Exercise 12 D
Question 1
Ans: In $\triangle P Q R$.
XY 11 QR
$P X=1 \mathrm{~cm}, X Q=3 \mathrm{~cm}, Y R=4.5 \mathrm{~cm}$ and
$Q R=9 \mathrm{~cm}$
(IMAGE TO BE ADDED)
$P Y=x \mathrm{~cm}$ and $X y=y \mathrm{~cm}$
if $X Y \| Q R$
So $P X Y \sim \triangle P Q R$
so $\frac{P X}{P Q}=\frac{P Y}{P R}=\frac{X Y}{Q R}$
$\begin{aligned}&\Rightarrow \frac{1}{1+3}=\frac{x}{x+4.5}=\frac{y}{9} \\&\Rightarrow \frac{1}{4}=\frac{x}{x+4.5}=\frac{y}{9} \\&\frac{x}{x+4.5}=\frac{1}{4} \Rightarrow 4 x=x+4.5 \\&\Rightarrow 4 x-x=4.5 \\&\Rightarrow 3 x=4.5\end{aligned}$
(i) $\Rightarrow x=\frac{4.5}{3}=1.5 \mathrm{~cm}$
and $\frac{y}{9}=\frac{1}{4}=4 y=9$
$\Rightarrow y=\frac{9}{4}=2.25 \mathrm{~cm}$
(ii) so $\mathrm{Py}=1.5 \mathrm{~cm}$ and $\mathrm{Xy}=2.25 \mathrm{~cm}$
(iii) if $\triangle P XY \sim \triangle P Q R$
So area of $\triangle P X Y$
area of $\triangle P Q R=\frac{P X^{2}}{P Q^{2}}$
$\Rightarrow \frac{A}{\text { area } \triangle P Q R}=\frac{1^{2}}{4^{2}}=\frac{1}{16}$
So Area $\triangle P Q R=16 A \mathrm{~cm}^{2}$
(iv) Now area of figure X.4RQ = Area of $\triangle P Q R$ - area of $\triangle P \times 4$
$=16 \mathrm{~A}-\mathrm{A}=15 \mathrm{~A} \mathrm{~cm}^{2}$
Question 2
Ans: In the higure two triangles are similar
i-e. $\triangle A B C \sim \triangle A P Q, P Q$ is not parallel
To $B C, P C=4 \mathrm{~cm}, A Q=3 \mathrm{~cm}, Q B=12 \mathrm{~cm}$,
$\begin{gathered}B C=15 \mathrm{~cm} \\A P=x, \text { and } P Q=y\end{gathered}$
(image to be added)
Then $\frac{A Q}{A C}=\frac{A P}{A B}=\frac{P Q}{B C}$
$\begin{aligned}&\Rightarrow \frac{3}{x+4}=\frac{x}{3+12}=\frac{y}{15} \\&\Rightarrow \frac{3}{x+4}=\frac{x}{15} \Rightarrow x^{2}+4 x=45 \\&\Rightarrow x^{2}+4 x-45=0 \\&\Rightarrow x^{2}+9 x-5 x-45=0\end{aligned}$
$\Rightarrow x(x+9)-5(x+9)=0$
$\Rightarrow(x+9)(x-5)=0$
Either $x+9=0$ then $x=-9$ which is not possible being negative
or $x-5=0$ then $x=5$
So $\quad A P=5$
and $\frac{3}{x+4}=\frac{y}{15}$
$\Rightarrow \frac{3}{5+4}=\frac{y}{15}$
$\Rightarrow y=\frac{3 \times 15}{9}=5$
is $\triangle A P Q \sim \triangle A B C$
$\text {If } \frac{\text { area } \triangle A P Q}{\text { area } \triangle A D C}=\frac{P Q^{2}}{BC^{2}}$
$=\frac{(5)^{2}}{(15)^{2}}=\frac{25}{225}=\frac{1}{9}$
So
Ratio between the areas of $\triangle A P Q$ and $\triangle A B C$ $=1: 9$
Question 3
Ans: $\triangle A B C \approx \triangle D E F$
$A B=2, D E=4$
(image to be added)
(a) if $\Delta s$ are similar
So $\frac{\text { area } \triangle A B C}{\text { arer } \triangle D E F}=\frac{A B^{2}}{D E^{2}}=\frac{(2)^{2}}{(4)^{2}}=\frac{4}{16}=\frac{1}{4}$
So Ratio between their areas $=1: 4$
(b)$\triangle A B C: \triangle O E F=16: 25$
$\Rightarrow \frac{\text { ara } D A B C}{\text { area } \triangle D E F}=\frac{16}{25}$
$\Rightarrow \frac{\text { area } \triangle A B C}{\text { area } \triangle D E F}=\frac{A B^{2}}{D E^{2}}$
So $\frac{A B^{2}}{D E^{2}}=\frac{16}{25}=\left(\left.\frac{4}{5}\right)^{2}\right.$
So $\frac{A B}{D E}=\frac{4}{5}$
Question 4
Ans: In $\triangle A B C$
