3 Marks Question - Maths STD 10 Questions - Vidyadip

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21 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks

If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar:

Answer

This criterion is referred to as the AAA
(Angle-Angle-Angle) criterion of similarity of two triangles.
This theorem can be proved by taking two triangles $ABC$ and $DEF$ such that $\angle A =\angle D , \angle B =\angle E$ and $\angle C =\angle F$ (see Fig. 6.24)
Cut $DP = AB$ and $DQ = AC$ and join $PQ$.
So,
$\Delta ABC \cong \Delta DPQ$
(Why ?)
This gives
$\angle B =\angle P =\angle E \text { and } PQ \| EF$ .....(How?)
Therefore,
$\frac{ DP }{ PE }=\frac{ DQ }{ QF }$ .....(Why?)
i.e.,
$\frac{ AB }{ DE }=\frac{ AC }{ DF }$ .....(Why?)
Similarly,
$\frac{ AB }{ DE }=\frac{ BC }{ EF } \text { and so } \frac{ AB }{ DE }=\frac{ BC }{ EF }=\frac{ AC }{ DF } \text {. }$

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Question 23 Marks

In Figure, $\frac{{QR}}{{QS}} = \frac{{QT}}{{PR}}$ and $\angle$1 = $\angle$2. Show that $\triangle PQS \sim \triangle TQR$.

Answer

Given: In figure, $\frac{{QR}}{{QS}} = \frac{{QT}}{{PR}}$and $\angle$ 1 = $\angle$ 2
To prove: $\triangle PQS \sim \triangle TQR$
Proof: In $\triangle $ PQR $\because $ $\angle$ 1= $\angle$ 2
$\therefore $ PR = QP (1).......[ $\because $ sides opposite to equal angle of a triangle are equal]
Now, $\frac{{QR}}{{QS}} = \frac{{QT}}{{PR}}$ ......given
$ \Rightarrow \frac{{QR}}{{QS}} = \frac{{QT}}{{QP}}$ (2).......Using(1)
Again in $\triangle PQS$ and $\triangle TQR$
$\because \frac{{QR}}{{QS}} = \frac{{QT}}{{QP}}...........From(2)$
$\therefore \frac{{QS}}{{QR}} = \frac{{QP}}{{QT}}$ and $\angle SQP = \angle RQT$
$\therefore \triangle PQS \sim \triangle TQR$...........SAS similarity criterion

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Question 33 Marks

If AD and PM are medians of triangles ABC and PQR, respectively where $\triangle $ ABC $ \sim $ $\triangle $PQR, Prove that $\frac{{AB}}{{PQ}} = \frac{{AD}}{{PM}}$

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Question 43 Marks

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of $\triangle$PQR (see figure). Show that $\triangle A B C \sim \triangle P Q R$.

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Question 53 Marks

In the figure, $E$ is the point on side $CB$ produced on an isosceles triangle $ABC$ with $AB = AC$. If $AD$ $ \bot $ $BC$ and EF$ \bot $ $AC,$ prove that $\triangle $ABD $ \sim $ $\triangle $ECF.

Answer

$E$ is the point on side $C B$ produced on an isosceles triangle $A B C$ with $A B=A C \cdot A D-B C \perp$ and $E F \perp A C$. with $A B=A C$. Also, $A D \perp B C$ and $E F \perp A C$.
To prove: $\triangle ABD \sim \triangle ECF$
Proof: In $\triangle A B D$ and $\triangle E C F$,
$\therefore A B=A C$........Given
$\therefore \angle A C B=\angle A B C$......Angle opposite to equal sides of a triangle are equal
$\Rightarrow \angle A B C=\angle A C B$
$\Rightarrow \angle A B D=\angle E C F$.........(1)
$\angle A D B=\angle E F C$.
(2) [Each equal to $90^{\circ} In$ view of (1) and (2)]
$\triangle ABD \sim \triangle ECF$
..AA similarity criterion

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Question 63 Marks

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that $\frac{{AO}}{{BO}} = \frac{{CO}}{{DO}}$.

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Question 73 Marks

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

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Question 83 Marks

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

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Question 93 Marks

The diagonals of a quadrilateral ABCD intersect each other at the point O such that $\frac{{AO}}{{BO}} = \frac{{CO}}{{DO}}$. Show that ABCD is a trapezium.

