2 Marks Questions - Maths STD 10 Questions - Vidyadip

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Question 12 Marks

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Answer

This theorem can be proved by taking a line $DE$ such that $\frac{ AD }{ DB }=\frac{ AE }{ EC }$ and assuming that $DE$ is not parallel to $BC$ $($see Fig. $6.12).$
If $DE$ is not parallel to $BC$, draw a line $DE ^{\prime}$ parallel to $BC$.
So,
$\frac{ AD }{ DB }=\frac{ AE ^{\prime}}{ E ^{\prime} C } \quad \text { (Why ?) } $
$\text{Therefore,} $
$\frac{ AE }{ EC }=\frac{ AE ^{\prime}}{ E ^{\prime} C } \quad \text { (Why?) }$
Adding $1$ to both sides of above, you can see that $E$ and $E ^{\prime}$ must coincide. $($Why $?)$
Let us take some examples to illustrate the use of the above theorems.

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Question 22 Marks

In the figure, $ABC$ and $AMP$ are two right triangles, right angled at $B$ and $M$ respectively. Prove that:

$\triangle ABC \sim \triangle AMP$
$\frac{{CA}}{{PA}} = \frac{{BC}}{{MP}}$

Answer

Given: In the figure, $ABC$ and $AMP$ are two right triangles, right angled at $B$ and $M$ respectively, To prove:

$\triangle $$ABC$ $ \sim $ $\triangle $$AMP$
$\frac{{CA}}{{PA}} = \frac{{BC}}{{MP}}$

Proof:

In $\triangle $ABC $ \sim $ $\triangle $AMP
$\angle$ABC = $\angle$$AMP (1) ........ [$Each equal to $90^\circ$$]$
$\angle$BAC=$\angle$$MAP (2).........[$Common angle$]$
In view of $(1)$ and $(2)$
$\triangle $$ABC$ $ \sim $ $\triangle $$AMP ..........AA$ similarity criterion
$\triangle $$ABC$ $ \sim $ $\triangle $$AMP.........$ Proved above in$(i)$
$\therefore $ $\frac{{CA}}{{PA}} = \frac{{BC}}{{MP}}$$.........$Corresponding sides of two similar triangles are proportional.

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Question 32 Marks

In the figure, $ABC$ and $AMP$ are two right triangles, right angled at $B$ and $M$ respectively. Prove that:

$\triangle ABC \sim \triangle AMP$
$\frac{{CA}}{{PA}} = \frac{{BC}}{{MP}}$

Answer

Given: In the figure, $ABC$ and $AMP$ are two right triangles, right angled at $B$ and $M$ respectively,
To prove:

$\triangle $ABC $ \sim $ $\triangle $AMP
$\frac{{CA}}{{PA}} = \frac{{BC}}{{MP}}$

Proof:

In $\triangle $$ABC$ $ \sim $ $\triangle $$AMP$
$\angle$ABC = $\angle$$AMP (1) ........ [$Each equal to $90^\circ$$]$
$\angle$BAC=$\angle$$MAP (2).........[$Common angle$]$
In view of $(1)$ and $(2)$
$\triangle $$ABC$ $ \sim $ $\triangle $$AMP ..........AA$ similarity criterion
$\triangle $$ABC$ $ \sim $ $\triangle $$AMP.........$Proved above in$(i)$
$\therefore $ $\frac{{CA}}{{PA}} = \frac{{BC}}{{MP}}$.........Corresponding sides of two similar triangles are proportional.

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Question 42 Marks

$E$ is a point on side $AD$ produced of a parallelogram $ABCD$ and $BE$ intersects $CD$ at $F$. Prove that $\Delta A B E \sim \Delta C F B.$

Answer

In $ \Delta$'s $ABE$ and $CFB$, we have

$ \angle$$AEB =$ $ \angle$$CBF [$Alternate angles$]$
$ \angle$$A =$ $ \angle$$C [$Opposite angles of a parallelogram$]$
Thus, by $AA$-criterion of similarity, we have,
$ \Delta A B E \sim \Delta C F B.$

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Question 52 Marks

$E$ is a point on side $AD$ produced of a parallelogram $ABCD$ and $BE$ intersects $CD$ at $F$. Prove that $\Delta A B E \sim \Delta C F B.$

Answer

In $\Delta$'s $ABE$ and $CFB$, we have

$\angle$ $AEB =\angle$ $CBF$ $[$Alternate angles$]$
$\angle$ $A$ =$\angle$ $C$ $[$Opposite angles of a parallelogram$]$
Thus, by $AA$-criterion of similarity, we have,
$\Delta A B E \sim \Delta C F B.$

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Question 62 Marks

In the figure, altitudes $AD$ and $CE$ of $\triangle$ $ABC$ intersect each other at the point $P$. Show that: $\vartriangle PDC \sim \vartriangle BEC$

Answer

In $\triangle $$PDC$ and $\triangle $$BEC$, we have
$\angle$$PDC =$$\angle$$BEC ..........(1) [$Each equal to $90^\circ$$]$
$\angle$$DCP =$ $\angle$$BEC ..........(2) [$Common angle$]$
In view of $(1)$ and $(2),$
$\triangle $$PDC$ $ \sim $ $\triangle $$BEC [AA$ similarity criterion$]$

