Chapter 06 Triangles Exercise-03
EXERCISE 6.3
1. State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
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Solution
(i) $\angle A=\angle P=60^{\circ}$
$\angle B=\angle Q=80^{\circ}$
$\angle C=\angle R=40^{\circ}$
Therefore, $\triangle ABC \tilde{A} \phi \ddot{E} \dagger \hat{A} 1 \dfrac{1}{4} \Delta PQR$ [By AAA similarity criterion]
$ \dfrac{AB}{QR}=\dfrac{BC}{RP}=\dfrac{CA}{PQ} $
(ii)
$\therefore \triangle ABC \sim \triangle QRP$
[By SSS similarity criterion]
(iii)The given triangles are not similar as the corresponding sides are not proportional.
$MNQP=MLQR=12$
2. In Fig. 6.35, $\triangle ODC \sim \Delta OBA, \angle BOC=125^{\circ}$ and $\angle CDO=70^{\circ}$. Find $\angle DOC, \angle DCO$ and $\angle OAB$.
Fig. 6.35
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Solution
DOB is a straight line.
$\therefore \angle DOC+\angle COB=180^{\circ}$
$\Rightarrow \angle DOC=180^{\circ}-125^{\circ}$
$=55^{\circ}$
In $\triangle DOC$,
$\angle DCO+\angle CDO+\angle DOC=180^{\circ}$
(Sum of the measures of the angles of a triangle is $180^{\circ}$.)
$\Rightarrow \angle DCO+70^{\circ}+55^{\circ}=180^{\circ}$
$\Rightarrow \angle DCO=55^{\circ}$
It is given that $\triangle ODC \angle \hat{A} 1 / 4 \triangle OBA$.
$\therefore \angle OAB=\angle OCD$ [Corresponding angles are equal in similar triangles.]
$\Rightarrow \angle OAB=55^{\circ}$
3. Diagonals $A C$ and $B D$ of a trapezium $A B C D$ with $A B|| D C$ intersect each other at the point O. Using a similarity criterion for two triangles, show that $\dfrac{OA}{OC}=\dfrac{OB}{OD}$.
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#missing4. In Fig. 6.36, $\dfrac{QR}{QS}=\dfrac{QT}{PR}$ and $\angle 1=\angle 2$. Show that $\Delta PQS \sim \Delta TQR$.
Fig. 6.36
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Solution
In $\triangle PQR, \angle PQR=\angle PRQ$
$\therefore PQ=PR(i)$
Given, $\dfrac{QR}{QS}=\dfrac{QT}{PR}$
Using $(i)$, we obtain
$\dfrac{QR}{QS}=\dfrac{QT}{QP}$
In $\triangle PQS$ and $\triangle TQR$.
$ \begin{aligned} & \dfrac{QR}{QS}=\dfrac{QT}{QP} \quad[\text{ Using }(i i)] \\ & \angle Q=\angle Q \\ & \therefore \triangle PQS \sim \triangle TQR \quad \text{ [SAS similarity criterion] } \end{aligned} $
5. $\quad S$ and $T$ are points on sides $PR$ and $QR$ of $\triangle PQR$ such that $\angle P=\angle RTS$. Show that $\Delta RPQ \sim \Delta RTS$.
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Solution
In $\triangle RPQ$ and $\triangle RST$,
$\angle RTS=\angle QPS$ (Given)
$\angle R=\angle R$ (Common angle)
$\therefore \Delta RPQ \propto 1 / 4 \Delta RTS$ (By AA similarity criterion)
6. In Fig. 6.37, if $\Delta \mathrm{ABE} \cong \triangle \mathrm{ACD}$, show that $\triangle \mathrm{ADE} \sim \triangle \mathrm{ABC}$.
Fig. 6.37
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Solution
It is given that $\triangle ABE \cong \triangle ACD$.
