Tag: relations between the areas of triangles

Questions Related to relations between the areas of triangles

If $\triangle ABC\sim \triangle DEF$ and $AB:DE=3:4$, then the ratio of area of triangles taken in order is 

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. $\dfrac{9}{16}$

  2. $\dfrac{16}{9}$

  3. $\dfrac{15}{9}$

  4. $\dfrac{9}{15}$

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Correct Option: A
Explanation:
 In similar triangle,
         Rratio of areas of triangle = square of ratio of corresponding sides
       Ratio of areas of triangle=${\dfrac {3^2}{4^2}}$
                                                 =$\dfrac{9}{16}$
In $\Delta A B C$, $P,Q,R$ are points on $\overline { B C } , \overline { C A } , \overline { A B }$ respectively, dividing them in the ratio $1 : 4,3 : 2$ and $3 : 7$. The points $S$ divides $AB$ in the ratio $1 : 3$.
Then $\frac { | \overline { A P } + \overline { B Q } + \overline { C R } | } { | \overline { C S } | } =$
maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. $\frac { 1 } { 5 }$

  2. $\frac { 2 } { 5 }$

  3. $\frac { 5 } { 2 }$

  4. $\frac { 7 } { 10 }$

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Correct Option: A

The areas of two similar triangles are $16cm^2$ and $36cm^2$ respectively. If the altitude of the first triangle is $3cm$, then the corresponding altitude of the other triangle is:

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. $4cm$

  2. $6.5cm$

  3. $4.5cm$

  4. $6cm$

Show answer
Correct Option: C
Explanation:
Let ${A} _{1}$ and ${A} _{2}$ be the areas of the similar triangles.

$\Rightarrow \dfrac{{A} _{1}}{{A} _{2}}=\dfrac{{s} _{1}^{2}}{{s} _{2}^{2}}$

$\Rightarrow \dfrac{16}{36}=\dfrac{{\left(3\right)}^{2}}{{s} _{2}^{2}}$ given $({s} _{1}=3 \ cm )$

$\Rightarrow {s} _{2}^{2}=\dfrac{36\times 9}{16}$

$\Rightarrow {s} _{2}=\dfrac{6\times 3}{4}=4.5 \ cm$  

State true or false:


The ratio of the areas of two triangles on the same base is equal to the ratio of their heights.

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. True

  2. False

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Correct Option: A
Explanation:

To prove: Ratio of the areas of two triangles of the same bases is equal to the ratio of their heights.
Proof:
Let us take triangle ${ T } _{ 1 }$ with height  ${ h } _{ 1 }$ and base ${ b } _{ 1 }$ and Triangle ${ T } _{ 2 }$ with height  ${ h } _{ 2 }$ and base ${ b } _{ 2 }$.

Area of Triangle ${ T } _{ 1 }$ $=\dfrac { 1 }{ 2 } \times \text{base}\times \text{height}\ =\dfrac { 1 }{ 2 } \times { b } _{ 1 }\times { h } _{ 1 }$

Area of Triangle ${ T } _{ 2 }$ $=\dfrac { 1 }{ 2 } \times \text{base}\times \text{height}\ =\dfrac { 1 }{ 2 } \times { b } _{ 2 }\times { h } _{ 2 }$

Ratio of area of two triangles,
$\dfrac { \text{Area of triangle} \ { T } _{ 1 } }{\text{ Area of triangle} \ { T } _{ 2 } } =\frac { \frac { 1 }{ 2 } \times { b } _{ 1 }\times { h } _{ 1 } }{ \frac { 1 }{ 2 } \times { b } _{ 2 }\times { h } _{ 2 } }$      
$ =\dfrac { h _{ 1 } }{ { h } _{ 2 } } $                                      (Base of two triangles are equal. So, $ { b } _{ 1 } = { b } _{ 2 }$) 
Proved.              

State true or false:

The ratio of the areas of two triangles of the same height is equal to the ratio of their bases.

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

To prove: Ratio of the areas of two triangles of the same height is equal to the ratio of their bases.
Proof:
Let us take triangle ${ T } _{ 1 }$ with height  ${ h } _{ 1 }$ and base ${ b } _{ 1 }$ and Triangle ${ T } _{ 2 }$ with height  ${ h } _{ 2 }$ and base ${ b } _{ 2 }$.