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Question 72 Marks

In the figure, altitudes $AD$ and $CE$ of $\triangle$$ABC$ intersect each other at the point $P$. Show that: $\vartriangle AEP \sim \vartriangle ADB$

Answer

In $\triangle $AEP and $\triangle $$ADB$, we have
$AEP=$ $\angle$ $ADP .........(1) [$Each equal to $90^\circ$$]$
$\angle$EAP=$\angle$$DAB ..... (2) [$Common angle$]$
In view of $(1)$ and $(2),$
$\triangle $$AEP$$ \sim $$\triangle $$ADB [AA$ similarity criterion$]$

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Question 82 Marks

In the figure, altitudes $AD$ and $CE$ of $\triangle$$ABC$ intersect each other at the point $P$. Show that: $\vartriangle ABD \sim \vartriangle CBE$

Answer

In $\triangle $$ABD$ and $\triangle $$CBE, ....... (1) [$Each equal to $90^\circ$$]$
$\angle$$ABC =$ $\angle$$CBE ...... (2) [$Common angle$]$
In view of $(1)$ and $(2),$
$\triangle $$ABD$ $ \sim $ $\triangle $$CBE.......[AA$ similarity criterion$]$

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Question 92 Marks

In the figure, altitudes $AD$ and $CE$ of $\triangle$$ABC$ intersect each other at the point $P$. Show that: $\vartriangle AEP \sim \vartriangle CDP$

Answer

In $\triangle $$AEP$ and $\triangle $$CDP,$
$\angle$$AEP =$ $\angle$$CDF ...... (1) [$Each equal to $90^\circ$$]$
$\angle$$EPA =$ $\angle$$DPC ...... (2) [$vert.opp. $\angle$$s]$
In view of $(1)$ and $(2),$
$\triangle AEP \sim \triangle CDP$$ .........[AA$ similarity criterion$]$

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Question 102 Marks

In the figure, if $\triangle$$ABE$ $\cong$ $\triangle$$ACD$. Show that $\triangle$$ADE ~$ $\triangle$$ABC$

Answer

Given: In the figure, $\triangle ABE \cong \triangle ACD$
To prove: $\triangle ADE \sim \triangle ABC$
Proof:
$\because \triangle ABE \cong \triangle ACD$$........[$Given$]$
$\therefore $ $AB = AC........[CPCT]$
$AE = AD ........(1)$
Also, $\angle$ $DAE=$ $\angle$ $BAC.......[$Common $\angle$ $].......(2)$
In view of $(1)$ and $[SAS$ similarity criterion$]$

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Question 112 Marks

In the figure, if $\triangle$$ABE$ $\cong$ $\triangle$$ACD$. Show that $\triangle$$ADE ~$ $\triangle$$ABC$.

Answer

Given: In the figure, $\triangle ABE \cong \triangle ACD$
To prove: $\triangle ADE \sim \triangle ABC$
Proof:
$\because \triangle ABE \cong \triangle ACD$$........[$Given$]$
$\therefore$ $AB = AC........[CPCT]$
$AE = AD ........(1)$
Also, $\angle$ $DAE=$ $\angle$ $BAC.......[$Common $\angle].......(2)$
In view of $(1)$ and $[SAS$ similarity criterion$]$

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Question 122 Marks

$S$ and $T$ are points on sides $PR$ and $QR$ of $\triangle PQR$ such that $\angle P = \angle R T S$. Show that $\triangle R P Q \sim \triangle R T S$.

Answer

According to questions it is given that $S$ and $T$ are points on sides $PR$ and $QR$ of $\triangle PQR$ such that $\angle P = \angle R T S$

To Prove $\triangle R P Q \sim \triangle R T S$
Proof In $\triangle RPQ$ and $\triangle RTS$, we have
$\angle P = \angle R T S$ $($given$)$
$\angle R = \angle R$ $($common$)$
$\therefore \quad \triangle R P Q \sim \triangle R T S$ $[$by $AA$-similarity$]$.

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Question 132 Marks

$S$ and $T$ are points on sides $PR$ and $QR$ of $\triangle PQR$ such that $\angle P = \angle R T S$. Show that $\triangle R P Q \sim \triangle R T S$.

Answer

According to questions it is given that $S$ and $T$ are points on sides $PR$ and $QR$ of $\triangle$ $PQR$ such that $\angle$$P = \angle R T S$

To Prove $\triangle R P Q \sim \triangle R T S$
Proof In $\triangle RPQ$ and $\triangle RTS$, we have
$\angle P = \angle R T S$ (given)
$\angle R = \angle R$ (common)
$\therefore \quad \triangle R P Q \sim \triangle R T S$ $[$by $AA$-similarity$]$.

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Question 142 Marks

Diagonal $AC$ and $BD$ of a trapezium $ABCD$ with $AB || DC$ intersect each other at point $O$. Using a similarity criterion for two triangles, show that $\frac { O A } { O C } = \frac { O B } { O D }$.

