Tag: areas of similar triangles

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

Two isosceles triangles have equal vertical angles and their areas are in the ratio $16:25$. Find the ratio of their corresponding heights.

  1. $4:5$
  2. $25:16$
  3. $5:4$
  4. $16:25$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\triangle ABC$ and $\triangle DEF$ be the given triangles in which $AB=AC, DE=DF$, $\angle A=\angle D$
and $\cfrac{Area\quad (\triangle ABC)}{Area\quad (\triangle DEF)}=\cfrac{16}{25}$
Draw $AL\bot  BC$ and $DM\bot  EF$
Now, $\cfrac{AB}{BC}=1$ and $\cfrac{DE}{DF}=1$  ($\because \quad AB=AC;\quad DE=DF$)
$\Rightarrow \cfrac{AB}{AC}=\cfrac{DE}{DF}$,
$\therefore$ $\ln \triangle ABC$ and $\triangle DEF$, we have
$\cfrac{AB}{DE}=\cfrac{AC}{DF}$ and $\angle A=\angle D$
$\Rightarrow$ $\triangle ABC\sim \triangle DEF$ [By SAS similarity axiom)
But, the ratio of the areas of two similar $\triangle s$ is the same as the ratio of the squares of their corresponding heights.
$\cfrac{Area\quad (\triangle ABC)}{Area\quad (\triangle DEF)}=\cfrac {{AL}^{2}}{{DM}^{2}}$
$\Rightarrow$ $\cfrac{16}{25}={ \left( \cfrac {AL}{DM}  \right)  }^{ 2 }$
$\Rightarrow$ $\cfrac{4}{5}$
$\therefore$ $AL:DM=4:5$, i.e., the ratio of their corresponding heights$=4:5$

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

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:

  1. $10$ $cm$
  2. $12$ $cm$
  3. $\cfrac{20}{3}$ $cm$
  4. $8$ $cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

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

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

  1. $11.2\ cm$
  2. $11.6\ cm$
  3. $11.4\ cm$
  4. $10.8\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
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)
$\Rightarrow \quad \cfrac { 64 }{ 121 } =\cfrac { { BC }^{ 2 } }{ { EF }^{ 2 } } \quad { \left{ \cfrac { BC }{ EF }  \right}  }^{ 2 }={ \left{ \cfrac { 8 }{ 11 }  \right}  }^{ 2 }$
$\Rightarrow \quad \cfrac { BC }{ EF } =\cfrac { 8 }{ 11 } \quad \Rightarrow \quad BC=\cfrac { 8 }{ 11 } \times EF$
$\Rightarrow \quad BC=\cfrac { 8 }{ 11 } \times 15.4cm=11.2cm$

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

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$^2$)

  1. $54$
  2. $36$
  3. $72$
  4. $96$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $\displaystyle ar\left ( \triangle DEF \right )=\frac{54\times 16}{9}=96: \text{cm}^{2}$

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

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:

  1. $6.4$ cm
  2. $10$ cm
  3. $8.8$ cm
  4. $3.2$ cm
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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:

  1. $\dfrac 13$
  2. $\dfrac 14$
  3. $\dfrac 15$
  4. $\dfrac 16$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

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

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

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

  1. $7:1$
  2. $1:7$
  3. $1:49$
  4. $1:14$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $= 1:49$.

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

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

  1. $a : b$
  2. $2a : 2b$
  3. $a^{2} : b^{2}$
  4. $\dfrac {1}{a} : \dfrac {1}{b}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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:
$k^2=\cfrac {area of \triangle A}{area of \triangle B}=(\cfrac 49)^2=\cfrac {16}{81}$

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

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?

  1. $4.9\ cm$
  2. $6.3\ cm$
  3. $7\ cm$
  4. $9\ cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $=4.9 \times \cfrac 97=6.3$

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

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

  1. $6.3\ cm$
  2. $5.4\ cm$
  3. $7.2\ cm$
  4. $4.5\ cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
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. 
$LSF=\cfrac {30}{18}=\cfrac 53$
$\cfrac {BC}{EF}=\cfrac 53 \Rightarrow \cfrac 9{EF}=\cfrac 53$
$\therefore EF=9 \times \cfrac 35 = 5.4cm$