Exercise 12 C
Question 1
Ans: $\triangle A B C, A B=6 \mathrm{~cm}, A C=3 \mathrm{~cm}$
M is mid point of $A B$
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So $A M=\frac{1}{2} A B=\frac{1}{2} \times 6=3 \mathrm{~cm}$
$M N \| B C$ is drown
So $N$ is mid-point of $A C$
So $A N=\frac{1}{2} A C=\frac{1}{2} \times 3=\frac{3}{2}=1.5 \mathrm{~cm}$
Question 2
Ans: Parallelogram $A B C D A B=6 \mathrm{~cm} A D=5 \mathrm{~cm}$
Diagonals $A C$ and $B D$ intersect each other at $O$
$E$ is mid-point as $B C$, $O E$ is Joined
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To prove:
(i)$D E \| A B \quad$ (ii) $D E=\frac{1}{2} A B$
Proof: In $\triangle A B C$
O is mid-point of $A C$
and $E$ is the mid-point of $B C$
So $O E \angle A B$ and $O E=\frac{1}{2} A B$
Hence proved
Question 3
Ans: ABC is a triangle , D is mid point of AE and E is mid point of AC
(i) If BC = 6cm find DE
(ii) IF $\angle D B C=140^{\circ}$ find $\angle A B E$
If D and E are the mid point of $\triangle A B C$
(i) SO $D E \| B C$ and $D E=\frac{1}{2} B C$
and $\triangle A B C \sim \triangle A D E$
So $\begin{aligned} & \frac{A D}{A B}=\frac{D E}{B C} \Rightarrow \frac{1}{2} \frac{A B}{A B}=\frac{D E}{6} \\ \Rightarrow & \frac{D E}{6}=\frac{1}{2} \\ \Rightarrow & D E=\frac{6}{2}=3 \mathrm{~cm} \end{aligned}$
(ii) $\angle D B C=140^{\circ} .$
But $\angle A D E=\angle D B C$
$\begin{aligned}&=140^{\circ} \\&\text { So } \angle A D E^{\circ}=140^{\circ}\end{aligned}$
Question 4
Ans: In trapezium $A B C D$
$A B \| D C$
$A D=B C=4 \mathrm{~cm}$
$\angle D A B=\angle C B A=60^{\circ}$
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(i) the length of $C D$
(ii) distance between $A B$ and $C D$
Draw DL and CM perpendicular on AB in $\triangle A D L$ and $\triangle B C M$
$\angle L=\angle M$
$\angle A=\angle B$
$D A=C B$
$\begin{aligned}\text { So } \triangle A D L & \cong M B C m \\\text { So } A L &=M B\end{aligned}$
Now in right $\triangle A D L$
since $\frac{D L}{A D} \Rightarrow \sin 60^{\circ}=\frac{D L}{4}$
$\Rightarrow \frac{3}{2}=\frac{D L}{4} \Rightarrow D L=\frac{4 \times \sqrt{3}}{2}=2 \sqrt{3} \mathrm{~cm}$
and $\tan 60^{\circ}=\frac{D L}{A L}=\sqrt{3}=\frac{2 \sqrt{3}}{A L}$
$\Rightarrow A L=\frac{2 \sqrt{3}}{\sqrt{3}}=2$
So MB = AL = 2cm
But CD = LM= AB -AL-MB
$=10-2-2=10-4=6 \mathrm{~cm}$
Question 5
Ans: (i) In ||gm ABCD , AB||CD
Diagonals AC and BD bisects each other at K now in Triangle ABC
if E is mid point of AD and K is mid point of AD
So EK or EKF ||CD or AB
$\begin{aligned} & \frac{A E}{A D}=\frac{A K}{A C}=\frac{E K}{C D} \\ \Rightarrow & \frac{6}{6+6}=\frac{8}{8+8}=\frac{5}{y} \\ \Rightarrow & \frac{5}{y}=\frac{6}{12} \Rightarrow \frac{5}{y}=\frac{1}{2} \Rightarrow y=5 \times 2=10 \end{aligned}$
So $y=10 \mathrm{~cm}$
similarly in $\triangle A C B$.