Answer

Given: The diagonals of a quadrilateral ABCD intersect each other at the point O such that $\frac{{AO}}{{BO}} = \frac{{CO}}{{DO}}$
To prove: ABCD is trapezium.
Construction: Through O draw a line OE||BA intersecting AD at E.
Proof: In $\triangle DBA$$\because OE||BA$

$\therefore \frac{{DO}}{{BO}} = \frac{{DE}}{{AE}} \Rightarrow \frac{{CO}}{{AO}} = \frac{{DE}}{{AE}}$
$\because \frac{{AO}}{{BO}} = \frac{{CO}}{{DO}}\,[Given]$
$\Rightarrow \frac{{DO}}{{BO}} = \frac{{CO}}{{AO}} \Rightarrow \frac{{AO}}{{CO}} = \frac{{AE}}{{DE}}$.........[Taking reciprocals]
$\therefore $In $\triangle ADC$
OE $\parallel$ CD ...........[By converse basic proportionality theorem]
But OE $\parallel$ BA
BA $\parallel$ CD........[By construction]
The quadrilateral ABCD is a Trapezium.

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Question 103 Marks

Give two different examples of each of the following:

similar figures.
non-similar figures.

Answer

Two examples of similar figures are:
1. Two equilateral triangles with sides 1 cm and 2 cm respectively
2. Two squares with sides 1 cm and 2 cm respectively
Now two examples of non-similar figures are:
1. Trapezium and square
2. Triangle and parallelogram

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Question 113 Marks

In the given figure, the line segment $XY$ is parallel to side $AC$ of $\triangle$ABC and it divides the triangle into two parts of equal areas. Find the ratio $\frac{\mathrm{A} \mathrm{X}}{\mathrm{AB}}$.

Answer

Given, $\triangle$ABC $\sim\triangle$FEG ….(1)
(i) Corresponding angles of similar triangles
$\Rightarrow$ $\angle$BAC = $\angle$EFG ….(2)
And $\angle$ABC = $\angle$FEG …(3)
$\Rightarrow$ $\angle$ACB = $\angle$FGE
$\Rightarrow \frac{1}{2}\angle ACB = \frac{1}{2}\angle FGE$
$\Rightarrow$ $\angle$ACD = $\angle$FGH and $\angle$BCD = $\angle$EGH ……(4)
Consider $\triangle$ACD and $\triangle$FGH
$\Rightarrow$ From (2) we have
$\Rightarrow$$\angle$DAC = $\angle$HFG
From (4) we have
$\Rightarrow$ $\angle$ACD = $\angle$EGH
Also, $\angle$ADC = $\angle$FGH
If the $\angle A=\angle F$, then by angle sum property of triangle 3 angle will also be equal.
By AAA similarity, in two triangles, if the angles are equal, then sides opposite to the equal angles are in the same ratio (or proportional) and hence the triangles are similar.
$\therefore\triangle$ADC $\sim\triangle$FHG
(ii) By Converse proportionality theorem
$\Rightarrow \frac{C D}{G H}=\frac{A C}{F G}$
(iii) Consider $\triangle$DCB and $\triangle$HGE
From eq(3) we have
$\Rightarrow$ $\angle$DBC = $\angle$HEG
From (4) we have
$\Rightarrow$ $\angle$BCD = $\angle$FGH
Also, $\angle$BDC = $\angle$EHG
$\therefore\triangle$DCB $\sim\triangle$HGE
Hence proved.

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Question 123 Marks

ABCD is a trapezium with AB || DC. E and F are two points on non-parallel sides AD and BC respectively, such that EF is parallel to AB. Show that $\frac{AE}{ED}=\frac{BF}{FC}$

Answer

Given, In trapezium ABCD,
$AB || DC$ and $EF || DC$
To prove $\frac{AE}{ED}=\frac{BF}{FC}$
Construction: Join AC to intersect EF at G.

Proof Since, $AB || DC$ and $EF || DC$
$EF || AB$ [since, lines parallel to the same line are also parallel to each other ]...... (i)
In $\triangle ADC$, $EG || DC$ [$\because $ $EF || DC$]
By using basic proportionality theorem,
$\frac{AE}{ED}=\frac{AG}{GC}$ ....(ii)
In $\triangle ABC$, $ GF || AB$ [$\because $ $EF || AB$ from (i)]
By using basic proportionality theorem ,
$\frac{CG}{AG}=\frac{CF}{BF}$ or $\frac{AG}{GC}=\frac{BF}{CF}$ [ On taking reciprocal of the terms]............. (iii)
From Equations (ii) and (iii), we get
$\frac{AE}{ED}=\frac{BF}{FC}$
Hence Proved.

