# Triangles : Exercise - 6.4 (Mathematics NCERT Class 10th)

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Q.1     Let $\Delta ABC \sim \Delta DEF$ and their areas be, respectively, 64 $c{m^2}$ and 121 cm2. If EF = 15.4 cm, Find BC.
Sol.

We have, ${{Area\left( {\Delta ABC} \right)} \over {Area\left( {\Delta DEF} \right)}} = {{B{C^2}}\over {E{F^2}}}$
$\Rightarrow$ ${{64} \over {121}} = {{B{C^2}} \over {{{\left( {15.4} \right)}^2}}}$
$\Rightarrow$ ${8 \over {11}} = {{BC} \over {15.4}}$
$\Rightarrow$ $BC = \left( {{8 \over {11}} \times 15.4} \right)$ cm = 11.2 cm

Q.2     Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.
Sol.

In $\Delta s$ AOB and COD, we have $\angle AOB = \angle COD$ [Vert. opp. $\angle s$]
and, $\angle OAB = \angle OCD$ [Alernate $\angle s$]
Therefore by AA-criterion of similarity, we have
$\Delta AOB \sim \Delta COD$
$\Rightarrow$ ${{Area\left( {\Delta AOB} \right)} \over {Area\left( {\Delta COD} \right)}} = {{A{B^2}} \over {D{C^2}}}$
$\Rightarrow$ ${{Area\left( {\Delta AOB} \right)} \over {Area\left( {\Delta COD} \right)}} = {{{{\left( {2DC} \right)}^2}} \over {D{C^2}}} = {4 \over 1}$
Hence, area $\left( {\Delta AOB} \right)$ : area $\left( {\Delta COD} \right)$ = 4 : 1

Q.3     In figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that ${{ar\left( {\Delta ABC} \right)} \over {ar\left( {\Delta DBC} \right)}} = {{AO} \over {DO}}$ Sol.

Given : Two $\Delta s$ ABC and DBC which stand on the same base but on the opposite sides of BC.
To prove : ${{Area\left( {\Delta ABC} \right)} \over {Area\left( {\Delta DBC} \right)}} = {{AO} \over{DO}}$
Construction : Draw $AE \bot BC$ and $DF \bot BC$
Proof : In $\Delta s$ AOE and DOF, we have $\angle AEO = \angle DFO = 90^\circ$
$\angle AOE = \angle DOF$ [Vertically opp. $\angle s$]
Therefore by AA-criterion of similarity, we have
$\Delta AOE \sim \Delta DOF$
$\Rightarrow$ ${{AE} \over {DF}} = {{AO} \over {OD}}$ ........... (1)
Now, ${{Area\left( {\Delta ABC} \right)} \over {Area\left( {\Delta DBC} \right)}} = {{{1 \over 2} \times BC \times AE} \over {{1 \over 2} \times BC \times DF}}$
$\Rightarrow$ ${{AE} \over {DF}} = {{AO} \over {OD}}$ [Using (1)]
Therefore ${{Area\left( {\Delta ABC} \right)} \over {Area\left( {\Delta DBC} \right)}} = {{AO} \over {OD}}$

Q.4     If the areas of two similar triangles are equal, prove that they are congruent.
Sol.

Given : Two $\Delta s$ ABC and DEF such that
$\Delta ABC \sim \Delta DEF$
and, Area ($\Delta ABC$) = Area ($\Delta DEF$)
To prove : $\Delta ABC \cong \Delta DEF$ Proof : $\Delta ABC \sim \Delta DEF$
$\Rightarrow$ $\angle A = \angle D$, $\angle B = \angle E$, $\angle C = \angle F$
and, ${{AB} \over {DE}} = {{BC} \over {EF}} = {{AC} \over {DF}}$
To establish $\Delta ABC \cong \Delta DEF$, it is sufficient to prove that
AB = DE, BC = EF and AC = DF
Now, Area $\left( {\Delta ABC} \right)$ = Area $\left( {\Delta DEF} \right)$
$\Rightarrow$ ${{Area\left( {\Delta ABC} \right)} \over {Area\left( {\Delta DEF} \right)}} = 1$
$\Rightarrow$ ${{A{B^2}} \over {D{E^2}}} = {{B{C^2}} \over {E{F^2}}} = {{A{C^2}} \over{D{F^2}}} = 1$
$\Rightarrow$ ${{AB} \over {DE}} = {{BC} \over {EF}} = {{AC} \over {DF}} = 1$
$\Rightarrow$ AB = DE, BC = EF, AC = DF
Hence, $\Delta ABC \cong \Delta DEF$.

Q.5      D, E and F are respectively the mid-points of sides AB, BC and CA of $\Delta ABC$. Find the ratio of the areas of $\Delta DEF$ and $\Delta ABC$.
Sol.

