Tag: similarity

Questions Related to similarity

If $\triangle ABC\sim \triangle  PQR,$  $ \cfrac{ar(ABC)}{ar(PQR)}=\cfrac{9}{4}$,  $AB=18$ $cm$ and $BC=15$ $cm$, then $QR$ is equal to:

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

  1. $10$ $cm$

  2. $12$ $cm$

  3. $\cfrac{20}{3}$ $cm$

  4. $8$ $cm$

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Correct Option: A
Explanation:

Given, $\triangle ABC \sim \triangle PQR$,

Then, $\dfrac{ar(ABC)}{ar(PQR)} = \dfrac{AB^2}{PQ^2} = \dfrac{BC^2}{QR^2} = \dfrac{AC^2}{PR^2}$
$\dfrac{9}{4} = \dfrac{BC^2}{QR^2}$

$\dfrac{9}{4} = \dfrac{15^2}{QR^2}$
$QR^2 = \dfrac{4 \times 225}{9}$
$QR^2 = 100$
$QR = 10 \ cm$

Let $\triangle ABC\sim \triangle DEF$ and their areas be, respectively $64\ {cm}^{2}$ and $121\ {cm}^{2}$. If $EF=15.4\ cm$, find $BC$.

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

  1. $11.2\ cm$

  2. $11.6\ cm$

  3. $11.4\ cm$

  4. $10.8\ cm$

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Correct Option: A
Explanation:

$\triangle ABC\sim \triangle DEF\quad $ (Given)
$\Rightarrow \cfrac { ar(ABC) }{ ar(DEF) } =\cfrac { { BC }^{ 2 } }{ { EF }^{ 2 } } $ (ratio of Areas of Similar triangles are equal to ratio of squares of corresponding sides)
$\Rightarrow \quad \cfrac { 64 }{ 121 } =\cfrac { { BC }^{ 2 } }{ { EF }^{ 2 } } \quad { \left{ \cfrac { BC }{ EF }  \right}  }^{ 2 }={ \left{ \cfrac { 8 }{ 11 }  \right}  }^{ 2 }$
$\Rightarrow \quad \cfrac { BC }{ EF } =\cfrac { 8 }{ 11 } \quad \Rightarrow \quad BC=\cfrac { 8 }{ 11 } \times EF$
$\Rightarrow \quad BC=\cfrac { 8 }{ 11 } \times 15.4cm=11.2cm$

If $\triangle ABC$ is similar to $\triangle DEF$ such that $BC=3$ cm, $EF=4$ cm and area of $\triangle ABC=54: \text{cm}^{2}.$ Find the area of $\triangle DEF.$ (in cm$^2$)

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

  1. $54$

  2. $36$

  3. $72$

  4. $96$

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Correct Option: D
Explanation:

Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides,
Therefore, $\displaystyle \frac{ar\left ( \triangle ABC \right )}{ar\left ( \triangle DEF \right )}=\frac{BC^{2}}{EF^{2}}$ 

$\Rightarrow $ $\displaystyle \frac{54}{ar\left ( \triangle DEF \right )}=\frac{3^{2}}{4^{2}}$ 
Thus $\displaystyle ar\left ( \triangle DEF \right )=\frac{54\times 16}{9}=96: \text{cm}^{2}$

The areas of two similar triangles are $121$ cm$^{2}$ and $64$ cm$^{2}$, respectively. If the median of the first triangle is $12.1$ cm, then the corresponding median of the other is:

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

  1. $6.4$ cm

  2. $10$ cm

  3. $8.8$ cm

  4. $3.2$ cm

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Correct Option: C
Explanation:

The ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding medians. Therefore,
$\displaystyle \frac{121}{64}=\frac{\left ( 12.1 \right )^{2}}{x^{2}},$ where $x$ is the median of the other $\triangle .$
$\Rightarrow $ $\displaystyle x^{2}=\frac{\left ( 12.1 \right )^{2}\times 64}{121}\Rightarrow x=\sqrt{\frac{121}{100}\times 64}$
   $\displaystyle =\frac{11}{10}\times 8=8.8$ cm.