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Question 133 Marks

O is any point inside a rectangle ABCD see Fig. Prove that $OB^2 + OD^2 = OA^2 + OC^2$
.

Answer

Through O, draw PQ || BC so that P lies on AB and Q lies on DC.
Now, PQ || BC
Therefore, PQ $\perp$ AB and PQ$\perp$ DC ($\angle$B $= 90^\circ$ and $\angle$C $= 90^\circ$)
So, $\angle$$BPQ = 90^\circ$ and $\angle$$CQP = 90^\circ$
Therefore, BPQC and APQD are both rectangles.
Now, from $\triangle$OPB
$OB^2 = BP^2 + OP^2 ...(1)$
Similarly, from $\triangle$OQD,
$OD^2 = OQ^2 + DQ^2 ...(2)$
From \triangle OQC, we have
$OC^2 = OQ^2 + CQ^2...(3)$
and from $\triangle$OAP, we have
$OA^2 = AP^2 + OP^2 ...(4)$
Adding (1) and (2),
$OB^2 + OD^2 = BP^2 + OP^2+ OQ^2 + DQ^2$
$= CQ^2 + OP^2 + OQ^2 + AP^2 (As BP = CQ and DQ = AP)$
$= CQ^2 + OQ^2 + OP^2 + AP^2$
$OB^2 + OD^2= OC^2 + OA^2$^ [From (3) and (4)]
Hence proved.

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Question 143 Marks

$BL$ and $CM$ are medians of $\triangle $$ABC$ right angled at A. Prove that $4(BL^2+ CM^2) = 5BC^2$

Answer

BL and CM are medians of a $\triangle $ ABC in which $\angle$ $A=90^0$^
From $\triangle $$ABC, BC^2 = AB^2 + AC^2 ....(i)$
From right angled $\vartriangle $ ABL,
$BL^2 = AL^2+ AB^2$
i.e $B{L^2} = {\left( {\frac{{AC}}{2}} \right)^2} + A{B^2}$
$\Rightarrow $
$4BL^2= AC^2+ 4AB^2 .....(ii)$
From right-angled $\triangle $CMA,
$CM^2 = AC^2+ AM^2$^
i.e $C{M^2} = A{C^2} + {\left( {\frac{{AB}}{2}} \right)^2}$[mid-point]
$C{M^2} = A{C^2} + \frac{{A{B^2}}}{4}$
$4C{M^2} = 4A{C^2} + A{B^2}$ .....(iii)
Adding (ii) and (iii), we get
i.e.$4(BL^2 + CM^{2)} = 5(AC^2+ AB^2) = 5BC^2$^ [From (i)]

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Question 153 Marks

In the given figure, $\angle$ ACB = 90° 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°)

Thus, By AA similarity criteria

$\triangle ADC \sim \triangle ACB$

Thus, $\frac {AD} {AC} = \frac {AC} {AB} $

$\Rightarrow AC^2 = AD ×AB$... (i)

Similarly, $$$\triangle CDB \sim \triangle ACB $

And, $\frac {DC} {BC} = \frac {BC} {AB} $

$\Rightarrow BC^2 = DB ×AB$... (ii)

Dividing (ii) by (i)

$\frac {BC^2 } {AC^2 } = \frac {DB} {AD} $

Hence Proved

$$

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Question 163 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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Question 173 Marks

If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar

Answer

This criterion is referred to as the SAS (Side-Angle-Side) similarity criterion for two triangles.
As before, this theorem can be proved by taking two triangles $ABC$ and DEF such that $\frac{ AB }{ DE }=\frac{ AC }{ DF }(<1)$ and $\angle A =\angle D$ (see Fig. 6.28). Cut DP = AB, DQ $= AC$ and join $PQ$.
$\text { Now, } PQ \| EF \text { and } \triangle ABC \cong \triangle DPQ $
$\text { So, } \angle A =\angle D , \angle B =\angle P \text { and } \angle C =\angle Q $
$\text { Therefore, } \Delta ABC \sim \Delta DEF $
$\text { We now take some examples to illustrate the use of these criteria. }$

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Question 183 Marks

If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similiar.