$P$ is a point on $A B$ such that
$A P: P B=1: 2$
and $Q$ is a point on $A C$ soch that $P Q \| B C$
(image to be added)
if $P Q \| B C$
so $\triangle A P Q \backsim \triangle A B^{\prime} C$
if $\frac{\text { area of } \triangle A P Q}{\text { area of } D A B C}=\frac{A P^{2}}{A B^{2}}$
$=\frac{(1)^{2}}{(1+2)^{2}}=\frac{(1)^{2}}{(3)^{2}}=\frac{1}{9}$
$\Rightarrow 9$ areas of $\triangle A P Q=$ area of $\triangle A B C$
subtracting area of $\triangle A P Q$ from both sides area of $\triangle A B C$ - area of $\triangle A P Q=9$ are of $\triangle A P Q$-area of $\triangle A P Q$
$\Rightarrow$ area of trap. $B P Q C=8$ area of $\triangle A P Q$
$\Rightarrow \frac{\text { area of } \triangle A P Q}{\text { area of trap. } B P Q C}=\frac{1}{8}$
So area of $\triangle A P Q$ : area as trap. $B P Q C=1: 8$
Question 5
Ans: Area of $\triangle A B C=36 \mathrm{~cm}^{2}$
and Area of $\triangle D E F=81 \mathrm{~cm}^{2}$
$E F=6.9 \mathrm{~cm}$
is $\triangle A B C 〜\triangle D E F$
so $\frac{\text { area of } \triangle A B C}{\text { area of } \triangle D E F}=\frac{B C^{2}}{E F^{2}}$
$\Rightarrow \frac{36}{81}=\frac{(B C)^{2}}{(6.9)^{2}} \Rightarrow \frac{(6)^{2}}{(9)^{2}}=\frac{(B C)^{2}}{(6.9)^{2}}$
$\Rightarrow \frac{B C}{C .9}=\frac{6}{9}$
$\Rightarrow B C=\frac{6.9 \times 6}{9}$
$\Rightarrow B C=4.6 \mathrm{~cm}$
Question 6
Ans: In $\triangle A D E, B C \| D E$
$A D=3 \mathrm{~cm} D B=2 \mathrm{~cm}$
(image to be added)
So $A B=A D-D B=3-2=1 \mathrm{~cm}$ Area of $\triangle A B C=10 \mathrm{~cm}$
if $\triangle A D E: B C || D E$
So $D A B C \backsim \triangle A D E$
So $\frac{\text { area } \triangle A B C}{\text { area } △ A D E}=\frac{A D^{2}}{A D^{2}}$
$\Rightarrow \frac{10}{\text { area } \triangle A D E}=\frac{(1)^{2}}{(3)^{2}}=\frac{1}{9}$
$\Rightarrow$ area $D A D E=10 \times 9=90 \mathrm{~cm}^{2}$
Question 7
Ans: In $\triangle P R S$ and $\triangle P Q R$
$\angle P R S=\angle P Q R$
$\angle R P S=\angle R P Q$
So $\triangle P R S \sim \triangle P Q R$
So $\frac{\text { area of } △P R S}{\text { area of } △ P Q R}=\frac{P S^{2}}{P R^{2}}$
$\Rightarrow \frac{2}{\text { area of } \triangle P Q R}=\frac{(2)^{2}}{(3)^{2}}=\frac{1}{9}$
$\Rightarrow$ anea of $\triangle P Q R=\frac{2 \times 9}{4}=\frac{9}{2}=4.5 \mathrm{~cm}^{2}$
Question 8
Ans: In $\triangle A B C: A D: D B=5: 4 \Rightarrow \frac{A D}{D B}=\frac{5}{4}$
$\begin{aligned}&\Rightarrow \frac{A D}{A D+D B}=\frac{5}{5+4} \\&\Rightarrow \frac{A D}{A B}=\frac{5}{9}\end{aligned}$
$\Rightarrow \frac{A D}{A D+D B}=$ $\Rightarrow \frac{A D}{A B}=$ and $D E \| B C$
(image to be added)
So $\triangle A D E \approx \triangle A B C$
So $\frac{\text { Area }(\triangle A D E)}{\text { Area }(\triangle A B C)}=\frac{A D^{2}}{A B^{2}}$
{Area of two similar triangles are proportional to the squares of their corresponding sides }
$=\frac{(5)^{2}}{(9)^{2}}=\frac{25}{81}$
So $\frac{\text { Area }(\triangle A D E)}{\text { Area }(\triangle A B C)}=\frac{25}{81}$
Question 9
Ans: In $\triangle P A B$ and $\triangle P Q R$.