Answer

Given $A$ trapezium $ABCD$ in which $A B \| D C$. The diagonals $AC$ and $BD$ intersect at $O$.
To Prove In $\triangle OAB$ and $\triangle OCD$, we have
$\angle O A B = \angle O C D$ [alternate angles, since $A B \| D C$]
and $\angle O B A = \angle O D C$ [alternate angles, since $A B \| D C$]
$\therefore \quad \triangle O A B \sim \Delta O C D$ $[$by $AA$-similarity$]$.
Hence, $\frac { O A } { O C } = \frac { O B } { O D }$

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Question 152 Marks

Diagonal $AC$ and $BD$ of a trapezium $ABCD$ with $AB || DC$ intersect each other at point $O$. Using a similarity criterion for two triangles, show that $\frac { O A } { O C } = \frac { O B } { O D }$.

Answer

Given $A$ trapezium $ABCD$ in which $A B \| D C$. The diagonals $AC$ and $BD$ intersect at $O$.
To Prove In $\triangle OAB$ and $\triangle OCD$, we have
$\angle O A B = \angle O C D$ [alternate angles, since $A B \| D C$]
and $\angle O B A = \angle O D C$ [alternate angles, since $A B \| D C$]
$\therefore \quad \triangle O A B \sim \Delta O C D$ $[$by $AA$-similarity$]$.
Hence, $\frac { O A } { O C } = \frac { O B } { O D }$

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Question 162 Marks

In Figure, $\triangle$$ODC$ $\sim$ $\triangle$$OBA$, $\angle$$BOC = 125^\circ$ and $\angle$$CDO = 70^\circ$.
Find $\angle$$DOC$, $\angle$$DCO$ and $\angle$$OAB$.

Answer

From the given figure,
$\angle DOC + 125^\circ = 180^\circ $[linear pair]
$\angle DOC = 55^\circ$
Now, in $\triangle$$DOC,$
$\angle DCO + \angle ODC + \angle DOC = 180^\circ$[angle sum property of a triangle]
$\angle DCO +70^\circ + 55^\circ=180^\circ$
$\angle DCO = 55^\circ$
Now, $\triangle ODC \cong \triangle OBA$ [given]
$\therefore \angle OAB = \angle OCD = 55^\circ$

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Question 172 Marks

In Figure, $\triangle$$ODC$ $\sim$ $\triangle$$OBA$, $\angle$$BOC = 125^\circ$ and $\angle$$CDO = 70^\circ$. Find $\angle$$DOC$, $\angle$$DCO$ and $\angle$$OAB$.

Answer

From the given figure,
$\angle DOC + 125^\circ = 180^\circ $$[$linear pair$]$
$\angle DOC = 55^\circ$
Now, in $\triangle$$DOC,$
$\angle DCO + \angle ODC + \angle DOC = 180^\circ$$[$angle sum property of a triangle$]$
$\angle DCO +70^\circ + 55^\circ=180^\circ$
$\angle DCO = 55^\circ$
Now, $\triangle ODC \cong \triangle OBA$ $[$given$]$
$\therefore \angle OAB = \angle OCD = 55^\circ$

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Question 182 Marks

State the pair of triangles in the figure below are similar. Write the similarity criterion used by you for answering the question and also write the pair of similar triangles in the symbolic form:

Answer

In triangle $DEF,$ we have
$\angle D+\angle E+\angle F=180^{\circ}$ $($Sum of angles of triangle$)$
$70^{\circ}+80^{\circ}+\angle \mathrm{F}=180^{\circ}$
$\angle F=30^{\circ}$
In $PQR,$ we have
$\angle P+\angle Q+\angle R=180^{\circ}$
$\angle P+80^{\circ}+30^{\circ}=180^{\circ}$
$\angle \mathrm{P}=70^{\circ}$
In triangle $DEF$ and $PQR$, we have
$\angle D=\angle P=70^{\circ}$
$\angle F=\angle Q=80^{\circ}$
$\angle F=\angle R=30^{\circ}$
Hence, $\triangle DEF ~ \triangle PQR (AAA$ similarity$)$

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Question 192 Marks

State the pair of triangles in the figure below are similar. Write the similarity criterion used by you for answering the question and also write the pair of similar triangles in the symbolic form:

Answer

In triangle $DEF$, we have
$\angle D+\angle E+\angle F=180^{\circ}$ $($Sum of angles of triangle$)$
$70^{\circ}+80^{\circ}+\angle \mathrm{F}=180^{\circ}$
$\angle F=30^{\circ}$
In $PQR$, we have
$\angle P+\angle Q+\angle R=180^{\circ}$
$\angle P+80^{\circ}+30^{\circ}=180^{\circ}$
$\angle \mathrm{P}=70^{\circ}$
In triangle $DEF$ and $PQR$, we have
$\angle D=\angle P=70^{\circ}$
$\angle F=\angle Q=80^{\circ}$
$\angle F=\angle R=30^{\circ}$
Hence, $\triangle DEF ~ \triangle PQR (AAA$ similarity$)$

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Question 202 Marks

State the pair of triangles in the figure below are similar. Write the similarity criterion used by you for answering the question and also write the pair of similar triangles in the symbolic form:

Answer

In triangle $ABC$ and $DEF$, we have
$AB = 2.5, BC = 3$
$\angle \mathrm{A}=80^{\circ}$
$EF = 6$
$DF = 5$
$\angle F=80^{\circ}$
$\frac{A B}{D F}=\frac{2.5}{5}=\frac{1}{2}$
And, $\frac{B C}{E F}=\frac{3}{6}=\frac{1}{2}$
$\angle A = \angle F$
Hence, $\vartriangle$$ABC$ $\sim \vartriangle$$DEF ($by $SAS$ Rule$)$.