Answer

This theorem can be proved by taking two triangles $ABC$ and $DEF$ such that $\frac{ AB }{ DE }=\frac{ BC }{ EF }=\frac{ CA }{ FD }(<1)($ see Fig. 6.26):
Cut $DP = AB$ and $DQ = AC$ and join $PQ$.
It can be seen that $\frac{ DP }{ PE }=\frac{ DQ }{ QF }$ and $PQ \| EF$ (How?)
So, $\angle P =\angle E$ and $\angle Q =\angle F$.
Therefore,
$\frac{ DP }{ DE }=\frac{ DQ }{ DF }=\frac{ PQ }{ EF }$
So,
$\frac{ DP }{ DE }=\frac{ DQ }{ DF }=\frac{ BC }{ EF } \quad \text { (Why?) }$
So,
$BC = PQ \quad \text { (Why?) }$
Thus,
$\Delta ABC \cong \Delta DPQ$
(Why ?)
So,
$\angle A =\angle D , \angle B =\angle E \text { and } \angle C =\angle F \text { (How ?) }$

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Question 193 Marks

If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar:

Answer

This criterion is referred to as the AAA
(Angle-Angle-Angle) criterion of similarity of two triangles.
This theorem can be proved by taking two triangles $ABC$ and $DEF$ such that $\angle A =\angle D , \angle B =\angle E$ and $\angle C =\angle F$ (see Fig. 6.24)
Cut $DP = AB$ and $DQ = AC$ and join $PQ$.
So,
$\Delta ABC \cong \Delta DPQ$
(Why ?)
This gives
$\angle B =\angle P =\angle E \text { and } PQ \| EF$ .....(How?)
Therefore,
$\frac{ DP }{ PE }=\frac{ DQ }{ QF }$ .....(Why?)
i.e.,
$\frac{ AB }{ DE }=\frac{ AC }{ DF }$ .....(Why?)
Similarly,
$\frac{ AB }{ DE }=\frac{ BC }{ EF } \text { and so } \frac{ AB }{ DE }=\frac{ BC }{ EF }=\frac{ AC }{ DF } \text {. }$

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Question 203 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,
$\frac{ AD }{ DB }=\frac{ AE ^{\prime}}{ E ^{\prime} C }\ ($Why$)$?
Therefore,
$\frac{ AE }{ EC }=\frac{ AE ^{\prime}}{ E ^{\prime} C } \ ($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 213 Marks

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Answer

Proof: We are given a triangle $\text{ABC}$ in which a line parallel to side $\text{BC}$ intersects other two sides $\text{AB}$ and $\text{AC}$ at $\text{D}$ and $\text{E}$ respectively $($see Fig. $6.10).$
We need to prove that $\frac{ AD }{ DB }=\frac{ AE }{ EC }$.
Let us join $\text{BE}$ and $\text{CD}$ and then draw $\text{DM} \perp \text{AC}$ and $\text{EN} \perp \text{AB}$.
Now, area of $\triangle \text{ADE} \left(=\frac{1}{2}\right.$ base $\times$ height $)=\frac{1}{2} \text{AD} \times \text{EN}$.
Recall from Class $IX,$ that area of $\triangle \text{ADE}$ is denoted as $\operatorname{ar}( \text{ADE} )$.
So, $\ce{{ar}( ADE )=\frac{1}{2} AD \times EN}$
Similarly,
$\ce{{ar}( BDE )=\frac{1}{2} DB \times EN ,}$
$\ce{{ar}( ADE )=\frac{1}{2} AE \times DM}$ and $\ce{{ar}( DEC )=\frac{1}{2} EC \times DM.}$
Therefore,
$\frac{\operatorname{ar}( ADE )}{\operatorname{ar}( BDE )}=\frac{\frac{1}{2} AD \times EN }{\frac{1}{2} DB \times EN }=\frac{ AD }{ DB }$
and $\frac{\operatorname{ar}( ADE )}{\operatorname{ar}( DEC )}=\frac{\frac{1}{2} AE \times DM }{\frac{1}{2} EC \times DM }=\frac{ AE }{ EC }$
Note that $\triangle \text{BDE}$ and $\text{DEC}$ are on the same base $\text{DE}$ and between the same parallels $\text{BC}$ and $\text{DE}$.
So, $\ce{{ar}( BDE )=\operatorname{ar}( DEC )}$
Therefore, from $(1), (2)$ and $(3),$ we have :
$\ce{\frac{ AD }{ DB }=\frac{ AE }{ EC }}$
Is the converse of this theorem also true $($For the meaning of converse, see Appendix $1)\ ?$ To examine this,
let us perform the following activity:

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