$K E \| A B$
So $\begin{aligned} \frac{C K}{C A} &=\frac{K F}{A B} \Rightarrow \frac{8}{16}=\frac{x}{10} \\ \Rightarrow x &=\frac{8 \times 10}{16}=5 \end{aligned}$
Hence $x=5 \mathrm{~cm}, y=10 \mathrm{~cm}$
(ii)In the figure, PQRS is a parallelogram
$\begin{aligned}&O L=L T=6, S T=4, \angle S=3 \\&P Q=y \text { and } P L=x\end{aligned}$
we have to find the value of $x$ and $y$ in $\triangle Q T R$
L is mid point of QT So QL = LT =6cm and PLS|| QR
So S is mid point of TR
So TS= SR = SR = 4
But PQ = SR
So PQ = 4 or y = 4
In $\triangle P L Q$ and $\triangle S L T$
$\begin{aligned}L Q &=L T \\P Q &=T S \\\angle P L Q &=\angle S L T \\\text { So } \triangle P L Q & \cong \triangle S L T \\\text { So } P L &=L S \\\Rightarrow x &=3\end{aligned}$
Hence $x=3, y=4$
(iii)In $\triangle A B D$.
$A L=L D=3$
So $L$ is mid-point of $A D$
if $\quad L M \| B D$
So $m$ is mid -point of $A B$
So $\triangle A L M \backsim \triangle A D B$
So $\begin{aligned} & \frac{A L}{A D}=\frac{L m}{B D} \Rightarrow \frac{3}{6}=\frac{9}{p} \\ \Rightarrow & \frac{q}{p}=\frac{1}{2} \end{aligned}$
$\Rightarrow \quad 2 9=p$..........(i)
Similarly in $\triangle B C D$
Question 6
Ans: $\triangle A B C$ is an isoscelies triangle in which $A B=A C=10 \mathrm{~cm}$ $B C=12 \mathrm{~cm}$
$P Q R S$ is a Rectangle inside the isosceles triangle $P Q=58$ $=y \mathrm{~cm}$ and $P S=Q R=2 x \mathrm{~cm}$
To prove:
Construction: From A, draw AL ⊥ BC
Which Bisects BC at L
Proof: If L is mid point of BC
So $B L=L C=\frac{12}{2}=6 \mathrm{~cm}$ Now in rignt $\triangle A B L$
$\begin{aligned}& A B^{2}=A L^{2}+B L^{2} \\\Rightarrow &(10)^{2}=A L^{2}+(6)^{2} \\\Rightarrow & A L^{2}=(10)^{2}-(6)^{2} \\\Rightarrow & A L^{2}=100-36=64=8^{2}\end{aligned}$
So $A L=8 \mathrm{~cm}$
Now in $\triangle B P Q$ and $\triangle B A L$
$\begin{aligned}&\angle Q=\angle L \\&\angle B=\angle B\end{aligned}$
So $\triangle B P Q$∽ $\triangle B A L$
$\begin{aligned} & \frac{P Q}{A L}=\frac{B Q}{B C} \Rightarrow \frac{y}{8}=\frac{6-x}{6} \\ \Rightarrow & 6 y=48-8 x \\ \Rightarrow & 8 x=48-6 y \end{aligned}$
$x=\frac{48-6 y}{8}=6-\frac{3 y}{4}$
Hence $x=6-\frac{3 y}{4}$
Question 7
Ans: In the figure PQ|| ST
To prove:
(i) $\triangle$ PQR and STR are similar
(ii) PR.RT = QR.RS
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proof :
(a) In $\triangle P Q R$ and $\triangle$ STR
$\begin{aligned}&\angle P R Q=\angle S R T \\&\angle Q P R=\angle R S T\end{aligned}$
So $\triangle P Q R ~ \triangle S T R$
(b) $\begin{aligned}& \frac{P R}{R S}=\frac{R Q}{R T} \\\Rightarrow & P R \times R T=Q R \times R S\end{aligned}$
Hence proved
Question 8
Ans: ABCD is a trapezium in which AB||DC and diagonals AC and BD intersect each other at E.