In $\Delta ABC$, a line is drawn parallel to $BC$ to meet sides $AB$ and $AC$ in $D$ and $E$ respectively. If the area of the $\Delta ADE$ is $\dfrac 19$ times area of the $\Delta ABC$, then the value of $\dfrac {AD}{AB}$ is equal to:

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

  1. $\dfrac 13$

  2. $\dfrac 14$

  3. $\dfrac 15$

  4. $\dfrac 16$

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Correct Option: A
Explanation:

By theorem on ratio of areas of similar triangles, we get

$\dfrac {A(\triangle ADE)}{A(\triangle ABC)} = \left(\dfrac {AD}{DB}\right)^2$

$\therefore \dfrac 19 = \dfrac {AD^2}{DB^2}$

$\therefore \dfrac {AD}{DB}= \dfrac 13$.

If the sides of two similar triangles are in the ratio $1:7$, find the ratio of their areas.

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

  1. $7:1$

  2. $1:7$

  3. $1:49$

  4. $1:14$

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Correct Option: C
Explanation:

We know that the relation between area of two similar triangle:
If two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding sides. 

Given, sides of two similar triangles are in the ratio $1:7$.
So, the ratio of their areas $= 1:49$.

The corresponding sides of two similar triangles are in the ratio $a : b$. What is the ratio of their areas?

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

  1. $a : b$

  2. $2a : 2b$

  3. $a^{2} : b^{2}$

  4. $\dfrac {1}{a} : \dfrac {1}{b}$

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Correct Option: C
Explanation:

Given two triangles are similar, then the ratio of the areas $=a^2:b^2$

Eg: The ratio of the sides of a similar triangle is $4:9$
Scale factor for the sides of these triangles $k=\cfrac 49$
$\therefore$ Ratio of area will be:
$k^2=\cfrac {area of \triangle A}{area of \triangle B}=(\cfrac 49)^2=\cfrac {16}{81}$

The ratio of areas of two similar triangles is $81 : 49$. If the median of the smaller triangle is $4.9\ cm$, what is the median of the other?

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

  1. $4.9\ cm$

  2. $6.3\ cm$

  3. $7\ cm$

  4. $9\ cm$

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Correct Option: B
Explanation:

Area of $\triangle ABC= \cfrac 12 \times base \times height$

In similar triangles, $\cfrac {base 1}{base 2}=\cfrac {height 1}{height 2}=\cfrac {side 1}{side}$
$\therefore \cfrac {Area 1}{Area 2}= (\cfrac {Median 1}{Median 2})^2$
Ratio of Medians $=\sqrt{\cfrac {81}{49}}=\cfrac 97 >1$ 
$\therefore$ Altitude of smaller triangle $=4.9 \times \cfrac 97=6.3$

$\triangle ABD \sim \triangle DEF$ and the perimeters of $\triangle ABC$ and $\triangle DEF$ are $30 cm$ and $18 cm$ respectively. If $BC = 9 cm$, calculate measure of $EF$.

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

  1. $6.3\ cm$

  2. $5.4\ cm$

  3. $7.2\ cm$

  4. $4.5\ cm$

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Correct Option: B
Explanation:

Similar triangles are triangle with similar shape but can have different sizes. 

Since there is something common about them, then to establish the relationship we have something called linear scale factor which is used to get the length of the other when the length of the other similar triangles is known. 
$LSF=\cfrac {30}{18}=\cfrac 53$
$\cfrac {BC}{EF}=\cfrac 53 \Rightarrow \cfrac 9{EF}=\cfrac 53$
$\therefore EF=9 \times \cfrac 35 = 5.4cm$

Two isosceles triangles have their corresponding angles equal and their areas are in the ratio $25 : 36$. Find the ratio of their corresponding heights

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

  1. $25 : 35$

  2. $36 : 25$

  3. $5 : 6$

  4. $6 : 5$

Show answer
Correct Option: C
Explanation:
We know, Ratios of areas of similar triangles is equal to ratio of squares of their corresponding sides.
Hence,

$\dfrac{Area \ of \ \triangle _1}{Area \ of \ \triangle _2}=\dfrac{(height \ of \ \triangle _1)^2}{(height \ of \ \triangle _2)^2}$

Taking square root on both sides,

$\dfrac{(height \ of \ \triangle _1)}{(height \ of \ \triangle _2)}=\sqrt{\dfrac{Area \ of \ \triangle _1}{Area \ of \ \triangle _2}}$

$\dfrac{(height \ of \ \triangle _1)}{(height \ of \ \triangle _2)}=\sqrt{\dfrac{25}{36}}=\dfrac{5}{6}$

Option C