$\therefore \text{AB=AC[By CPCT] (1) } $
And, $ \text{AD=AE[By CPCT]}(2) $
In $\triangle ADE$ and $\triangle ABC$,
$\dfrac{AD}{AB}=\dfrac{AE}{AC}$
[Dividing equation (2) by (1)]
$\angle A=\angle A[$ Common angle]
$\therefore \triangle ADE \sim \Delta ABC$ [By SAS similarity criterion]
7. In Fig. 6.38, altitudes $\mathrm{AD}$ and $\mathrm{CE}$ of $\triangle \mathrm{ABC}$ intersect each other at the point $P$. Show that:
Fig. 6.38
(i) $\triangle \mathrm{AEP} \sim \triangle \mathrm{CDP}$
(ii) $\triangle \mathrm{ABD} \sim \Delta \mathrm{CBE}$
(iii) $\triangle \mathrm{AEP} \sim \triangle \mathrm{ADB}$
(iv) $\triangle \mathrm{PDC} \sim \Delta \mathrm{BEC}$
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Solution
(i)
In $\triangle AEP$ and $\triangle CDP$,
$\angle AEP=\angle CDP($ Each 90 $)$
$\angle APE=\angle CPD$ (Vertically opposite angles)
Hence, by using AA similarity criterion,
$\triangle AEP \propto 1 / 4 \Delta CDP$
(ii)
In $\triangle ABD$ and $\triangle CBE$,
$\angle ADB=\angle CEB(.$ Each $.90^{\circ})$
$\angle ABD=\angle CBE$ (Common)
Hence, by using AA similarity criterion,
$\triangle ABD \propto 1 / 4 \triangle CBE$
(iii)
In $\triangle AEP$ and $\triangle ADB$,
$\angle AEP=\angle ADB(.$ Each $.90^{\circ})$
$\angle PAE=\angle DAB$ (Common)
Hence, by using AA similarity criterion,
$\triangle AEP \propto 1 / 4 \Delta ADB$
(iv)
In $\triangle PDC$ and $\triangle BEC$,
$\angle PDC=\angle BEC(.$ Each $.90^{\circ})$
$\angle PCD=\angle BCE$ (Common angle)
Hence, by using AA similarity criterion,
$\triangle PDC \propto 1 / 4 \triangle BEC$
8. $\mathrm{E}$ is a point on the side $\mathrm{AD}$ produced of a parallelogram $\mathrm{ABCD}$ and $\mathrm{BE}$ intersects $\mathrm{CD}$ at $\mathrm{F}$. Show that $\triangle \mathrm{ABE} \sim \Delta \mathrm{CFB}$.
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Solution
In $\triangle ABE$ and $\triangle CFB$,
$\angle A=\angle C$ (Opposite angles of a parallelogram)
$\angle AEB=\angle CBF$ (Alternate interior angles as $AE \| BC$ )
$\therefore \triangle ABE \propto 1 / 4 \triangle CFB$ (By AA similarity criterion)
9. In Fig. 6.39, $\mathrm{ABC}$ and $\mathrm{AMP}$ are two right triangles, right angled at $B$ and $M$ respectively. Prove that:
Fig. 6.39
(i) $\triangle \mathrm{ABC} \sim \Delta \mathrm{AMP}$
(ii) $\dfrac{\mathrm{CA}}{\mathrm{PA}}=\dfrac{\mathrm{BC}}{\mathrm{MP}}$
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Solution
In $\triangle ABC$ and $\triangle AMP$,
$\angle ABC=\angle AMP(.$ Each $.90^{\circ})$
$\angle A=\angle A$ (Common)
$\therefore \triangle ABC \tilde{A} \phi \ddot{E}+\hat{A} 1 / 4 \Delta AMP$ (By AA similarity criterion)
$\Rightarrow \dfrac{CA}{PA}=\dfrac{BC}{MP} \quad$ (Corresponding sides of similar triangles are proportional)
10. $\mathrm{CD}$ and GH are respectively the bisectors of $\angle \mathrm{ACB}$ and $\angle \mathrm{EGF}$ such that $\mathrm{D}$ and $\mathrm{H}$ lie on sides $\mathrm{AB}$ and $\mathrm{FE}$ of $\triangle \mathrm{ABC}$ and $\triangle \mathrm{EFG}$ respectively. If $\triangle \mathrm{ABC} \sim \Delta \mathrm{FEG}$, show that:
(i) $\dfrac{\mathrm{CD}}{\mathrm{GH}}=\dfrac{\mathrm{AC}}{\mathrm{FG}}$
(ii) $\triangle \mathrm{DCB} \sim \Delta \mathrm{HGE}$
(iii) $\Delta \mathrm{DCA} \sim \Delta \mathrm{HGF}$
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Solution
Answer :
It is given that $\triangle ABC \tilde{A} \phi \ddot{E} \dagger \hat{A} 114 \Delta FEG$.
