$A^{\prime}(2,2) \quad B^{\prime}(2,6)$ and $C^{\prime}(6,2)^{\prime}$ Respectively . O is the origin (center) of the enlargement triangle whose vertices are $A^{\prime}(2,2) \quad B^{\prime}(2,6)$ and $C^{\prime}(6,2)$
Question 5
Ans: (a) Coordinates of vertices are $P(1,2), Q(2,2), R(2,1)$ Scale factor 1: 4
and Center of enlargement is origin (0,0)
So $(x, y) \rightarrow(4 x, 4 y)$
So coordinates of the enlarged figure will be p(4,8) Q(8,8) and R(8,4)
(b) Coordinates of vertices are $A(0,0): B(3,2), C(4,0)$ scall factor is $1: 2$
$(x, y) \rightarrow(2 x, 2 y)$
(c) Coordinates of figure are $A(5,5), B(5,7) \cdot C(7,7)$ and $D(7,5)$ and enlargement factor is 1:4
So $(x, y) \rightarrow(4 x, 4 y)$
and center of enlargement is the point (6,6)
So Co-ordinates of the vertices of enlargement figure will be A' $(2,2) \quad B^{\prime}(10,2) \quad C^{\prime} \quad(10,10)$ and $D^{\prime}(2,10)$
(IMAGE TO BE ADDED)
(d) Coordinates of figures are $A(5,41, \times(5,8), 4(7,8)$ and $z(7,4)$ and enlargement factor is 1:2 And center of enlargement is the point (6,6)
The vertices of the enlargement figure A'X'Y'Z' will be $A^{\prime}(4,2)$, $x^{\prime}(4,10), y^{\prime}(8,10)$ and $z^{\prime}(8,2)$
(IMAGE TO BE ADDED)
Question 6
Ans: Take CK = 6 and take its mid point k'
Such that ck' = 3cm
Join ck, Lm, mc and NC
Draw K'L'||KL , K'N' ||KN, N'M||NM and L'M'||LM
Then figure K'L M'N is the image of figure
KLMN by the scale factor $\frac{1}{2}$ with centre $c$.
(IMAGE TO BE ADDED)
Question 7
Ans: Join PA, QB, RC and SD and produce them
When meet all of them at a point L. This is the required center of enlargement and its co-ordinates are (-2,3)
(IMAGE TO BE ADDED)
Question 8
Ans: Side of a square = 3cm
When it is enlarged the scale factor = 2
i.e K=2
Area of the square= $(\operatorname{side})^{2}=(3)^{2}=$ $9 \mathrm{~cm}^{2}$
$=(2)^{2} \times 9=4 \times 9=36 \mathrm{~cm}^{2}$
Question 9
Ans: Draw a square ABCD with each side 2cm
Join AC and BD and produced them both sides
Take A'B' =4cm and complete the required square A'B'C'D
with each side 4cm
Now area of square $A B C D=2 \times 2=4 \mathrm{~cm}^{2}$ and area of square $A'B' C'D'$ with each side $2 \times 2=4 \mathrm{~cm}$
$=4 \times 4=16 \mathrm{~cm}^{2}$
Hence area of the image of square ABCD = $16 \mathrm{~cm}^{2}$
(IMAGE TO BE ADDED)
Question 10
Ans: Area of triangle $=12 \mathrm{~cm}^{2}$
Area of enlarged triangle $=108 \mathrm{~cm}^{2}$
Scale factor =k and then scale factor of Area = $k^{2}$
Then $k^{2} \times 12=108$
$\begin{gathered}\Rightarrow k^{2}=\frac{108}{12} \\=9\end{gathered}$
So $k=\sqrt{9}=3$
Question 11
Ans: A parallelogram ABCE is given whose vertices are A(6,3) B(9,2) ,C (12,3) D(9,3) Scale factor is $\frac{1}{3}$
and center o(0,0) transforms parallelogram ABCD on to ||gm A'B'C'D
(IMAGE TO BE ADDED)
a (i) Join OA, OB, OC and OD
Take a point A' on OA such that OA'=$\frac{1}{3}$ OA
Similarly take points $B^{\prime}, C^{\prime}$ and $D^{\prime}$ Join $A^{\prime} B^{\prime}, B^{\prime} C^{\prime}, C^{\prime} D^{\prime}$ and $D^{\prime} A^{\prime}$ $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ is the required ||gm
(ii) Area of || gm $A B C D=A^{\prime} B^{\prime} \times$ perpendicular between $A^{\prime} B^{\prime}$ and $C^{\prime} D^{\prime}=1 \times 2=2 \mathrm{~cm}^{2}$
(b) Let AB = 3cm
Now ||gm ABCD is enlarged with a scale factor $\frac{2}{5}$
given now ||gm A"B "C"D"
length of side $A^{\prime \prime} B^{\prime \prime}=1.2 \mathrm{~cm}$
Join $O A, O B, O C$ and $O D$
Tale a point A" $\left(6 \times \frac{2}{5} ; 3 \times \frac{2}{5}\right)$ or $(2.4,1.2)$
Similarly point B'' (2.4, 1.2) C'' (3.6,3.6) and D''(4,5, 3.6) and Join A''B'', B''C'' , C''D'' , D''A'' respectively
$A^{\prime \prime} B^{\prime \prime} C^{\prime \prime} D^{\prime \prime}$ is the required parallelogram
on mearuring length of $A^{\prime \prime} B^{\prime \prime}=1.2 \mathrm{~cm}$.
