4 Marks Questions

Question 14 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 $A B C$ in which a line parallel to side $BC$ intersects other two sides $AB$ and $AC$ at $D$ and $E$ respectively $($see Fig. $6.10).$
We need to prove that $\frac{ AD }{ DB }=\frac{ AE }{ EC }$.
Let us join $BE$ and $CD$ and then draw $DM \perp AC$ and $EN \perp AB$.
Now, area of $\triangle ADE \left(=\frac{1}{2}\right.$ base $\times$ height $)=\frac{1}{2} AD \times EN$.
Recall from Class $IX,$ that area of $\triangle ADE$ is denoted as $\operatorname{ar}( ADE )$.
So,
$\operatorname{ar}( ADE )=\frac{1}{2} AD \times EN$
Similarly,
$\operatorname{ar}( BDE )=\frac{1}{2} DB \times EN ,$
$\operatorname{ar}( ADE )=\frac{1}{2} AE \times DM \text { and } \operatorname{ar}( DEC )=\frac{1}{2} EC \times DM .$
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 $\triangle BDE$ and $DEC$ are on the same base $DE$ and between the same parallels $BC$ and $DE$.
So,
$\operatorname{ar}( BDE )=\operatorname{ar}( DEC )$
Therefore, from $(1), (2)$ and $(3),$ we have :
$\frac{ AD }{ DB }=\frac{ AE }{ EC }$
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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Question 24 Marks

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

Answer

Given: $AD$ and $PM$ are median of triangles $ABC$ and $PQR$ respectively where $\triangle ABC \sim \triangle PQR$

To prove: $\frac{{AB}}{{PQ}} = \frac{{AD}}{{PM}}$
Proof: $\triangle ABC \sim \triangle PQR ........$Given
$\therefore \frac{{AB}}{{PQ}} = \frac{{BC}}{{QR}} = \frac{{CA}}{{RP}} .......(1).....[ \because$ Corresponding sides of two similar triangles are proportional$]$
and $\angle A = \angle P, \angle B = \angle Q, \angle C = \angle R, ..........(2) [ \because$ corresponding sides of two similar triangles are proportional$]$
But $BC = 2BD$ and $QR = 2QM.............. \because AD$ and $PM$ are medians
So, from$(1), \frac{{AB}}{{PQ}} = \frac{{2BD}}{{2QM}}$
$\Rightarrow \frac{{AB}}{{PQ}} = \frac{{BD}}{{QM}} ........(3)$
Also, $\angle ABD = \angle PQM .........(4)..........$ From $(2)$
$\therefore \triangle ABD \sim \triangle PQM .......SAS$ similarity criterion
$\therefore \frac{{AB}}{{PQ}} = \frac{{AD}}{{PM}} ........ [\because$ Corresponding sides of two similar triangles are proportional$]$

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Question 34 Marks

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.

Answer

Let $AB$ denoted the vertical pole of length $6m.$
$BC$ is the shadow of the pole on the ground $BC = 4m.$
Let $DE$ denote the tower.
$EF$ is shadow of the tower on the ground.
$EF = 28\ m.$
Let the height of the tower be $h\ m.$

In $\triangle ABC $and $\triangle DEF,$
$\angle B = \angle E ......[$Each equal to $90^\circ$ because pole and tower are standing vertical to the ground$]$
$\angle C = \angle F .....[$Same elevation$]$
$\angle A = \angle D \because $ shadows are cast at the same time
$\therefore \triangle ABC$ and $\triangle DEF,$
$\angle B= \angle E ............[$Each equal to $90^\circ$ because pole and tower are standing vertical to the ground$.]$
$\angle A= \angle D ( \because $ shadows are cast at the same time$)$
$\therefore \vartriangle ABC \sim \vartriangle DEF ......(AA$ similarity criterion$)$
$\therefore \frac{{AB}}{{DE}} = \frac{{BC}}{{EF}} ...........[ \because $ corresponding sides of two similar triangles are proportional$]$
$\Rightarrow \frac{6}{h} = \frac{4}{{28}}$
$\Rightarrow h = \frac{{6 \times 28}}{4} \Rightarrow h = 42$
Hence, the height of the tower is $42\ m$

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Question 44 Marks

Sides $AB$ and $AC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $PR$ and median $PM$ of another triangle $PQR.$ Show that $\Delta A B C \sim \Delta P Q R$.