$\therefore \angle A=\angle F, \angle B=\angle E$, and $\angle ACB=\angle FGE$ $\angle ACB=\angle FGE$
$\therefore \angle ACD=\angle FGH$ (Angle bisector)
And, $\angle DCB=\angle HGE$ (Angle bisector)
In $\triangle ACD$ and $\triangle FGH$,
$\angle A=\angle F$ (Proved above)
$\angle ACD=\angle FGH$ (Proved above)
$\therefore \triangle ACD \sim \Delta FGH$ (By AA similarity criterion)
$\Rightarrow \dfrac{CD}{GH}=\dfrac{AC}{FG}$
In $\triangle DCB$ and $\triangle HGE$,
$\angle DCB=\angle HGE$ (Proved above)
$\angle B=\angle E$ (Proved above)
$\therefore \triangle DCB \tilde{A} \phi \ddot{E}+A^{1} / 4 \Delta HGE$ (By AA similarity criterion)
In $\triangle DCA$ and $\triangle HGF$,
$\angle ACD=\angle FGH$ (Proved above)
$\angle A=\angle F$ (Proved above)
$\therefore \triangle DCA \sim \Delta HGF$ (By AA similarity criterion)
11. In Fig. 6.40, $E$ is a point on side $CB$ produced of an isosceles triangle $A B C$ with $AB=AC$. If $AD \perp BC$ and $EF \perp AC$, prove that $\triangle ABD \sim \triangle ECF$.
Fig. 6.40
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Solution
It is given that $A B C$ is an isosceles triangle.
$\therefore AB=AC$
$\Rightarrow \angle ABD=\angle ECF$
In $\triangle ABD$ and $\triangle ECF$, $\angle ADB=\angle EFC(.$ Each $.90^{\circ})$
$\angle BAD=\angle CEF$ (Proved above)
$\therefore \Delta ABD \sim \Delta ECF$ (By using $AA$ similarity criterion)
12. Sides $A B$ and $B C$ and median $A D$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $QR$ and median $PM$ of $\triangle PQR$ (see Fig. 6.41). Show that $\triangle ABC \sim \Delta PQR$.
Fig. 6.41
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Solution
Median divides the opposite side.
$ \therefore \quad BD=\dfrac{BC}{2} \text{ and } QM=\dfrac{QR}{2} $
Given that,
$ \dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{AD}{PM} $
$\Rightarrow \dfrac{AB}{PQ}=\dfrac{\dfrac{1}{2} BC}{\dfrac{1}{2} QR}=\dfrac{AD}{PM}$
$\Rightarrow \dfrac{AB}{PQ}=\dfrac{BD}{QM}=\dfrac{AD}{PM}$
In $\triangle ABD$ and $\triangle PQM$,
$ \begin{aligned} & \dfrac{AB}{PQ}=\dfrac{BD}{QM}=\dfrac{AD}{PM} \\ & \therefore \triangle ABD \tilde{A} \phi \ddot{E} \dagger \hat{A} 1 / 4 \Delta PQM \text{ (By SSS similarity criterion) } \\ & \Rightarrow \angle ABD=\angle PQM \text{ (Corresponding angles of similar triangles) } \end{aligned} $
In $\triangle A B C$ and $\triangle P Q R$,
$\angle ABD=\angle PQM$ (Proved above) $\dfrac{AB}{PQ}=\dfrac{BC}{QR}$
$\therefore \triangle ABC \tilde{A} \phi \ddot{E}+\hat{A} 11 / 4 \triangle PQR$ (By SAS similarity criterion)
13. $D$ is a point on the side $BC$ of a triangle $ABC$ such that $\angle ADC=\angle BAC$. Show that $CA^{2}=CB . CD$.
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Solution
In $\triangle ADC$ and $\triangle BAC$,
$\angle ADC=\angle BAC$ (Given)
$\angle ACD=\angle BCA$ (Common angle)
$\therefore \triangle ADC \tilde{A} \phi \ddot{E}+\hat{A} 11 / 4 \Delta BAC$ (By AA similarity criterion)
We know that corresponding sides of similar triangles are in proportion.
$\therefore \dfrac{CA}{CB}=\dfrac{CD}{CA}$
$\Rightarrow CA^{2}=CB \times CD$
14. Sides $A B$ and $A C$ and median $A D$ of a triangle $A B C$ are respectively proportional to sides $PQ$ and $PR$ and median $PM$ of another triangle $PQR$. Show that $\triangle ABC \sim \Delta PQR$.
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Solution
Given that,
$ \dfrac{AB}{PQ}=\dfrac{AC}{PR}=\dfrac{AD}{PM} $
Let us extend $A D$ and $P M$ up to point $E$ and $L$ respectively, such that $A D=D E$ and $P M=M L$. Then, join $B$ to $E, C$ to $E, Q$ to $L$, and $R$ to $L$.
We know that medians divide opposite sides.