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Question 212 Marks

State the pair of triangles in the figure below are similar. Write the similarity criterion used by you for answering the question and also write the pair of similar triangles in the symbolic form:

Answer

In triangle $MNL$ and $QPR$, we have
$\angle M=\angle Q=70^{\circ}$
but
$\frac{M N}{P Q}=\frac{2.5}{5}=\frac{1}{2}$
$\frac{\mathrm{ML}}{\mathrm{QR}}=\frac{5}{10}=\frac{1}{2}$
$\Rightarrow \frac{\mathrm{MN}}{\mathrm{PQ}} = \frac{\mathrm{ML}}{\mathrm{QR}}$
Therefore, $\vartriangle$$MNL$ $\sim$ $\vartriangle$$QPR$ are similar.

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Question 222 Marks

$D$ is a point on the side $BC$ of a triangle $ABC$ such that $\angle ADC =$$\angle BAC$.Show that $CA^2= CB$$\cdot$$CD$.

Answer

Given:$\triangle\mathrm{ABC}$
where $\angle$$ADC =$ $\angle$$BAC$
To Prove : $CA^2= CB.CD$
Proof: In $\triangle$$ABC$ and $\triangle$$DAC$, we have

$\angle$$ADC =$ $\angle$$BAC$ and $\angle$$C =$ $\angle$$C$
Therefore, by $AA$-criterion of similarity, we obtain
$\Delta A B C \sim \Delta D A C$
$\Rightarrow \quad \frac { A B } { D A } = \frac { B C } { A C } = \frac { A C } { D C }$
$\Rightarrow \quad \frac { C B } { C A } = \frac { C A } { C D }$
$\Rightarrow CA^2= CB.CD$

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Question 232 Marks

$D$ is a point on the side $BC$ of a triangle $ABC$ such that $\angle$$ADC =$ $\angle$$BAC.$ Show that $CA^2= CB \cdot CD.$

Answer

Given:$\triangle\mathrm{ABC}$
where $\angle$$ADC =$ $\angle$$BAC$
To Prove : $CA^2= CB.CD$
Proof: In $\triangle$$ABC$ and $\triangle$$DAC$, we have

$\angle$$ADC =$ $\angle$$BAC$ and $\angle$$C =$ $\angle$$C$
Therefore, by $AA$-criterion of similarity, we obtain
$\Delta A B C \sim \Delta D A C$
$\Rightarrow \quad \frac { A B } { D A } = \frac { B C } { A C } = \frac { A C } { D C }$
$\Rightarrow \quad \frac { C B } { C A } = \frac { C A } { C D }$
$\Rightarrow$ $CA^2= CB.CD$

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Question 242 Marks

State the pair of triangles in the figure below are similar. Write the similarity criterion used by you for answering the question and also write the pair of similar triangles in the symbolic form:

Answer

From the triangle, $\frac{A B}{Q R}=\frac{B C}{P R}=\frac{A C}{P Q}=0.5$
Hence the corresponding sides are propotional. Thus the corresponding angles will be equal.
The triangles $ABC$ and $QRP$ are similar i.e, $\vartriangle ABC \sim \vartriangle QRP$ by $SSS$ similarity.

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Question 252 Marks

State the pair of triangles in the figure below are similar. Write the similarity criterion used by you for answering the question and also write the pair of similar triangles in the symbolic form:

Answer

From the figure:
$\angle$$A =$ $\angle$$P = 60^\circ$
$\angle$$B =$ $\angle$$Q = 80^\circ$
$\angle$$C =$ $\angle$$R = 40^\circ$
Therefore, $\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$ $[$By $AAA$ similarity$]$
Now corresponding sides of triangles will be propotional,
$\frac{A B}{PQ}=\frac{B C}{QR}=\frac{C A}{RP}$

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Question 262 Marks

Prove that the line joining the mid points of any two sides of a triangle is parallel to the third side.

Answer

Given: A $\triangle ABC$ in which $D$ and $E$ are the mid-points of sides $AB$ and $AC$ respectively. $DE$ is the line joining $D$ and $E$.
To prove:$DE$ $\parallel$ $BC$
Proof:
$\because $ $D$ is the mid-point of $AB$
$\therefore $ $AD=DB$
$\therefore $ $\frac{{AD}}{{DB}} = 1$$.....(1)$
$\because $ $E$ is the mid-point of $AC$

$\therefore $ $AE=EC$
$\therefore $ $\frac{{AE}}{{EC}} = 1$$.....(2)$
From $(1)$ and $(2)$ $\frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}$
$\therefore \frac{{AE}}{{EC}} = 1......(2)$....By converse of basic proportionality theorem

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Question 272 Marks

Prove that a line draw through the mid point of one side of a triangle parallel to another side bisects the third side.

