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
33\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}}
34
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
35\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
36Proof : \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.
37Therefore 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}}}
38Proof : 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)
39Proof : \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
40 \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



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