Exercise 12 B
Question 1
Ans:
(i) In $\Delta$ (a), third angle $=180^{\circ}-\left(75^{\circ}+55^{\circ}\right)=180^{\circ}-130^{\circ}=50^{\circ}$
So Three angles are $75^{\circ}, 55^{\circ}, 150^{\circ}$
In $\triangle(b)$ , third angle is = $180^{\circ}\left(55^{\circ}+80^{\circ}\right)=180^{\circ}-135^{\circ}=45^{\circ}$
Three angles are $45^{\circ}, 80^{\circ}, 55^{\circ}$
We see that triangle (a) is different
(ii) In $\Delta$ (a) third angle is = $180^{\circ}-\left(100^{\circ}+45^{\circ}\right)$
$=180^{\circ}-145=35^{\circ}$
So Three angles will be = $100^{\circ}, 35^{\circ}, 45^{\circ}$
In $\Delta$ (c) third angle is = $180^{\circ}-\left(25^{\circ}+45^{\circ}\right)$
$=180^{\circ}-70^{\circ}=110^{\circ}$
So Three angles will be $110^{\circ}, 25^{\circ}, 45^{\circ}$
we see that triangle (c) is different
(iii) In Δ (a) two sides are equal
In Δ(b), no two sides are equal
In Δ (c) two sides are equal
soΔ (b)$ is different
Question 2
Ans: (i) In the first pair of triangle
The ratio in Corresponding sides are
$\frac{A B}{4 z}=\frac{B C}{x y}=\frac{A C}{x z}$
$\Rightarrow \frac{6}{9}=\frac{4.5}{6.75}=\frac{7}{10.5}$
$\Rightarrow \frac{2}{3}=\frac{450}{675}=\frac{70}{105}$
$\Rightarrow \frac{2}{3}=\frac{2}{3}=\frac{2}{3}$
Yes , it is the same ration
So $\frac{A B}{y z}=\frac{B C}{x y}=\frac{A C}{x z}=\frac{2}{3}$
(ii) In the second pair of triangles
The ratio in the corresponding sides are
$\frac{\mathrm{Lm}}{\mathrm{RP}}=\frac{\mathrm{mN}}{\mathrm{RQ}}=\frac{\mathrm{LN}}{\mathrm{PQ}}$
$\Rightarrow \frac{16}{2.4}=\frac{8}{1.2}=\frac{14}{2.1}$
$\Rightarrow \frac{16 \times 10}{24}=\frac{8 \times 10}{12}=\frac{14 \times 10}{21}$
$\Rightarrow \frac{20}{3}=\frac{20}{3}=\frac{20}{3}=\frac{2}{3}$
Yes , it is the same into
So $\frac{ Lm}{R P}=\frac{m N}{R Q}=\frac{L N}{P Q}=\frac{2}{3}$
Question 4
Ans: (i) In $\triangle A B C, D$ and E are the points on the sides AB and AC respectively such that AD=$3, B D=4.51 A E$ =4 and CE = 6
(IMAGE TO BE ADDED)
$\begin{aligned}&\text { Now } \frac{A D}{D B}=\frac{3}{4.5}=\frac{2}{3} \\&\text { and } \frac{A E}{E C}=\frac{4}{6}=\frac{2}{3} \\&\text { uf } \frac{A D}{D B}=\frac{A E}{E C}\end{aligned}$
So DE ||BC
(ii) In $\triangle A B C, D$ and $E$ are the points on the sides AB and AC respectively such that
AB= $=12 \mathrm{~cm}, A D=8 \mathrm{~cm}, A E=12 \mathrm{~cm}$ and $A C=18 \mathrm{~cm}$
(IMAGE TO BE ADDED)
$D B=A B-A D=12-8=12-8=4 \mathrm{~cm}$
$E C=A C-A E=18-12=6 \mathrm{~cm} .$
Now $\frac{A D}{O B}=\frac{8}{4}=\frac{2}{1}$
and $\frac{A E}{E C}=\frac{12}{6}=\frac{2}{1}$
if $\frac{A D}{D B}=\frac{A E}{E C}$
so $D E \| D C$
(iii) In $\triangle A B C, D$ and $E$ are the points on the sides'
AB and AC respectively such that AB = 5.6cm
AD = 1.4 cm , AC = 7.2cm and AE = 1.8cm
(IMAGE TO BE ADDED)