Since, D and E are the mid-points of the sides AB and BC of $\Delta ABC$ respectively. Therefore DE || AC
$\Rightarrow$ DE || FC ............. (1)
Since, D and F are the mid-points of the sides AB and AC of $\Delta ABC$ respectively.
Therefore DF || BC $\Rightarrow$ DF || EC ............ (2)
From (1) and (2), we can say that DECF is a parallelogram.
Similarly, ADEF is  a parallelogram
Now, in $\Delta s$ DEF and ABC, we have
$\angle DEF = \angle A$ [Opp. $\angle s$ of ||gm ADEF]
and, $\angle EDF = \angle C$ [Opp. $\angle s$ of ||gm DECF]
Therefore by AA-criterion of similarity, we have
$\Delta DEF \sim \Delta ABC$
$\Rightarrow$ ${{Area\left( {\Delta DEF} \right)} \over {Area\left( {\Delta ABC} \right)}} = {{D{E^2}} \over {A{C^2}}}$
$= {{{{\left( {{1 \over 2}AC} \right)}^2}} \over {A{C^2}}} = {1 \over 4}$ [Because DE =======  ]
Hence, Area $\left( {\Delta DEF} \right)$ : Area $\left( {\Delta ABC} \right)$ = 1 : 4.

Q.6      Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Sol.

Given : $\Delta ABC \sim \Delta PQR$, AD and PM are the medians of $\Delta s$ ABC and PQR respectively.
To prove : ${{Area\left( {\Delta ABC} \right)} \over {Area\left( {\Delta PQR} \right)}} = {{A{D^2}}\over {P{M^2}}}$ Proof : Since $\Delta ABC \sim \Delta PQR$
${{Area\left( {\Delta ABC} \right)} \over {Area\left( {\Delta PQR} \right)}} = {{{{\left( {AB} \right)} ^2}} \over {{{\left( {PQ} \right)}^2}}}$ ................ (1)
But, ${{sideAB} \over {sidePQ}} = {{medianAD} \over {medianPM}}$ .............. (2)
From (1) and (2), we have
${{Area\left( {\Delta ABC} \right)} \over {Area\left( {\Delta PQR} \right)}} = {{A{D^2}} \over {P{M^2}}}$

Q.7    Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of the diagonals.
Sol.

Given : A square ABCD. Equilateral $\Delta s$ BCE and have been drawn on side and the diagonals AC respectively.
To prove : Area $\left( {\Delta BCE} \right)$ ${1 \over 2}\left( {Area\Delta ACF} \right)$ Proof : $\Delta BCE \sim \Delta ACF$
[Being equilateral so similar by AAA criterion of similarity]
$\Rightarrow$ ${{Area\left( {\Delta BCE} \right)} \over {Area\left( {\Delta ACF} \right)}} = {{B{C^2}} \over {A{C^2}}}$
$\Rightarrow$ ${{Area\left( {\Delta BCE} \right)} \over {Area\left( {\Delta ACF} \right)}} = {{B{C^2}} \over {{{\left( {\sqrt 2 BC} \right)}^2}}}$
[Because Diagonal = $\sqrt 2$ side $\Rightarrow$ AC = $\sqrt 2$ BC]
$\Rightarrow$ ${{Area\left( {\Delta BCE} \right)} \over {Area\left( {\Delta ACF} \right)}} = {1 \over 2}$

Tick the correct answer and justify :
Q.8    ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is
(a) 2 : 1                      (b) 1 : 2
(c) 4 : 1                      (d) 1 : 4
Sol.

Since $\Delta ABC$ and $\Delta BDE$ are equilateral, they are equiangular and hence
$\Delta ABC \sim \Delta BDE$ $\Rightarrow$ ${{Area\left( {\Delta ABC} \right)} \over {Area\left( {\Delta BDE} \right)}} = {{B {C^2}} \over {B{D^2}}}$
$\Rightarrow$ ${{Area\left( {\Delta ABC} \right)} \over {Area\left( {\Delta BDE} \right)}} = {{{{\left( {2BD} \right)}^2}} \over {B{D^2}}}$
[Because D is the mid-point of BC therefore BC = 2BD]
$\Rightarrow$ ${{Area\left( {\Delta ABC} \right)} \over {Area\left( {\Delta BDE} \right)}} = {4 \over 1}$
Therefore (C) is the correct answer.

Q.9      Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio.
(A) 2 : 3                       (B) 4 : 9
(C) 81 : 16                  (D) 16 : 81
Sol.

Since the ratio of the areas of two similar triangles equal to the ratio of the squares of any two corresponding sides. Therefore,
ratio of areas =  ${\left( 4 \right)^2}:{\left( 9 \right)^2} = 16:81$
Therefore (D) is the correct answer