To prove: AE.DF = BE.CE
Proof: In $\triangle A E B$ and $\triangle C E D$
$\angle A E B=\angle C E D$
$\angle E A B=\angle E C D$
and $\angle E P A=\angle E D C$
So $\triangle A E B \backsim \triangle C E D$
So $\frac{A E}{C E}=\frac{B E}{D E}$
So $A E \cdot D E=B E \cdot C E$
Question 9
Ans: In $\triangle A B C$
$A L \perp B C$ and $B M \perp A C$ are drawn To prove:
To prove:
(i) $\triangle A L C \sim \triangle B M C$
(ii) $\mathrm{Cm} \cdot \mathrm{CA}=\mathrm{Cl} \cdot \mathrm{CB}$
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proof: In $\triangle A L C$ and $\triangle B M C$
$\begin{aligned}\angle L &=\angle m \\\angle C &=\angle C\end{aligned}$
So $\triangle A L C ~ \triangle B M C$
So $\frac{\mathrm{CM}}{\mathrm{CL}}=\frac{\mathrm{CB}}{\mathrm{CA}}$
$\begin{aligned}&\Rightarrow \mathrm{Cm} \cdot \mathrm{CA}=\mathrm{CB} \cdot \mathrm{CL} \\&\Rightarrow \mathrm{CM} \cdot \mathrm{CA}=\mathrm{CL} \cdot \mathrm{CB}\end{aligned}$
Hence proved
Question 10
Ans: $\triangle A B C$ is a right angled Triangle $A D \perp B C$ is drawn
To prove:
(i) $\triangle A B D \backsim \triangle A B C$
(ii) $\triangle A C D=\triangle A B C$
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proof :
(i) In $\triangle A B D$ and $\triangle B C$
$\begin{aligned}&\angle B=\angle B \\&\angle A D B=\angle B A C \\&\text { SO } \triangle A B D \sim \triangle A B C\end{aligned}$
(ii) Similarly in $\triangle A C D$ and $\triangle A B C$
$\angle C=\angle C$
$\angle A D C=\angle B A C$
So $\triangle A C D ∽ \triangle A B C$
Hence proved
Question 11
Ans:
$\begin{aligned} \angle B A D &=\angle C A E \\ \angle B &=\angle D \end{aligned}$
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To prove:
(i) $\frac{A B}{A D}=\frac{A C}{A E}$
(ii) $\angle A D B=\angle A E C$
Proof: $\angle B A D=\angle C A E$
Adding $\angle C A D$ both sides.
$\begin{aligned} & \angle B A D+\angle C A D=\angle C A D+\angle C A E \\ \Rightarrow & \angle B A C=\angle D A E \end{aligned}$
Now in $\triangle A B C$ and $\triangle A D E$
$\begin{aligned} \angle B A C &=\angle D A E \\ \angle B &=\angle D \end{aligned}$
(i) So $\triangle A B C \backsim \triangle A D E$
So $\frac{A B}{A D}=\frac{A C}{A E}$
or $\frac{A B}{A C}=\frac{A D}{A E}$
(ii) Now in $\triangle A B D$ and $\triangle A E C$
$\begin{aligned}&\angle B A D=\angle C A E \\&\frac{A B}{A C}=\frac{A D}{A E} \Rightarrow A\end{aligned}$
So $\triangle A B D \sim \triangle A E C$
So $\angle A D B=\angle A E C$ Hence proved
Question 12
Ans: In the figure $A B C D$ and $A E F G$ are two squares
To prove
(i) AF : $A C=A C: A D$
(ii) triangles ACF and ADG are similar
proof : If AC and AF are the diagonals of two squares ABCD and AEFG respectively
So $\triangle A D C \sim \triangle A G F$
