Tag: areas of similar triangles

Questions Related to areas of similar triangles

If in $\displaystyle \Delta ABC$ and $\displaystyle \Delta DEF,\frac { AB }{ DE } =\frac { BC }{ FD } $, then they will be similar if :

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

  1. $\displaystyle \angle B=\angle E$

  2. $\displaystyle \angle A=\angle D$

  3. $\displaystyle \angle B=\angle D$

  4. $\displaystyle \angle A=\angle F$

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Correct Option: C
Explanation:

If two sides of a triangle are proportional to the corresponding two sides in another triangle, and their included angles are equal, then the two triangles are similar by SAS rule.

If $\quad \dfrac { AB }{ DE } = \dfrac { BC }{ FD } $, then for two triangles ABC and DEF to be similar, the included angle must be equal. In this case, the included angles are $\quad \angle B\quad and\quad \angle D$.

Ratio of areas of two similar triangles is equal to :

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

  1. ratio of squares of the corresponding altitudes

  2. ratio of squares of corresponding medians.

  3. Either (A) or (B)

  4. (A) and (B) both

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Correct Option: D
Explanation:

Ratio of areas of two similar triangles is equal to ratio of squares of the corresponding altitudes and ratio of squares of corresponding medians. This means that if the ratio of either altitude or median is given and asked to find ratio of areas , then it will be the ratio of squares of the corresponding altitudes or ratio of squares of corresponding medians.

If $\Delta ABC\sim \Delta DEF$ such that area of $\Delta ABC$ is $9 cm^2$ and area of $\Delta DEF$ is $16 cm^2$ and $BC=1.8 cm$, then EF is

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

  1. 2.4 cm

  2. 1.35 cm

  3. 2.1 cm

  4. 3.2 cm

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Correct Option: A
Explanation:
$ar(\triangle ABC)=9cm^2,\,ar(\triangle DEF)=16cm^2$ and $BC=1.8cm$

$\triangle ABC\sim\triangle DEF$                [ Given ]

$\Rightarrow$  $\dfrac{ar(\triangle ABC)}{ar(\triangle DEF)}=\dfrac{(BC)^2}{(EF)^2}$                    [ Area of similar triangle theorem ]

$\Rightarrow$  $\dfrac{9}{16}=\dfrac{(1.8)^2}{(EF)^2}$
Taking square root on both sides,

$\Rightarrow$  $\dfrac{3}{4}=\dfrac{1.8}{EF}$

$\Rightarrow$  $EF=\dfrac{1.8\times 4}{3}$

$\Rightarrow$  $EF=\dfrac{7.2}{3}$

$\Rightarrow$  $EF=2.4\,cm$


Two similar triangles have

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

  1. equal sides

  2. equal areas

  3. equal angles

  4. None of these

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Correct Option: C
Explanation:

Two similar triangles have equal angles.

The sides of two similar triangles are in the ratio $4:9$ Areas of these triangles are in the ratio

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

  1. $3 : 5$

  2. $4 : 9$

  3. $81 : 16$

  4. $16 : 81$

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Correct Option: D
Explanation:

If two triangles are similar to each other, then the ratio of the area of this triangle will be equal to the square of the ratio of the corresponding sides of this triangle.

$\therefore$ The ratio between area of these triangle$=\dfrac{(4)^2}{(9)^2}=\dfrac{16}{81}$

The areas of two similar triangles are $\displaystyle 9\ { cm }^{ 2 }$ and $\displaystyle 16\ { cm }^{ 2 }$, respectively. The ratio of their corresponding heights is

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

  1. $3 : 4$

  2. $4 : 3$

  3. $2 : 3$

  4. $4 : 5$

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Correct Option: A
Explanation:
In similar traingles: -

${(\dfrac{{h1}}{{h2}})^2} = \dfrac{{S1}}{{S2}}$

Where h1 and h2 are the heights 

and S1, S2 are the areas of similar traingles

${{\rm{(}}\dfrac{{h1}}{{h2}})^2} = \dfrac{{9c{m^2}}}{{16c{m^2}}}$

$\dfrac{{h1}}{{h2}} = \sqrt {\dfrac{9}{{16}}} $

$\dfrac{{h1}}{{h2}} = \dfrac{3}{4}\ or\  {3:4}$

Triangle A has a base of x and a height of 2x. Triangle B is similar to triangle A, and has a base of 2x. What is the ratio of the area of triangle A to triangle B?

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

  1. 1:2

  2. 2:1

  3. 2:3

  4. 1:4

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Correct Option: A
Explanation:
1 to 4: Since you know that triangle B is similar to triangle A, you can set up a proportion to represent the relationship between the sides of both triangles:
$\dfrac{base}{height}=\dfrac{x}{2x}=\dfrac{2x}{?}$
By proportional reasoning, the height of triangle B must be 4x. Calculate the area of each triangle with the area formula:
Triangle A: $A=\dfrac{b\times h}{2}=\dfrac{(x)(2x)}{2}=x^2$
Triangle B: $A=\dfrac{b\times h}{2}=\dfrac{(2x)(4x)}{2}=4x^2$
The ratio of the area of triangle A to triangle B is 1 to 4.

In $\displaystyle \Delta ABC\sim \Delta DEF$ and their areas are $\displaystyle { 36cm }^{ 2 }$ and $\displaystyle { 64cm }^{ 2 }$ respectively.If side AB=3 cm. Find DE.

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

  1. 3 cm

  2. 2 cm

  3. 5 cm

  4. 4 cm

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Correct Option: D
Explanation:

In $\displaystyle \Delta ABC\sim \Delta DEF$
$\displaystyle \frac { ar.\left( \Delta ABC \right)  }{ ar.\left( \Delta DEF \right)  } =\frac { { AB }^{ 2 } }{ { DE }^{ 2 } } =\frac { { AC }^{ 2 } }{ { DF }^{ 2 } } =\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } $
$\displaystyle \frac { 36 }{ 64 } =\frac { { AB }^{ 2 } }{ { DE }^{ 2 } } $
$\displaystyle \frac { 6 }{ 8 } =\frac { 3 }{ DE } $
$\displaystyle DE=\frac { 8\times 3 }{ 6 } =\frac { 24 }{ 6 } =4cm$
Therefore, D is the correct answer.

The areas of two similar triangles are $121 cm^2$ and $81 cm^2$ respectively. Find the ratio of their corresponding heights.

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

  1. $\dfrac{11}{9}$

  2. $\dfrac{10}{9}$

  3. $\dfrac{9}{11}$

  4. $\dfrac{9}{10}$

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Correct Option: A
Explanation:

Given the areas of two similar triangles are $121$ sq cm and $81$ sq cm

We know that, the ratio of areas of two similar triangles is equal to the ratio of the squares of the corresponding heights.

The ratio of area of triangles $= \dfrac{121}{81}=\dfrac{(11)^{2}}{(9)^{2}}$
Then ratio of height of triangle $=\sqrt{\left [ \dfrac{11}{9} \right ]^{2}}=\dfrac{11}{9}$

What is the ratio of the heights of two isosceles triangles which have equal vertical angles, and of which the areas are in the ratio of $9 : 16$?

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

  1. $4.5:8$

  2. $3:4$

  3. $4:3$

  4. $8:4.5$

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Correct Option: B