Answer

Guven: A $DABC$ in which $D$ is the mid-point of $AB$ and $DE$ $\parallel$ $BC$.
To prove: $E$ is the mid-point of $AC$
Proof:
$\therefore DE||BC$
$\therefore \frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}$ $(1)......... [$By basic proportionality theorem$]$
$\because $ $D$ is the midpoint of $AB$
$\therefore AD = DB\,\,\therefore \frac{{AD}}{{DB}} = 1$
$\therefore \frac{{AE}}{{EC}} = 1......From(1)$
$\therefore AE = EC$,
$\therefore $ $E$ is the mid-point of $AC$.

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Question 282 Marks

In figure $A, B$ and $C$ are points on $OP, OQ$ and $OR$ respectively such that $AB || PQ$ and $AC || PR$. Show that $BC || QR$.

Answer

In $\triangle OPQ,$$\because AB||PQ$
$\therefore \frac{{OA}}{{AP}} = \frac{{OB}}{{BQ}}$$......(1)$
By basic proportionality theorem
In $\triangle OPR,\,\,\,\because AV||PR$ $\therefore \frac{{OA}}{{AP}} = \frac{{OC}}{{CR}}$$.......[$By basic proportionality theorem$]$
From $(1)$ and $(2)$
$\frac{{OB}}{{BQ}} = \frac{{OC}}{{OR}}$
$\therefore BC||QR$$......[$By converse basic proportionality theorem$]$

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Question 292 Marks

In figure $A, B$ and $C$ are points on $OP, OQ$ and $OR$ respectively such that $AB || PQ$ and $AC || PR$. Show that $BC || QR$.

Answer

In $\triangle OPQ,$$\because AB||PQ$
$\therefore \frac{{OA}}{{AP}} = \frac{{OB}}{{BQ}}$$......(1)$
By basic proportionality theorem
In $\triangle OPR,\,\,\,\because AV||PR$
$\therefore \frac{{OA}}{{AP}} = \frac{{OC}}{{CR}}$$.......[$By basic proportionality theorem$]$
From $(1)$ and $(2)$
$\frac{{OB}}{{BQ}} = \frac{{OC}}{{OR}}$
$\therefore BC||QR$$......[$By converse basic proportionality theorem$]$

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Question 302 Marks

In figure, $DE || OQ$ and $DF || OR$. Show that $EF || QR$.

Answer

In $\triangle PQO$ $\because DE||OQ$
$\therefore \frac{{PD}}{{DO}} = \frac{{PE}}{{EQ}}$ $....... (1) [$By basic proportionality theorem$]$
In $\triangle PRO\,\because DF||OR$
$\therefore \frac{{PD}}{{DO}} = \frac{{PF}}{{FR}}$$....... (2) [$By basic proportionality theorem$]$
from $(1)$ and $(2)$, $\frac{{PE}}{{EQ}} = \frac{{PF}}{{FR}}$
$\therefore $ $EF||QR$ $...... [$By converse of basic proportionality theorem$]$

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Question 312 Marks

In figure, $DE || OQ$ and $DF || OR$. Show that $EF || QR$.

Answer

In $\triangle PQO$ $\because DE||OQ$
$\therefore \frac{{PD}}{{DO}} = \frac{{PE}}{{EQ}}$$ ....... (1) $[By basic proportionality theorem]
In $\triangle PRO\,\because DF||OR$
$\therefore \frac{{PD}}{{DO}} = \frac{{PF}}{{FR}}$$....... (2) $[By basic proportionality theorem]
from $(1)$ and $(2)$, $\frac{{PE}}{{EQ}} = \frac{{PF}}{{FR}}$
$\therefore $ $EF||QR$ ...... [By converse of basic proportionality theorem]

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Question 322 Marks

In figure $DE || AC$ and $DF || AE$. Prove that $\frac{{BF}}{{FE}} = \frac{{BE}}{{EC}}$

Answer

In $\triangle ABE,$ we have $DF||AE,$then
$\frac{{BD}}{{AD}} = \frac{{BF}}{{FE}}\,$ $[$By BPT$] ...... (i)$
In $\triangle ABC,$ we have $DE||AC,$ then
$\frac{{BD}}{{AD}} = \frac{{BE}}{{EC}}\,$ $[$By BPT$] ...... (2)$
From $(i)$ and $(2)$, We get
$\frac{{BF}}{{FE}} = \frac{{BE}}{{EC}}$

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Question 332 Marks

In the given figure, $L M \| C B$ and $\mathrm { LN } \| \mathrm { CD }$. Prove that $\frac { A M } { A B } = \frac { A N } { A D }$.