Therefore, $BD=DC$ and $QM=MR$
Also, $AD=DE$ (By construction)
And, $PM=ML$ (By construction)
In quadrilateral $A B E C$, diagonals $A E$ and $B C$ bisect each other at point $D$.
Therefore, quadrilateral ABEC is a parallelogram.
$\therefore A C=B E$ and $A B=E C$ (Opposite sides of a parallelogram are equal)
Similarly, we can prove that quadrilateral $PQLR$ is a parallelogram and $PR=QL, PQ=LR$
It was given that
$ \begin{aligned} & \dfrac{A B}{P Q}=\dfrac{A C}{P R}=\dfrac{A D}{P M} \\ & \Rightarrow \dfrac{A B}{P Q}=\dfrac{B E}{Q L}=\dfrac{2 A D}{2 P M} \\ & \Rightarrow \dfrac{A B}{P Q}=\dfrac{B E}{Q L}=\dfrac{A E}{P L} \end{aligned} $
$\therefore \triangle ABE \tilde{A} \phi \ddot{E} \dagger \hat{A} 1 / 4 \Delta PQL$ (By SSS similarity criterion)
We know that corresponding angles of similar triangles are equal.
$\therefore \angle BAE=\angle QPL$.
Similarly, it can be proved that $\triangle AEC \tilde{A} \phi \ddot{E} \dagger \hat{A} 11 / 4 \Delta PLR$ and
$\angle CAE=\angle RPL \ldots$
Adding equation (1) and (2), we obtain
$\angle BAE+\angle CAE=\angle QPL+\angle RPL$
$\Rightarrow \angle CAB=\angle RPQ$
In $\triangle ABC$ and $\triangle PQR$,
$\dfrac{AB}{PQ}=\dfrac{AC}{PR}$
(Given)
$\angle CAB=\angle RPQ$ [Using equation (3)]
$\therefore \triangle ABC \tilde{A} \phi \ddot{E} \dagger \hat{A} 1 \dfrac{1}{4} \Delta PQR$ (By SAS similarity criterion)
15. A vertical pole of length $6 m$ casts a shadow $4 m$ long on the ground and at the same time a tower casts a shadow $28 m$ long. Find the height of the tower.
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Solution
Let $A B$ and $C D$ be a tower and a pole respectively.
Let the shadow of $B E$ and $D F$ be the shadow of $A B$ and $C D$ respectively.
At the same time, the light rays from the sun will fall on the tower and the pole at the same angle.
Therefore, $\angle DCF=\angle BAE$
And, $\angle DFC=\angle BEA$
$\angle CDF=\angle ABE$ (Tower and pole are vertical to the ground)
$\therefore \triangle ABE \sim \Delta CDF$ (AAA similarity criterion) $\Rightarrow \dfrac{AB}{CD}=\dfrac{BE}{DF}$
$\Rightarrow \dfrac{AB}{6 m}=\dfrac{28}{4}$
$\Rightarrow AB=42 m$
Therefore, the height of the tower will be 42 metres.
16. If $A D$ and $P M$ are medians of triangles $A B C$ and $P Q R$, respectively where $\Delta ABC \sim \Delta PQR$, prove that $\dfrac{AB}{PQ}=\dfrac{AD}{PM}$.
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Solution
It is given that $\triangle A B C \tilde{A} \phi \ddot{E} \dagger \hat{A} 1 / 4 \triangle P Q R$
We know that the corresponding sides of similar triangles are in proportion.
$\therefore \dfrac{AB}{PQ}=\dfrac{AC}{PR}=\dfrac{BC}{QR}$
Also, $\angle A=\angle P, \angle B=\angle Q, \angle C=\angle R$
Since AD and PM are medians, they will divide their opposite sides.
$$ \begin{equation*} \therefore \quad BD=\dfrac{BC}{2} \text{ and } QM=\dfrac{QR}{2} \tag{3} \end{equation*} $$
From equations ( 1 ) and (3), we obtain
$$ \begin{equation*} \dfrac{AB}{PQ}=\dfrac{BD}{QM} \tag{4} \end{equation*} $$
In $\triangle ABD$ and $\triangle PQM$,
$\angle B=\angle Q[$ Using equation (2)] $\dfrac{AB}{PQ}=\dfrac{BD}{QM}$
$\therefore \triangle ABD \tilde{A} \phi \ddot{E}+\hat{A} 11 / 4 \triangle PQM$ (By SAS similarity criterion)
$\Rightarrow \dfrac{AB}{PQ}=\dfrac{BD}{QM}=\dfrac{AD}{PM}$