So $\quad D B=A B-A D=5.6-1.4=4.2 \mathrm{~cm}$
$\begin{aligned}&E C=A C-A E=7.2-1.8=5.4 \mathrm{~cm} \\&\text { Now } \frac{A D}{D B}=\frac{1.4}{4.2}=\frac{1}{3} \\&\frac{A E}{E C}=\frac{1.8}{5.4}=\frac{1}{3}\end{aligned}$
If $\frac{A D}{D B}=\frac{A E}{E C}$
So $O E \| B C$
Question 5
Ans: In the figure $\triangle A B C$ and $\triangle D E F$ are similar
$\begin{aligned}&A B=x, A C=7, B C=y \\&D E=S, D F=3 \text { and } E F=8\end{aligned}$
If Δ are similarly
corresponding sides are proportional
$\text { So } \begin{aligned}& \frac{A B}{D E}=\frac{A C}{D F}=\frac{B C}{E F} \\\Rightarrow & \frac{x}{5}=\frac{7}{3}=\frac{y}{8}\end{aligned}$
$\frac{x}{5}=\frac{7}{3}$
$\Rightarrow x=\frac{5 \times 7}{3}=\frac{35}{3}=11 \frac{2}{3}$
and $\frac{y}{8}=\frac{7}{3}$
$\Rightarrow y=\frac{8 \times 7}{3}=\frac{56}{3}=18 \frac{2}{3}$
Question 6
Ans: if $\triangle A B O \sim \triangle D C O$
So Their Corresponding sides are Proportional
So $\frac{A B}{C D}=\frac{A O}{O D}=\frac{B O}{O C}$
But $A B=3 \mathrm{~cm}, C D=2 \mathrm{~cm}, O C=3.2 \mathrm{~cm}$ and $O D=2.4 \mathrm{~cm}$
(IMAGE TO BE ADDED)
So $\frac{3}{2}=\frac{A O}{2.4}=\frac{B O}{3.2}$
So $\frac{A 0}{2.4}=\frac{3}{2}$
$\Rightarrow A O=\frac{3 \times 2.4}{2}=3.6$
and $\frac{BO}{3.2}=\frac{3}{2}$
$\Rightarrow B O=\frac{3 \times 3.2}{2}=4.8$
Hence $O A=3.6 \mathrm{~cm}$ and $O B=4.8 \mathrm{~cm}$
Question 7
Ans: (IMAGE TO BE ADDED)
$\triangle A B C \backsim \triangle A D E$
$A D: D B=2: 3$
$D E=5 \mathrm{~cm}, A L \perp D E$ and $A M \perp B C$
$A L=x$
If $\triangle A B C \backsim \triangle A D E$
So $\frac{A B}{A D}=\frac{A C}{A E}=\frac{B C}{D E}$
$\begin{aligned}&\Rightarrow \frac{2+3}{2}=\frac{A C}{A E}=\frac{B C}{5} \\&\text { (i) } \frac{5}{2}=\frac{B C}{3}\end{aligned}$
$\Rightarrow \quad B C=\frac{5 \times 5}{2}=\frac{25}{2} \mathrm{~cm}=12.5 \mathrm{~cm}$
(ii) If $\triangle A D E \sim \triangle A B C$
So $\frac{A L}{A M}=\frac{A D}{A B}$
(In similar triangles altitudes are also proportional)
$\begin{aligned} \Rightarrow \frac{x}{A m}=\frac{2}{5} & \Rightarrow A M=\frac{5 x}{2} \\ \text { So } A M &=\frac{5 x}{2} \end{aligned}$
Question 8
Ans: In trapezium $A B C D, A B \| D C$
$A B=4 \mathrm{~cm} \quad B C=3 \mathrm{~cm} \text { and, } C D=6 \mathrm{~cm}$
(IMAGE TO BE ADDED)
(i) In the figure
$\triangle E D C$ and $\triangle E A B$
$\begin{aligned}&\angle E=\angle E \\&\angle E D C=\angle E A B \\&\angle E C D=\angle E B A \\&\text { SO } \triangle E D C-\triangle E A B\end{aligned}$
(ii) Let EB = x, then, EC = x+ 3
If $\triangle E A B \backsim \triangle E D C$
$\begin{aligned} & \operatorname{so} \frac{E B}{E C}=\frac{A B}{D C} \\ \Rightarrow & \frac{x}{x+3}=\frac{4}{6} \end{aligned}$
$\Rightarrow 6 x=4 x+12$
$\Rightarrow 6 x-4 x=12 \Rightarrow 2 x=12$
$\Rightarrow x=\frac{12}{2}=6$
So $E B=6 \mathrm{~cm}$