Answer

According to question it is given that In $\triangle ALM$, $L M \| C B$
$\therefore \quad \frac { A B } { A M } = \frac { A C } { A L }$ Therefore, by Thales' theorem
$\Rightarrow \frac { A M } { A B } = \frac { A L } { A C }$ $..........(i)$
In $\triangle ALN$, $L N \| C D$
$\therefore \quad \frac { A C } { A L } = \frac { A D } { A N }$ Therefore by Thales theorem
$\Rightarrow \frac { A L } { A C } = \frac { A N } { A D }$ $...........(ii)$
From $(i)$ and $(ii)$ we get
$\frac { A M } { A B } = \frac { A N } { A D }$

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Question 342 Marks

In the given figure, $L M \| C B$ and $\mathrm { LN } \| \mathrm { CD }$. Prove that $\frac { A M } { A B } = \frac { A N } { A D }$.

Answer

According to question it is given that In $\triangle ALM$, $L M \| C B$
$\therefore \quad \frac { A B } { A M } = \frac { A C } { A L }$
Therefore, by Thales' theorem
$\Rightarrow \frac { A M } { A B } = \frac { A L } { A C }$ $..........(i)$
In $\triangle ALN$, $L N \| C D$
$\therefore \quad \frac { A C } { A L } = \frac { A D } { A N }$
Therefore by Thales theorem
$\Rightarrow \frac { A L } { A C } = \frac { A N } { A D }$ $...........(ii)$
From $(i)$ and $(ii)$ we get
$\frac { A M } { A B } = \frac { A N } { A D }$

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Question 352 Marks

$E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\triangle$$PQR$. For $PQ = 1.28\ cm$, $PR = 2.56\ cm, PE = 0.18\ cm$ and $PF = 0.36\ cm,$ state whether $EF || QR.$

Answer

We have
$PQ = 1.28\ cm, PR = 2.56\ cm, PE = 0.18\ cm, PF = 0.36\ cm$
$EQ = PQ - PE = 1.28 - 0.18 = 1.10\ cm$
and $FR = PR - PF = 2.56 - 0.36 = 2.20\ cm$
Now we can find
$\frac{{PE}}{{EQ}} = \frac{{0.18}}{{1.10}} = \frac{9}{{55}}$
$\frac{{PF}}{{PR}} = \frac{{0.36}}{{2.20}} = \frac{9}{{55}}$ $\therefore \frac{{PE}}{{EQ}} = \frac{{PF}}{{FR}}$
$\therefore EF ||QR$ (By converse of basic proportionality theorem)

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Question 362 Marks

$E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\triangle$$PQR$. For $PQ = 1.28\ cm, PR = 2.56\ cm, PE = 0.18\ cm$ and $PF = 0.36\ cm$, state whether $EF || QR$.

Answer

We have
$PQ = 1.28\ cm, PR = 2.56\ cm, PE = 0.18\ cm, PF = 0.36\ cm$
$EQ = PQ - PE = 1.28 - 0.18 = 1.10\ cm$
and $FR = PR - PF = 2.56 - 0.36 = 2.20\ cm$
Now we can find
$\frac{{PE}}{{EQ}} = \frac{{0.18}}{{1.10}} = \frac{9}{{55}}$
$\frac{{PF}}{{PR}} = \frac{{0.36}}{{2.20}} = \frac{9}{{55}}$
$\therefore \frac{{PE}}{{EQ}} = \frac{{PF}}{{FR}}$
$\therefore EF ||QR$ (By converse of basic proportionality theorem)

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Question 372 Marks

It is given that E and F are points on the sides $PQ$ and $PR$ respectively of a $\triangle$$PQR$. For $PE = 4\ cm, QE = 4.5\ cm, PF = 8\ cm$ and $RF = 9\ cm$, state whether $EF||QR$.

Answer

From the given information, we can have
$\frac{{PE}}{{EQ}} = \frac{4}{{4.5}} = \frac{{40}}{{45}} = \frac{8}{9}.....(I)$
$\frac{{PF}}{{RF}} = \frac{8}{9}....(II)$
From $(I)$ and $(II)$, it is clear that $\frac{{PE}}{{QE}} = \frac{{PF}}{{RF}}$
Therefore, $EF||QR$ (By converse of basic proportionality theorem)

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Question 382 Marks

It is given that $E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\triangle$$PQR$. For $PE = 4\ cm, QE = 4.5\ cm, PF = 8\ cm$ and $RF = 9\ cm$, state whether $EF||QR$.

Answer

From the given information, we can have
$\frac{{PE}}{{EQ}} = \frac{4}{{4.5}} = \frac{{40}}{{45}} = \frac{8}{9}.....(I)$
$\frac{{PF}}{{RF}} = \frac{8}{9}....(II)$
From $(I)$ and $(II)$, it is clear that $\frac{{PE}}{{QE}} = \frac{{PF}}{{RF}}$
Therefore, $EF||QR$ (By converse of basic proportionality theorem)

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Question 392 Marks

$E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\triangle$$PQR$. For $PE = 3.9\ cm, EQ = 3\ cm, PF = 3.6\ cm$ and $FR = 2.4\ cm$ case, state whether $EF || QR$.