(IMAGE TO BE ADDED)
Question 12
Ans: Scale in model of a van is the real van = 1: 40
Length of the real van = 8cm
So length of the model = $\frac{8 \times 1}{40} m$
$=\frac{8 \times 100}{40}=20 \mathrm{~cm}$
Question 13
Ans: Scale of the plan of a class room $=1: 50$ Length ias class room $=8 \mathrm{~m}$ and breadth $=6 \mathrm{~m}^{\circ}$
So length of the drawing of the room
$=\frac{8 \times 1}{50} \times 100 \mathrm{~cm}=16 \mathrm{~cm}$
and breadth $=\frac{6 \times 1}{50} \times 100$
$=12 \mathrm{~cm}$
Question 14
Ans: Scale of the plan of the gymnasium of the school = 1: 7
In the plan'
Length = 28cm and breadth = 16cm
So length of actual gymnasium = $\frac{28 \times 75}{1} \mathrm{~cm}$
$=\frac{28 \times 75}{100} m=21 m$
and breadth $=\frac{16 \times 75}{100}=12 \mathrm{~m}$
So Dimension of gymnasium $=21 \mathrm{~m}$ by $12 \mathrm{~m}$
Question 15
Ans: model of a Eiffel. Tower $=16 \mathrm{~cm}$ high
But real height $=320 \mathrm{~m}$
So scale $=16: 320 \times 100$ $=1.20 \times 100$ $=1: 2000$
Question 16
Ans: Dimension of the model of a multi story building are $1.2 \mathrm{~m} \times 75 \mathrm{~cm} \times 2 \mathrm{~m}$
$=120 \mathrm{~cm} \times 75 \mathrm{~cm} \times 200 \mathrm{~cm}$
Scale $=1: 30$, then
Actual length $=\frac{120 \times 30}{1} \mathrm{~cm}$
$=\frac{120 \times 30}{100}=36 \mathrm{~m}$
Breadth $=\frac{375 \times 30}{100}=\frac{45}{2}=22.5 \mathrm{~m}$
$\text { Height }=\frac{200 \times 30}{100}=60 \mathrm{~m}$
Hence dimensions are $36 \mathrm{~m} \times 22.5 \mathrm{~m} \times 60 \mathrm{~m}$.
Question 17
Ans: Scale of map $=1: 2500$
Dimension of a Triangular plot are
$A B=3 \mathrm{~cm}, B C=4 \mathrm{~cm} \text { and } \angle A B C=90^{\circ}$
(IMAGE TO BE ADDED)
(i) So Actual length of $A B$
$\begin{aligned}&=\frac{3 \times 2500}{1} \mathrm{~cm} \\&=\frac{3 \times 2500}{100}=75 \mathrm{~m}\end{aligned}$
and Length of BC= $\frac{4 \times 2500}{100}=100 \mathrm{~m}$
(ii) Actual area $=\frac{1}{2} \times B C \times A B$
$=\frac{1}{2} \times 75 \times 100 \mathrm{~m}^{2}$
$=3750 \mathrm{~m}^{2}$
$=\frac{3750}{1000 \times 1000}=0003750 \mathrm{~km}^{2}$
$=0.00375 \mathrm{~km}^{2}$
Question 18
Ans: Scale of the map $=1: 200000$
It means that 1cm on the map will represent 200000 $\mathrm{cm} 2$ on the ground
(i) so $1 \mathrm{~cm}=200000 \mathrm{~cm}$
$=\frac{20000}{100 \times 1000} \mathrm{~km}=2 \mathrm{~km}$
(ii) if $1 \mathrm{~cm}$ on the map $=2 \mathrm{~km}$ on the 9 round and $1 \mathrm{~cm}^{2}$ $=2 \times 2=4 \mathrm{~km}^{2}$ on the ground
(iii) Area of a plot on the ground $=20 \mathrm{~km}^{2}$
So Area of plot to be represented an the map will be $=\frac{20}{4}=5 \mathrm{~cm}^{2}$
Question 19
Ans: Scale of the model of a snip $=1: 200$
(i) Length of model $=4 \mathrm{~m}$
$\text { so } \text { Actual length }=\frac{4 \times 200}{1}=800 \mathrm{~m}$
(ii) Area of deck = $160000 \mathrm{~m}^{2}$
So Area an the model $=\frac{160000}{200 \times 200}=4 \mathrm{~m}^{2}$
(iii) volume of model $=200$ litres
$=\frac{200}{1000}=0.2 \mathrm{~m}^{2}$
So volume of actual ship $=1 \mathrm{~m}^{3}=1000 \mathrm{~L}$
$=0.2 \times(200)^{3} \mathrm{~m}^{3}$
$=\frac{2}{10} \times 200 \times 200 \times 200=1600000 \mathrm{~m}^{3}$
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