$\begin{aligned}\angle A B P &=\angle Q \\\angle P &=\angle P\end{aligned}$
So $\triangle P A B \sim \triangle P Q R$
In right $\triangle P Q R, P Q=8 \mathrm{~cm}$ and $P R=10 \mathrm{~cm}$
But $P R^{2}=P Q^{2}+Q R^{2}$
$\begin{aligned}&\Rightarrow(10)^{2}=(8)^{2}+(Q R)^{2} \\&\Rightarrow 100=64+(Q R)^{2} \\&\Rightarrow Q R^{2}=100-64=36=(6)^{2}\end{aligned}$
So $Q R=6 \mathrm{~cm}$
if $\triangle P A B \sim \triangle P Q R$
So $\frac{A B}{Q R}=\frac{P B}{P Q} \Rightarrow \frac{2}{6}=\frac{P B}{8}$
$\Rightarrow P B=\frac{2}{6} \times 8=\frac{8}{3} \mathrm{~cm}$
Again $\triangle P A B \sim \triangle P Q R$
So
$\frac{\operatorname{arca}(\Delta P A B)}{\operatorname{arcr}(\triangle P Q R)}=\frac{A B^{2}}{QR^{2}}$
{Area of similar triangles are proportional to the squares of their corresponding sides}
$=\frac{(2)^{2}}{(6)^{2}}=\frac{4}{36}=\frac{1}{9}$
So area $(\triangle P Q R)=9$ area $(\triangle P A B)--(i)$ subtracting area ( $\triangle P A B$ ) from both sides,
Area $(\triangle P Q R)$ - are $(\triangle P A B)=9$ area $(\triangle P A B)$-area ( $ \triangle P A B$ )
$\begin{aligned}\Rightarrow & \text { area quad. } A Q R B=8 \text { area }(\triangle P A B) \\& \frac{\text { area }(\text { quad } \cdot A Q R B)}{\text { area }(\triangle P Q R)}=\frac{8 \text {area}(\triangle P A B)}{9 \text { area }(\triangle P A B)}=\frac{8}{9}\end{aligned}$
So Area quad. $A Q R B$ : area $(\triangle P Q R)=8: 9$
Question 10
Ans: $\triangle A B C \sim \triangle P Q R$
Area $(\triangle A B C)=64 \mathrm{~cm}$ and arer $(\triangle P Q R)=121 \mathrm{~cm}$
$Q R=15.4 \mathrm{~cm}$
if triangles are similar.
$\text { So } \frac{\text { area }(\triangle A B C)}{\text { area }(\triangle P Q R)}=\frac{B C^{2}}{QR^{2}} \text {. }$
$\Rightarrow \frac{64}{121}=\frac{(B C)^{2}}{(15.4)^{2}} \Rightarrow \frac{(8)^{2}}{(11)^{2}}=\frac{(B C)^{2}}{(15.4)^{2}}$
$\Rightarrow \frac{B C}{15.4}=\frac{8}{11}$
$\Rightarrow B C=\frac{8}{11} \times 15.4$
$\Rightarrow B C=11.2 \mathrm{~cm}$
Question 11
Ans: DE and F are the mid- point of the sides BC, CA and AB of △ABC
DE,EF and FD are joined
If E and F are mid points of AC and AB
So EF ||BC and = $\frac{1}{2} B C$
$\Rightarrow \frac{E F}{B C}=\frac{1}{2}$............(i)
similanly $D$ and $E$ are the mid-point of $B C$ and $A C$
if $D E \| A B$ and $=\frac{1}{2} A B$
$\Rightarrow \frac{D E}{A B}=\frac{1}{2}$.......(ii)
and $D$ and $F$ are mid-point of $B C$ and $A B$
So $D F \| A C$ and $=\frac{1}{2} A C$
$\Rightarrow \frac{DF}{A C}=\frac{1}{2}$.......(iii)
From (i) (ii) and (iii)
$\frac{E F}{D C}=\frac{D E}{A B}=\frac{D F}{A C} \quad \text { (each }=\frac{1}{2} \text { ) }$
So $\triangle D E F \backsim \triangle A B C$
$\text { So } \frac{\text { area } \triangle D E F}{\text { area } \triangle A B C}=\frac{E F^{2}}{BC^{2}}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$
So area $\triangle D E F:$ area $\triangle A B C=1: 4$
Question 12
Ans: $A B C D$ is a trapezium in which $A B \| D C$ and $A B=2 D C$ $A C$ and $B D$ intersect each other at $O$
In $\triangle A O B$ and $\triangle C O D$.