Question 9
Ans: If the mean and the lamp post are perpendicular On the ground
Height of man PQ = 1.8m
And his shadow QN = 1.5m
(IMAGE TO BE ADDED)
Let LM be the lamp post and LM=xm
Now in △ L M N
So $\triangle P N Q \backsim \triangle L N M$
$\text { if } \begin{aligned}\frac{P Q}{L m} &=\frac{N Q}{N m} \\\Rightarrow \frac{1.8}{x} &=\frac{1.5}{1.5+5} \\\Rightarrow \frac{1.8}{x} &=\frac{1.5}{6.5}\end{aligned}$
$\Rightarrow \quad x=\frac{1.8 \times 6.5}{1.5}=7.8$
So height of the lamp post = 7.8m
Question 10
Ans: Two lines ACE and BCD intersect each other at C
AC $=4 \mathrm{~cm} \cdot C E=6 \mathrm{~cm}$
$B C=3 \mathrm{~cm}$ and $C D=8 \mathrm{~cm}$
(i) In $\triangle A B C$ and $\triangle C D E$
$\begin{aligned}&\frac{B C}{A C}=\frac{3}{4} \text { and } \frac{C E}{C D}=\frac{6}{8}=\frac{3}{4} \\&\angle A C B=\angle D C E\end{aligned}$
So $\triangle A B C \sim \triangle C D E$ because $\frac{B C}{A C}=\frac{C E}{C D}=\frac{3}{4}$ and $\angle A C B=\angle D C E$
(ii) $\angle D=\angle A=40^{\circ}$
(iii) $\quad A B=x \mathrm{~cm}$
and $\frac{A C}{C D}=\frac{A B}{O E}=\frac{4}{8}=\frac{1}{2}$
$\begin{aligned}&\Rightarrow \frac{x}{D E}=\frac{1}{2} \\&\Rightarrow D E=2 x \\&\text { So } D E=2 x\end{aligned}$
Question 11
Ans: In right Angle $\triangle A B C, \angle C=90$
p is any point on $A B$
(IMAGE TO BE ADDED)
From P, PQ is drawn parallel to AC which meets BC at Q
AC = 2.5cm, BC = 6cm, PQ=1cm
In $\triangle A B C$ is $P Q \| A C$
(i) So $\triangle A B C \backsim \triangle P B Q$
$\therefore A C=\frac{B C}{B Q} \Rightarrow \frac{2.5}{1}=\frac{6}{B Q}$
$\Rightarrow B Q=\frac{6 \times 1}{2.5}=\frac{C}{2.5}=2.4 \mathrm{~cm}$
(ii) In right $\triangle P Q B$
$\begin{aligned}B P^{2} &=P Q^{2}+B Q^{2} \\=(1)^{2}+\left(2.41^{2}\right.&=1+5.76 \\&=6.75=(2.6)^{2} \\& \text { So } D P=2.6 \mathrm{~cm}\end{aligned}$
Question 12
Ans: Let in one triangle ABC,
$\angle A=72^{\circ} \quad \angle B=80^{\circ}$
Then $\angle C=180^{\circ}-\left(72^{\circ}+80^{\circ}=180^{\circ}-152^{\circ}\right.$
$=28^{\circ}$
and in the other triangles $P Q B$
$\angle P=28^{\circ}, \angle Q=72^{\circ}$
$=\angle P=28^{\circ}, \angle Q=72^{\circ}$
So $\angle R=180^{\circ}-\left(28^{\circ}+72^{\circ}\right)=180^{\circ}-100^{\circ}=80^{\circ}$
if $\angle A=\angle Q, \angle B=\angle R$ and $\angle C=\angle P$
So $\triangle A D C〜\triangle P Q R$
Question 13
Ans: In the figure, $A D=2.7 \mathrm{~cm}, A E=2.5 \mathrm{~cm} B E=1.1 \mathrm{~cm}$ and
$B C=3.2 \mathrm{~cm}$
(i) In $\triangle A B C$ and $\triangle A D E$
$\begin{aligned}&\angle A=\angle A \\&\angle A B C=\angle A D E \\&\text { So } \triangle A B C \sim \triangle A D E\end{aligned}$
(ii) $\text { So } \frac{B C}{P E}=\frac{A B}{A D}=\frac{A C}{A E}$
$\Rightarrow \frac{5.2}{D E}=\frac{(1.1+2.5)}{2.7}$