Answer

We have
$\frac{{PE}}{{EQ}} = \frac{{3.9}}{3} = \frac{{1.3}}{1}.....(I)$
$\frac{{PF}}{{FR}} = \frac{{3.6}}{{2.4}} = \frac{3}{2} = \frac{{1.5}}{1}....(II)$
From $(I)$ and $(II)$,
we get
$\frac{{PE}}{{EQ}} \ne \frac{{PF}}{{FR}}$
Therefore, $EF$ is not parallel to $QR$. (By converse of basic proportionality theorem)

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Question 402 Marks

$E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\triangle$ $PQR$. For $PE = 3.9\ cm, EQ = 3\ cm, PF = 3.6\ cm$ and $FR = 2.4\ cm$ case, state whether $EF || QR$.

Answer

We have
$\frac{{PE}}{{EQ}} = \frac{{3.9}}{3} = \frac{{1.3}}{1}.....(I)$
$\frac{{PF}}{{FR}} = \frac{{3.6}}{{2.4}} = \frac{3}{2} = \frac{{1.5}}{1}....(II)$
From $(I)$ and $(II),$
we get
$\frac{{PE}}{{EQ}} \ne \frac{{PF}}{{FR}}$
Therefore, $EF$ is not parallel to $QR.$ $($By converse of basic proportionality theorem$)$

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Question 412 Marks

In figure $(i)$ and $(ii), DE || BC$.
Find $EC$ in $(i)$ and $AD$ in $(ii)$

Answer

In $\triangle ABC,$

$\because DE||BC$
$\therefore \frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}$ ..... By Basic Proportionality theorem
$\Rightarrow \frac{{1.5}}{3} = \frac{1}{{EC}}$
$\Rightarrow EC = \frac{3}{{1.5}}$$EC = 2cm$
In $\vartriangle ABC,\because DE||BC$
$\therefore \frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}......$By Basic Proportionality theorem
$\therefore \frac{{AD}}{{DB}} = \frac{{1.8}}{{5.4}}$
$\Rightarrow \frac{{7.2 \times 1.8}}{{5.4}}$
$\Rightarrow$ $AD = 2.4cm$

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Question 422 Marks

In figure $(i)$ and $(ii), DE || BC$.
Find $EC$ in $(i)$ and $AD$ in $(ii)$

Answer

In $ \triangle ABC,$

$ \because DE||BC$
$ \therefore \frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}$ ..... By Basic Proportionality theorem
$ \Rightarrow \frac{{1.5}}{3} = \frac{1}{{EC}}$
$ \Rightarrow EC = \frac{3}{{1.5}}$ $EC = 2\ cm$
In $ \vartriangle ABC,\because DE||BC$
$ \therefore \frac{{AD}}{{DB}} = \frac{{AE}}{{EC}}......$ By Basic Proportionality theorem
$ \therefore \frac{{AD}}{{DB}} = \frac{{1.8}}{{5.4}}$
$ \Rightarrow \frac{{7.2 \times 1.8}}{{5.4}}$
$ \Rightarrow$ $AD = 2.4\ cm$

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Question 432 Marks

State whether the following quadrilaterals are similar or not:

Answer

The given quadrilateral $PQRS$ and $ABCD$ are not similar because their corresponding sides are proportional, i.e. $1:2$, but their corresponding angles are not equal. Or we can say that $PQRS$ is a rhombus and $ABCD$ is a square, so they cannot be similar.

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Question 442 Marks

In Figure, $OA$ $\cdot$ $OB = OC$ $\cdot$ $OD$. Show that $\angle$ $A =$ $\angle$$C$ and $\angle$$B =$ $\angle$$D$

Answer

In $\triangle $$AOD$ and $\triangle $$BOC$,
$OA$ $ \times$ $OB = OC$ $ \times$ $OD$
i.e $\frac{{OA}}{{OC}} = \frac{{OD}}{{OB}}$
And $\angle$ $AOD =$ $\angle$ $BOC [$Vertically opposite Angles$]$
$\therefore $ $\triangle $$AOD$ $ \sim $$\triangle $$BOC [$By $SAS]$
$\therefore $ $\angle$$A =$ $\angle$$C$ and $\angle$$B =$ $\angle$$D [$Corresponding angles of similar $\triangle $ $]$

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Question 452 Marks

In Figure, $OA$ $\cdot$ $OB = OC$ $\cdot$ $OD$. Show that $\angle$ $A =$ $\angle$$C$ and $\angle$$B =$ $\angle$$D$

Answer

In $\triangle $$AOD$ and $\triangle $$BOC$,
$OA$ $ \times$ $OB = OC$ $ \times$ $OD$
i.e $\frac{{OA}}{{OC}} = \frac{{OD}}{{OB}}$
And $\angle$ $AOD =$ $\angle$ $BOC [$Vertically opposite Angles$]$
$\therefore $ $\triangle $$AOD$ $ \sim $$\triangle $$BOC [$By $SAS]$
$\therefore $ $\angle$$A =$ $\angle$$C$ and $\angle$$B =$ $\angle$$D [$Corresponding angles of similar $\triangle $ $]$

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Question 462 Marks

Observe the Fig. given below and then find $\angle$$P$.