$\angle A O B=\angle C O D$
$\angle A B O=\angle O D C$
So $\triangle A O B \sim \triangle C O D$
if $\frac{\text { area }(D A O B)}{\text { area }(\triangle C O D)}=\frac{A B^{2}}{D C^{2}}=\frac{(2 D C)^{2}}{(D C)^{2}}$
$\frac{4 D C^{2}}{D C^{2}}=\frac{4}{1}$
So area $(\Delta A O Q)=4$ area $(\Delta C OD)$
Hence proved.
Question 13
Ans: $\triangle A B C$ and $\triangle D E F$ are similar
and area $(\triangle A B C)=$ area $(\triangle D E F)$
(image to be added)
$\frac{\text { area }(△ A B C)}{\text { area }(△D E F)}=1$.........(i)
$\frac{\text { area }(\triangle A B C)}{\text { area }(\triangle D E F)}=\frac{B C^{2}}{E F^{2}}=\frac{A B^{2}}{D E^{2}}=\frac{A C^{2}}{D F^{2}}$.........(ii)
From (i.) and (ii)
$\begin{aligned}& \frac{B C^{2}}{E F^{2}}=\frac{A B^{2}}{D E^{2}}=\frac{A C^{2}}{D F^{2}}=1 \\\Rightarrow & \frac{B C^{2}}{E F^{2}}=1 \Rightarrow B C^{2}=E F^{2} \\\Rightarrow & B C=E F\end{aligned}$
similarly $A B=D E$ and $A C=D F$
Hence $\triangle A B C \cong \triangle D E F$
Question 14
Ans: In $\triangle A B C$ ard $\triangle D E F$,
$A P \perp B C$ and $D Q \perp E F$
if $\triangle A B C \backsim \triangle D E F$
So $\frac{A B}{D E}=\frac{B C}{E F}=\frac{A C}{D F}$ and $\angle B=\angle E$
Now in $\triangle A B P$ and $\triangle P E Q$
$\begin{aligned}&\angle B=\angle E \quad(\text ) \\&\angle P=\angle Q\end{aligned}$
So $\triangle A B P \sim \triangle D E Q$
$\begin{aligned}&\text { So } \frac{A B}{D E}=\frac{A P}{D C} \\&\text { But } \frac{A B}{D E}=\frac{B C}{E F} \\&\text { So } \frac{A P}{D C}=\frac{B C}{E F}\end{aligned}$
Hence proved
Question 15
Ans: In $\triangle A B C, P$ and $Q$ are the points on $A B$ and $A C$ such that
$P Q \quad || B C$
and area $(\triangle A P Q)=$ area (quad. $P B C Q$ )
$\begin{aligned} \Rightarrow \text { area }(\Delta A P Q)+\text { area }(\Delta A P Q) &=\text { area }(\text { quad }) P B C O) \\ &+\text { area }(\Delta A P Q) \end{aligned}$
$\Rightarrow 2$ area $(A P Q)=$ area $(\triangle A B C)$
$\Rightarrow$ area $(\triangle A P Q)=\frac{1}{2}$ area $(\triangle A B C)$
$\Rightarrow \frac{\text { area }(\triangle A P Q)}{\text { area }(\triangle A B C)}=\frac{1}{2}$
is. $P Q \| B C$
So $\triangle A P Q \backsim \triangle A B C$
So $\frac{\text { area }(\triangle A P Q)}{\text { area }(\triangle A B C)}=\frac{A P^{2}}{A B^{2}}=\frac{1}{2}$
So $\frac{A P^{2}}{A B^{2}}=\frac{1}{2} \Rightarrow \frac{A P}{A B}=\frac{1}{\sqrt{2}}$
$\begin{aligned}& \frac{A B-B P}{A B}=\frac{1}{\sqrt{2}} \\\Rightarrow & \frac{A B}{A B}=\frac{B P}{A B}=\frac{1}{\sqrt{2}} \\& 1-\frac{B P}{A B}=\frac{1}{\sqrt{2}}\end{aligned}$
$\Rightarrow \frac{B P}{A B}=1-\frac{1}{\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}}$
Hence $\frac{B P}{A B}=\frac{\sqrt{2}-1}{\sqrt{2}}$
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