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

$E$ 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 42 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 52 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 62 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 72 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 82 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 92 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 102 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 112 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 122 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 132 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 142 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 152 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 162 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 172 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 182 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 192 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 202 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 212 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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Question 222 Marks

A ladder is placed against a wall such that its foot is at a distance of $2.5\ m$ from the wall and its top reaches a window at a height of $6\ m$ above the ground. Find the length of the ladder.

Answer

Let $AB$ be the ladder and $AC$ be the wall with the window at $A$

Also, $BC = 2.5\ m$ and $CA = 6\ m$
From Pythagoras Theorem, we have:
$ \mathrm{AB}^2=\mathrm{BC}^2+\mathrm{CA}^2 $
$ =(2.5)^2+(6)^2 $
$=6.25+36 $
$ =42.25$
taking square root on both sides.we get
$AB = 6.5$
Thus, length of the ladder is $6.5m$

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Question 232 Marks

In the given figure, $\angle$ $ACB = 90^\circ$ and $CD$ $\perp$ $AB$. prove that $\frac { B C ^ { 2 } } { A C ^ { 2 } } = \frac { B D } { A D }$.

Answer

In $\triangle $$ACD$ and $\triangle $ $ABC$ $\angle A = \angle A $ (Common) $\angle ADC = \angle ACB$
$($ each $90^\circ)$Thus, By $AA$ similarity criteria $\triangle ADC \sim \triangle ACB$
Thus, $\frac {AD} {AC} = \frac {AC} {AB} $
$\Rightarrow AC^2 = AD \times AB$$... (i)$
Similarly, $\triangle CDB \sim \triangle ACB $ And, $\frac {DC} {BC} = \frac {BC} {AB} $
$\Rightarrow BC^2 = DB \times AB$$... (ii)$ Dividing $(ii)$ by $(i)$ $\frac {BC^2 } {AC^2 } = \frac {DB} {AD} $ Hence Proved

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Question 242 Marks

If a line intersects the sides $AB$ and $AC$ of a $\triangle$$ABC$ at $D$ and $E$ respectively and is parallel to $BC$, prove that $\frac{A D}{A B}=\frac{A E}{A C}$ (see figure).

Answer

$DE || BC$ (Given)
therefore $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$ $($by BPT$)$
or $\frac{D B}{A D}=\frac{E C}{A E}$ $($by taking reciprocal on both sides$)$
or $\frac{D B}{A D}+1=\frac{E C}{A E}+1$ $($add $1$ on both sides$)$
we get $\frac{A B}{A D}=\frac{A C}{A E}$
So $\frac{A D}{A B}=\frac{A E}{A C}$
Hence proved

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