Now $\angle B A F=\angle B A C-\angle F A C$
$\angle G A C=\angle G A F-\angle F A C$
But $\angle B A C=\angle G A F=45^{\circ}$
(Diagonals bisects the opposite angles of a square)
So $\angle B A F=45^{\circ}-\angle F A C$ .........(i)
and $\angle G A C=45^{\circ}-\angle F A C$...........(ii)
From (i) and (ii)
So $\angle B A F=\angle G A C$
Similarly $\angle D A G=\angle D A C-\angle G A C=45^{\circ}-\angle G A C$
and $\angle F A C=\angle B A C-\angle B A F=45^{\circ}-\angle B A F$
But $\angle G A C=\angle B A F$
So $\angle D A G=\angle F A C$
Now in $\triangle A D G$ and $\triangle A C F$
$\angle D A G=\angle F A C$
So From (A)
$\triangle A D G \sim \triangle A C F$
So
$\begin{aligned} & \frac{A C}{A F}=\frac{A D}{A C} \\ \Rightarrow & \frac{A F}{A G}=\frac{A C}{A D} \end{aligned}$
$\Rightarrow A F: A G=A C: A D$ Hence proved
Question 13
Ans: In a square $A B C D$ diagonals $A C$ and $B D$ bisect each other at O
Bisector of $\angle C A B$ meets $B D$ is $X$ and $B C$
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To prove: $\triangle A CY \sim \triangle A B X$
$\angle C A Y=\angle B A Y$
$\angle A C Y=\angle A B X$
So $\triangle A C Y \sim \triangle A B X$ Hence proved
Question 14
Ans: In rectangle PQRS , X is a point on PQ such that $SX^{2}$
$=P x \cdot P Q$
To prove: $\Delta P \times S \sim \triangle X S R$
Proof: If $S \times{ }^{2}=P \times \cdot P Q$
$\Rightarrow \frac{s x}{p Q}=\frac{p x}{s x} \Rightarrow \frac{s x}{s R}=\frac{p x}{s x}$
Now In $\Delta$ PXS and $\Delta X S R$
$\frac{S X}{S R}=\frac{P X}{S X}$
$\angle PX S=\angle R S X$
So $\triangle$ PXS $\sim$ △XSR Hence proved
Question 15
Ans: $\triangle A B C, D$ is a point on BC such that $\angle A D C=\angle B A C$
To prove $: \frac{B C}{C A}=\frac{C A}{C D}$
Proof: In ISABC and $\triangle A D C$
$\begin{aligned}&\angle B A C=\angle A D C \\&\angle C=\angle C \\&\text { SO } \triangle A B C \sim \triangle A D C \\&\text { SO } \frac{B C}{A C}=\frac{A C}{C D} \\&\Rightarrow \frac{B C}{C A}=\frac{C A}{C D}\end{aligned}$
Question 16
Ans: Given: In $\triangle P Q R, P Q=P R$
Z is a point on PR such that Q R^{2}=P R \times R Z
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To prove: $Q Z=Q R$
Proof: $\quad Q R^{2}=P R \times R Z$
$\Rightarrow \frac{P R}{Q R}=\frac{Q R}{R Z}$
Now In $\triangle B Q R$ and $\triangle Q R Z$
$\frac{P R}{Q R}=\frac{Q R}{R Z}$
$\angle R=\angle R$
So $\triangle P O R \sim \triangle Q R Z$
$\frac{Q R}{R Z}=\frac{P R}{Q R}=\frac{P Q}{Q Z}$
But $\frac{P Q}{Q Z}=\frac{P R}{Q Z}$
So $\frac{P R}{Q R}=\frac{P R}{Q Z} \Rightarrow Q R=Q Z$
or QZ = QR Hence proved
Question 17
Ans: In $\triangle A B C, A D$ and $B E$ are perpendicular to the sides $B C$ and $A C$ respectively
To prove:
(i) $\triangle A D C \backsim \triangle B E C$.