Answer

Given : In $\Delta A B C \text { and } \Delta P Q R$ The $AD$ and $PM$ are their medians,
such that $\frac { A B } { P Q } = \frac { A D } { P M } = \frac { A C } { P R }$
To prove $: \Delta A B C \sim \Delta P Q R$
Construction : Produce $AD$ to $E$ such that $AD = DE$ and produce $PM$ to $N$ such that $PM = MN.$ Join $CE$ and $RN.$
Proof : In $\Delta A B D \text { and } \Delta E D C$
$AD=DE$
$\angle A D B = \angle E D C ($vertically opposite angles$)$
$BD=DC\text{(as AD is a median)}$
$\therefore \quad \Delta A B D \equiv \Delta E D C ($By $SAS$ congruency$)$
or, $AB=CE ($By $CPCT)$
Similarly, $PQ = RN $
$\frac { A B } { P Q } = \frac { A D } { P M } = \frac { A C } { P R } ($Given$)$
or, $\frac { C E } { R N } = \frac { 2 A D } { 2 P M } = \frac { A C } { P R }$
or $\frac{CE}{RN}=\frac{AE}{PN}=\frac{AC}{PR}$
So $∆ACE \sim ∆PRN$
$\angle 3=\angle 4$
$Similarly \angle 1=\angle 2$
$\angle 1+\angle3=\angle2+\angle4$
So $\angle A=\angle P\text{ and}$
$\frac{AB}{PQ}=\frac{AC}{PR}\text{(given)}$
Hence $∆ABC\sim ∆PQR$

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Question 54 Marks

$CD$ and $GH$ are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that $D$ and $H$ lie on sides $AB$ and $FE$ of $\triangle ABC$ and $\triangle EFG$ respectively. If $\triangle ABC \sim \triangle FEG,$ show that:

$\frac{C D}{G H}=\frac{A C}{F G}$
$\triangle DCB \sim\triangle HGE$
$\triangle DCA \sim\triangle HGF$

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 $3^{rd}$ 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 64 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

Since $XY \parallel AC$
$ \therefore \angle BXY = \angle BAC$
$ \angle BYX = \angle BCA [$Corresponding angles$]$
$ \therefore \triangle BXY \cong \triangle BAC [AA $ similarity$]$
$ \therefore \frac{{ar\left( {\triangle BXY} \right)}}{{ar\left( {\triangle BAC} \right)}} = \frac{{B{X^2}}}{{B{A^2}}} $
But $ ar( \triangle BXY)=ar(XYCA)$
$ \therefore 2( \triangle BXY)=ar( \triangle BXY)+ar(XYCA)$
$= ar( \triangle BAC)$
$\therefore \frac{{ar\left( {\triangle BXY} \right)}}{{ar\left( {\triangle BAC} \right)}} = \frac{1}{2} $
$ \therefore \frac{{B{X^2}}}{{B{A^2}}} = \frac{1}{2} $
$ \Rightarrow \therefore \frac{{BX}}{{BA}} = \frac{1}{{\sqrt 2 }} $
$ \therefore \frac{{BA - BX}}{{BA}} = \frac{{\sqrt 2 - 1}}{{\sqrt 2 }} $
$ \Rightarrow \frac{{AX}}{{AB}} = \frac{{\sqrt 2 - 1}}{{\sqrt 2 }} = \frac{{2 - \sqrt 2 }}{2}$

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Question 74 Marks

In Fig. $CM$ and $RN$ are respectively the medians of $\triangle ABC$ and $\triangle PQR$. If $\triangle ABC$ $\sim\triangle PQR,$ prove that:

$\triangle AMC \sim\triangle PNR$
$\frac{C M}{R N}=\frac{A B}{P Q}$
$\triangle CMB \sim\triangle RNQ$

Answer

$\triangle ABC \sim\triangle PQR ($Given$)$
So, $\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{C A}{R P} ...(1) ($corresponding sides of similar triangles are proportional$)$
and $\angle A = \angle P, \angle B = \angle Q$ and $\angle C = \angle R ...(2)$
But $AB = 2\ AM$ and $PQ = 2\ PN ($As $CM$ and $RN$ are medians$)$
So, from $(1),$
i.e., $\frac{A M}{P N}=\frac{C A}{R P} ...(3)$
Also, $∠ MAC = ∠ NPR [$From $(2)] ...(4)$
So, from $(3)$ and $(4),$
$\triangle AMC \sim\triangle PNR (SAS$ similarity criterion$) ...(5)$
From $(5), \frac{\mathrm{CM}}{\mathrm{RN}}=\frac{\mathrm{CA}}{\mathrm{RP}} ...(6) ($corresponding sides of similar triangles are proportional$)$
But $\frac{C A}{R P}=\frac{A B}{P Q} [$From $(1)] ...(7)$
Therefore, $\frac{C M}{R N}=\frac{A B}{P Q} [$From $(6)$ and $(7)] ...(8)$
Again, $\frac{A B}{P Q}=\frac{B C}{Q R} [$From $(1)]$
Therefore $\frac{\mathrm{CM}}{\mathrm{RN}}=\frac{\mathrm{BC}}{\mathrm{QR}} [$From $(8)] ...(9$)
Also, $\frac{C M}{R N}=\frac{A B}{P Q}=\frac{2 B M}{2 Q N}$
i.e.,$ \frac{\mathrm{CM}}{\mathrm{RN}}=\frac{\mathrm{BM}}{\mathrm{QN}} ...(10)$
i.e., $\frac{\mathrm{CM}}{\mathrm{RN}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{BM}}{\mathrm{QN}}$ $[$From $(9)$ and $(10)]$
Therefore, $\triangle CMB \sim\triangle RNQ (SSS$ similarity criterion$)$