$\Rightarrow \frac{5.2}{D E}=\frac{3.6}{2.7} \Rightarrow D E=\frac{5.2 \times 2.7}{3.6}=3.9$
Hence $D E=3.9 \mathrm{~cm}$
Question 14
Ans: ABC is a triangle AB. EF and CD are parallel line
AB = 15cm, EG = 5cm. BC = 10cm and DC = 18cm
We have to find EF and AC
(i) In $\triangle E F G$ and $\triangle C G D$
$\angle E F G=\angle G D C$
So $\triangle E F G \backsim \triangle C G D$
So $\frac{E G}{C h}=\frac{E F}{C D}$
$\Rightarrow \frac{5}{10}=\frac{E F}{18} \Rightarrow E F=\frac{5 \times 10}{10}=9$
So $E F=9 \mathrm{~cm}$
(ii) similarly in $\triangle A B C$ and $\triangle E C F$
$\begin{aligned}& \angle C=\angle C \\& \angle C A B=\angle C E F \\\text { So } \triangle A B C〜\triangle E C F \\\text { SO } & \frac{A C}{E C}=\frac{A B}{E F} \\\Rightarrow & \frac{A C}{10+5}=\frac{15}{9} \\\Rightarrow & \frac{A C}{15}=\frac{15}{9} \\\Rightarrow & A C=\frac{15 \times 15}{9}=\frac{225}{9}=25 \\& S \circ A C=25 \mathrm{~cm}\end{aligned}$
Question 15
Ans: $\triangle A B C \backsim \triangle P Q R$ and Let
In $\triangle P Q R , P Q=4 \mathrm{~cm} \cdot Q K=7 \mathrm{~cm}$ and $R P=8 \mathrm{~cm}$
and let in $\triangle A B C, A B=6 \mathrm{~cm}, B C=x$ and $C A=y$
Then $\frac{P Q}{A B}=\frac{Q R}{B C}=\frac{R P}{C A}$
$\begin{aligned}&\Rightarrow \frac{4}{6}=\frac{7}{x}=\frac{8}{y} \\&\text { Now } \frac{7}{x}=\frac{4}{6} \\&\Rightarrow x=\frac{7 \times 6}{4}=\frac{21}{2}=10.5 \\&\text { and } \frac{8}{y}=\frac{4}{6}\end{aligned}$
$\Rightarrow y=\frac{6 \times 8}{4}=12$
So other two corresponding sides are 10.5 cm and 12 cm
Question 16
Ans: (a) In $\triangle A B C ,D E \| C B$
so $\triangle A B C \backsim \triangle A D E$
Now $A D=3 \mathrm{~cm}, D B=4 \mathrm{~cm}, E C=12 \mathrm{~cm}$
$\text { let } A E=x \mathrm{~cm}$
if In $\triangle A B C, D E \| B C$
$\begin{aligned}& \frac{A D}{D B}=\frac{A E}{E C} \Rightarrow \frac{3}{4}=\frac{A E}{12} \\\Rightarrow & \frac{3}{4}=\frac{x}{12} \\\Rightarrow & x=\frac{3 \times 12}{4}=9\end{aligned}$
So $A E=9 \mathrm{~cm}$
(b) In $\triangle A B C, D E \| B C$
So $\frac{A D}{D B}=\frac{A E}{E C}$
$\begin{aligned}&\Rightarrow \frac{2.4}{E D}=\frac{2.7}{4.5} \\&\Rightarrow B D=\frac{2.4 \times 4.5}{2.7} \\&\text { So } B D=4 \mathrm{~cm}\end{aligned}$
(c) In $\triangle A B C, D E \| B C$
$\begin{aligned}\text { So } & \frac{A D}{D B}=\frac{A E}{E C} \\\Rightarrow & \frac{2 . C}{6.5}=\frac{3}{E C} \\\Rightarrow & E C=\frac{6.5 \times 3}{2.6} \\\Rightarrow & E C=\frac{15}{2}=7.5 \\& \text { So } E C=7.5 \mathrm{~cm}\end{aligned}$
(d) In $\triangle A B C, D E \| B C$
SO $\triangle A B C ~ \triangle A D E$
So $\begin{aligned} & \frac{A D}{A D}=\frac{A E}{A C} \\=& \frac{2}{6}=\frac{A E}{9} \\ \Rightarrow & A E=\frac{2 \times 9}{6}=3 \end{aligned}$
So $C E=A C-A E=9-3=6 \mathrm{~cm}$
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