(ii) $C A \cdot C E=C B \cdot C D$
(iii) $\triangle A B C \backsim \triangle D E C$
(iv) $C D \cdot A B=C A \cdot D E$
Proof:
(i) In $\triangle A D C$ and $\triangle B E C$
$\begin{aligned}&\angle C=\angle C \\&\angle A D C=\angle B E C\end{aligned}$
So $\triangle A D C \backsim \triangle B E C$
So $\frac{C A}{C B}=\frac{C D}{C E}$
(ii) So $C A \cdot C E=C B \cdot C D$
Again in $\triangle A B C$ and $\triangle D E C$
$\angle C=\angle C$
$\frac{C A}{C D}=\frac{C B}{C E}$
So $\triangle A B C \sim \triangle D E C$
SO $\frac{C B}{C D}=\frac{A B}{D E}$
(iv) $\begin{aligned} \Rightarrow & C A \cdot D E=C D \cdot A B \\ & \operatorname{or} C D \cdot A B=C A \cdot D E \end{aligned}$
Hence proved
Question 18
Ans: In quadrilateral ABCD
PQR and S are the points of trisection of the sides AB, BC ,CD and DA respectively and are adjacent to A and C
To Prove: PQRS is a parallelogram
construction: Join diagonals $A C$
proof: In $\triangle A D C$
$A P=\frac{1}{3} A B$ and $C Q=\frac{1}{3} B C$ or $\frac{A P}{A B}=\frac{C Q}{B C}$ $=\frac{1}{3}$
So $P Q \| A C$ and $P Q=\frac{2}{3} A C$ ............(i)
Similarly in $\triangle A D C$
$A S=\frac{1}{3} A D$ and $C R=\frac{1}{3} C D$ or $\frac{A S}{A D}=\frac{C R}{C D}$
$=\frac{2}{3}$
So
RS $||A C$ and $R S=\frac{2}{3} \mathrm{AC}$........(ii)
from (i) and (ii)
PQRS is a parallelogram Hence proved
Question 19
Ans: In $\triangle A B C, A D \perp B C$
$B D=4, A D=6, C D=9$
To prove :
(i) $\triangle A D B \sim \triangle C O A$
(ii) $\angle B A C$ is a right angle
Proof : In $\triangle A D B$ and $\triangle C D A$
$\begin{aligned}\therefore & \angle A D B=\angle A D C \\& \frac{A D}{B D}=\frac{6}{4}, \frac{C D}{A D}=\frac{9}{6} \\\Rightarrow & \frac{A D}{B D}=\frac{3}{2} \text { and } \frac{C D}{A D}=\frac{3}{2}\end{aligned}$
So $\frac{A D}{B D}=\frac{C D}{A D}$
So $\triangle A D B \backsim \triangle C P A$
So $\angle A B D=\angle C A D$ and $\angle D A D=\angle A C D$
So $\angle B A D+\angle C A B=\angle A B D+\angle A C D$
$\Rightarrow \angle B A C=\angle A B D+\angle A C D$
But $\angle B A C+\angle A B D+\angle A C D=180^{\circ}=\angle D A C=90^{\circ}$
Question 20
Ans: $\triangle A B C \backsim \triangle D E F$
$A L \perp B C$ and $D M \perp E F$
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To prove: $\frac{A B}{D E}=\frac{B C}{E F}=\frac{A C}{D F}=\frac{A L}{D M}$
Proof: if $\triangle A B C \sim \triangle D E F$
So $\angle B=\angle E$ and $\frac{A B}{D E}=\frac{B C}{E F}$.............(i)
Now in $\triangle A B C$ and $\triangle D E M$
$\angle B=\angle E$
$\angle L=\angle M$
$\triangle A B L$ and $\triangle D E M$
So $\frac{A B}{D E}=\frac{A L}{D M}$...........(ii)
From (i) and (ii)
$\frac{A B}{D E}=\frac{B C}{E F}=\frac{A L}{D M}$ Hence proved
Question 21
Ans: In parallelogram $A B C D$
$B D$ is diagonal and $E$ is any point on $B C$
$A E$ is Joined which intersect $B D$ at $F$
To prove: DF.EF $=F B \cdot F A$
proof: In$\triangle AFD and $\triangle B F E$
$\begin{aligned}&\angle A F D=\angle B F E \\&\angle A D F=\angle F B E\end{aligned}$
So $\triangle A F D \sim \triangle B F E$
So $\frac{D F}{F B}=\frac{F A}{F E}$
So $D F \cdot E F=F B \cdot FA$
Question 22
Ans: X is any point in $\triangle D E F$
XD , XE and SF are joined . P is a point on DX , PQ || DE is drawn meeting XE in Q and QR||EF is drawn meeting XF in R
To prove: PR||DF
Proof: If P is any point on DX and PQ||DE
So $\frac{X P}{P D}=\frac{X Q}{Q E}$......(i)
if $Q R \| E F$
So $\frac{X Q}{Q E}=\frac{X R}{R F}$
from (i)and (ii)
$\frac{X P}{P D}=\frac{X R}{R F}$
So $P R \| D F$ Hence proved