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Question 84 Marks

Anil's height is 90 cm. He is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of his shadow after 4 seconds.

Answer

First, we find the distance Anil has walked from the base of the lamp-post (EB):
Distance = Speed × Time
Distance (EB) = 1.2 m/s × 4 s = 4.8 m
Now, we use the property of similar triangles.
In $\triangle A B D$ and $\triangle C E D:$
$\angle B=\angle E=90^{\circ}$ (Both Anil and the lamp-post are perpendicular to the ground).
$\angle D=\angle D$ (Common angle).
By the AA similarity criterion, $\triangle A B D \sim \triangle C E D$.
The ratio of corresponding sides of similar triangles is equal:
$\frac{A B}{C E}=\frac{B D}{E D}$
We know that $E D=E B+B D$. Let the length of the shadow $(B D)$ be $x$.
$\frac{0.9}{3.6}=\frac{x}{4.8+x}$
Simplifying the fraction on the left side:
$\frac{1}{4}=\frac{x}{4.8+x}$
Now, we cross-multiply to solve for $x:$
$\begin{array}{l}1(4.8+x)=4 x \\ 4.8+x=4 x \\ 4.8=3 x \\ x=\frac{4.8}{3} \\ x=1.6\end{array}$
Therefore, the length of Anil's shadow after 4 seconds is 1.6 meters.

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Question 94 Marks

State and prove Thales theorem.

Answer

In $\triangle A B C$ if a line DE is drawn parallel to BC, intersecting AB at D and AC at E, then the theorem states that:
$\frac{A D}{D B}=\frac{A E}{E C}$
In $\triangle A B C$, a line DE is parallel to BC $(D E \| B C)$ and it intersects AB at D and AC at E.
To Prove:$\frac{A D}{D B}=\frac{A E}{E C}$
Construction:
Draw BE and CD.
Draw $D M \perp A C$ and $E N \perp A B$.
Proof: The area of a triangle is given by the formula: Area $=\frac{1}{2} \times$ base $\times$ height.
Consider $\triangle A D E$ and $\triangle B D E:$
$\begin{array}{l}\text { Area }(\triangle A D E)=\frac{1}{2} \times A D \times E N \\ \text { Area }(\triangle B D E)=\frac{1}{2} \times D B \times E N\end{array}$
Dividing the areas:
$\frac{\operatorname{Area}(\triangle A D E)}{\operatorname{Area}(\triangle B D E)}=\frac{\frac{1}{2} \times A D \times E N}{\frac{1}{2} \times D B \times E N}=\frac{A D}{D B}$ ...(1)
Now consider $\triangle A D E$ and $\triangle C D E:$
$\begin{array}{l}\operatorname{Area}(\triangle A D E)=\frac{1}{2} \times A E \times D M \\ \operatorname{Area}(\triangle C D E)=\frac{1}{2} \times E C \times D M\end{array}$
Dividing the areas:
$\frac{\operatorname{Area}(\triangle A D E)}{\operatorname{Area}(\triangle C D E)}=\frac{\frac{1}{2} \times A E \times D M}{\frac{1}{2} \times E C \times D M}=\frac{A E}{E C}$ ...(2)
Notice that $\triangle B D E$ and $\triangle C D E$ are on the same base DE and between the same parallel lines BC and DE. Therefore, their areas are equal.
$\operatorname{Area}(\triangle B D E)=\operatorname{Area}(\triangle C D E)$ ...(3)
From equations (1), (2), and (3), we can conclude that:
$\frac{A D}{D B}=\frac{A E}{E C}$
Hence, the theorem is proved.

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