Area of Triangle ${ T } _{ 1 }$ $=\dfrac { 1 }{ 2 } \times\ \text{ base}\times \text{height}\ =\dfrac { 1 }{ 2 } \times { b } _{ 1 }\times { h } _{ 1 }$

Area of Triangle ${ T } _{ 2 }$ $=\dfrac { 1 }{ 2 } \times\ \text{ base}\times \text{height}\ =\dfrac { 1 }{ 2 } \times { b } _{ 2 }\times { h } _{ 2 }$

Ratio of area of two triangles,
$\dfrac { \text{Area of triangle} \  { T } _{ 1 } }{\text{ Area of triangle} \  { T } _{ 2 } } =\dfrac { \frac { 1 }{ 2 } \times { b } _{ 1 }\times { h } _{ 1 } }{ \frac { 1 }{ 2 } \times { b } _{ 2 }\times { h } _{ 2 } } \ $
$ =\dfrac { { b } _{ 1 } }{ { b } _{ 2 } } $                      (Height of two triangles are equal So,$ { h } _{ 1 } = { h } _{ 2 }$) 
Hence, proved.

The areas of two similar triangles are $12$ ${cm}^{2}$ and $48$ ${cm}^{2}$. If the height of the smaller one is $2.1$ $cm$, then the corresponding height of the bigger one is:

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. $4.41$ $cm$

  2. $8.4$ $cm$

  3. $4.2$ $cm$

  4. $0.525$ $cm$

Show answer
Correct Option: C
Explanation:

Areas of two similar triangles are $12 cm^2$ and $48 cm^2$
For similar triangles the ratio of areas is equal to the ratio of square of corresponding heights
Hence, $\dfrac{A _1}{A _2} = \dfrac{(h _1)^2}{(h _2)^2}$

$\dfrac{12}{48} = \dfrac{(2.1)^2}{(h _2)^2}$

$(h _2)^2= 4 \times (2.1)^2$

$h _2 = 2 \times 2.1$

$h _2 = 4.2 cm$

A vertical stick of length $6m$ casts a shadow $4m$ long on the ground and at the same time a tower casts a shadow $28m$ long. Find the height of the tower.

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. $42m$

  2. $48m$

  3. $62m$

  4. $52m$

Show answer
Correct Option: A
Explanation:

The length of stick corresponds to the length of tower and shadow of stick corresponds to the shadow of tower.
Thus, $\dfrac{Length\ Stick}{Shadow\ Stick} = \dfrac{Length\ Tower}{Shadow\ Tower}$


$\dfrac{6}{4} = \dfrac{Length\ Tower}{28}$
$Length\ Tower = \dfrac{6 \times 28}{4}$
$Length\ Tower = 42$ m

The corresponding sides of two similar triangles are in the ratio $2$ to $3$. If the area of the smaller triangle is $12$ the area of the larger is

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. $24$

  2. $27$

  3. $18$

  4. $8$

Show answer
Correct Option: B
Explanation:

Area of similar triangles are in the ratio of square of the corresponding sides.
Hence, $\dfrac{\text{Area of smaller triangle}}{\text{Area of larger triangle}} = \frac{2^2}{3^2}$
$\Rightarrow \dfrac{12}{\text{Area of larger triangle}} = \dfrac{4}{9}$
$\text{Area of larger triangle}  = 27$

If  in $\triangle ABC$  and $\triangle EDA,$ $\displaystyle BC\bot AB,AE\bot AB$ and $\displaystyle DE\bot AC$ then $\displaystyle DE.BC=AD.AB$ 

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

In $\displaystyle \Delta ABC$ and $\displaystyle \Delta EDA$,
We have
$\displaystyle \angle ABC=\angle ADE$ [Each equal to $\displaystyle { 90 }^{ o }$]
$\displaystyle \angle ACB=\angle EAD$ [Alternate angles]
$\displaystyle \therefore $ By AA Similarity
$\displaystyle \Delta ABC\sim \Delta EDA$
$\displaystyle \Rightarrow \frac { BC }{ AB } =\frac { AD }{ DE } $
$\displaystyle \Rightarrow DE.BC=AD.AB$.
Hence proved.

If the ratio of the corresponding sides of two similar triangles is 2 : 3, then the ratio of their corresponding altitude is :

maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

  1. 3 : 2

  2. 16 : 81

  3. 4 : 9

  4. 2 : 3

Show answer
Correct Option: D
Explanation:

In two similar triangles, if the corresponding sides are in a particular ratio, then altitudes will also be in the same ratio.
Hence the ratio of the altitudes will be 2 : 3.