Answer

In $\triangle$$ABC$ and $\triangle$$PQR$,
$\frac{\mathrm{AB}}{\mathrm{RQ}}=\frac{3.8}{7.6}=\frac{1}{2}, \frac{\mathrm{BC}}{\mathrm{QP}}=\frac{6}{12}=\frac{1}{2} $ and $\frac{\mathrm{CA}}{\mathrm{PR}}=\frac{3 \sqrt{3}}{6 \sqrt{3}}=\frac{1}{2}$
That is, $\frac{\mathrm{AB}}{\mathrm{RQ}}=\frac{\mathrm{BC}}{\mathrm{QP}}=\frac{\mathrm{CA}}{\mathrm{PR}}$
So,$\triangle \mathrm{ABC} \sim \triangle \mathrm{RQP}$ $(SSS$ similarity criterion$)$
Therefore, $\angle$$C =$ $\angle$$P ($Corresponding angles of similar triangles$)$
But $\angle$$C = 180^\circ –$ $\angle$$A –$ $\angle$$B ($Angle sum property of triangle$)$
$\angle$$C = 180^\circ – 80^\circ – 60^\circ$
$\angle$$C = 40^\circ$
So, $\angle$$P = 40^\circ$

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Question 472 Marks

Observe the Fig. given below and then find $\angle$$P$.

Answer

In $\triangle$$ABC$ and $\triangle$$PQR$,
$\frac{\mathrm{AB}}{\mathrm{RQ}}=\frac{3.8}{7.6}=\frac{1}{2}, \frac{\mathrm{BC}}{\mathrm{QP}}=\frac{6}{12}=\frac{1}{2} $ and $\frac{\mathrm{CA}}{\mathrm{PR}}=\frac{3 \sqrt{3}}{6 \sqrt{3}}=\frac{1}{2}$
That is, $\frac{\mathrm{AB}}{\mathrm{RQ}}=\frac{\mathrm{BC}}{\mathrm{QP}}=\frac{\mathrm{CA}}{\mathrm{PR}}$
So,$\triangle \mathrm{ABC} \sim \triangle \mathrm{RQP}$ $(SSS $similarity criterion$)$
Therefore, $\angle$$C =$ $\angle$$P ($Corresponding angles of similar triangles$)$
But $\angle$$C = 180^\circ –$ $\angle$$A – $$\angle$$B ($Angle sum property of triangle$)$
$\angle$$C = 180^\circ – 80^\circ – 60^\circ$
$\angle$$C = 40^\circ$
So, $\angle$$P = 40^\circ$

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Question 482 Marks

In Figure, if $PQ || RS$, prove that $\triangle POQ \sim \triangle SOR$.

Answer

From the given figure we have,
$PQ || RS ($Given$)$
So, $\angle$$P =$ $\angle$$S ($Alternate angles$)$
and $\angle$$Q =$ $\angle$$R ($Alternate angles$)$
Also, $\angle$$POQ =$ $\angle$$SOR ($Vertically opposite angles$)$
Therefore, $\triangle POQ \sim \triangle SOR$ $(AAA$ similarity criterion$)$

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Question 492 Marks

In Fig. $\frac { P S } { S Q } = \frac { P T } { T R }$ and $\angle P S T = \angle P R Q.$ Prove that $ \triangle$$PQR$ is an isosceles triangle.

Answer

According to the question,we are given that,
$\frac { P S } { S Q } = \frac { P T } { T R }$

$\Rightarrow \quad S T \| Q R$ $[$By using the converse of Basic Proportionality Theorem$]$
$\Rightarrow \quad \angle P S T = \angle P Q R$ $[$Corresponding angles$]$
$\Rightarrow \quad \angle P R Q = \angle P Q R$ $[ \because \angle P S T = \angle P R Q ( \text { Given } ) ]$
$\Rightarrow$ $PQ = PR [$ $\because$ Sides opposite to equal angles are equal$]$
$\Rightarrow \quad \Delta$ $PQR$ is isosceles.

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Question 502 Marks

In Fig. $\frac { P S } { S Q } = \frac { P T } { T R }$ and $\angle P S T = \angle P R Q.$ Prove that $ \triangle$$PQR$ is an isosceles triangle.

Answer

According to the question,we are given that,
$\frac { P S } { S Q } = \frac { P T } { T R }$

$\Rightarrow \quad S T \| Q R$ $[$By using the converse of Basic Proportionality Theorem$]$
$\Rightarrow \quad \angle P S T = \angle P Q R$ [Corresponding angles]
$\Rightarrow \quad \angle P R Q = \angle P Q R$ $[ \because \angle P S T = \angle P R Q ( \text { Given } ) ]$
$\Rightarrow$ $PQ = PR [$ $\because$ Sides opposite to equal angles are equal$]$
$\Rightarrow \Delta$ $PQR$ is isosceles.

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