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

  1. $25 : 35$
  2. $36 : 25$
  3. $5 : 6$
  4. $6 : 5$
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

In similar triangles $\triangle ABC$ and $\triangle FDE, DE = 4 cm, BC = 8 cm$ and area of $\triangle FDE = 25 cm^2$. What is the area of $\Delta ABC$?

  1. 144 cm$^2$
  2. 121 cm$^2$
  3. 100 cm$^2$
  4. 81 cm$^2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

  $DE=4\,cm,\,BC=8\,cm$ and $ar(\triangle FDE)=25\,cm^2$                  [ Given ]


$\Rightarrow$  $\triangle ABC\sim\triangle FDE$             [ Given ]

$\Rightarrow$  $\dfrac{ar(\triangle ABC)}{ar(\triangle FDE)}=\dfrac{(BC)^2}{(DE)^2}$                       [ By area of similar triangle theorem ]

$\Rightarrow$  $\dfrac{ar(\triangle ABC)}{25}=\dfrac{(8)^2}{(4)^2}$

$\Rightarrow$  $ar(\triangle ABC)=\dfrac{64}{16}\times 25$

$\Rightarrow$  $ar(\triangle ABC)=4\times 25$
$\therefore$  $ar(\triangle ABC)=100\,cm^2$

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 $81\ cm^{2}$ and $49\ cm^{2}$. If the altitude of the bigger triangle is $4.5\ cm$, find the corresponding altitude of the smaller triangle.

  1. $3 cm$
  2. $2.5 cm$
  3. $4 cm$
  4. $3.5 cm$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given: Area of $two$ similar triangle $81{cm}^{2}$ and $49{cm}^{2}$

Altitude of bigger triangle $=4.5cm$
For similar triangle,
${\text{Ratio on sides}}^{2}=\text {Ratio of their Area}$
$\therefore$ $\cfrac { { 4.5 }^{ 2 } }{ { x }^{ 2 } } =\cfrac { 81 }{ 49 } $
$\cfrac { 45\times 45 }{ { x }^{ 2 }\times 100 } =\cfrac { 81 }{ 49 } $
$100{x}^{2}=25\times 49$
${x}^{2}=\cfrac{25\times 49}{100}$
${x}^{2}=\cfrac{49}{4}$
$x=\cfrac{7}{2}$
$x=3.5cm$

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

If $\triangle ABC$ and $\triangle PQR$ are similar and $\dfrac {BC}{QR} = \dfrac {1}{3}$ find $\dfrac {area (PQR)}{area (BCA)}$

  1. $9$
  2. $3$
  3. $\dfrac {1}{3}$
  4. $\dfrac {1}{9}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\triangle ABC$ & $\triangle PQR$ are similar.

$\therefore \cfrac{AB}{PQ}=\cfrac{BC}{QR}=\cfrac{1}{3}$
$\therefore \cfrac{\text{Area}(PQR)}{\text{Area}(BCA)}={(\cfrac{QR}{BC}})^{2}$
$\cfrac { { Area }(PQR) }{ { Area }(BCA) } ={ (\cfrac { 3 }{ 1 } ) }^{ 2 }=9$

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

What is the ratio of the areas of two similar triangles whose corresponding sides are in the ratio 15:19?

  1. $\sqrt{15} : \sqrt{19}$
  2. $15 : 19$
  3. $225 : 361$
  4. $125 : 144$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know ratio  of areas of two similar triangles is the ratio of the square of their corresponding sides.

Ratio of sides $\dfrac{15}{19}$
Ratio of areas $={ \left( \dfrac { 15 }{ 19 }  \right)  }^{ 2 }=\dfrac { 225 }{ 361 } $
So option $C$ is correct.

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 100 $cm^2$ and 64 $cm^2$. If the median of greater side of first triangle is 13 cm, find the corresponding median of the other triangle.

  1. 20 cm

  2. 15 cm

  3. 10 cm

  4. 16 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given area of two similar triangles are $100$ sq cm and $64$ sq cm
The areas of two Similar-Triangles are in the ratio of the squares of the corresponding medians
The ratio of area of triangle $=\dfrac{100}{64}=\dfrac{25}{16}$
Median of greater triangle is $13$ cm and let other median is $x$ cm
$\therefore \dfrac{(13)^{2}}{(x)^{2}}=\dfrac{25}{16}$

$\Rightarrow \dfrac{169}{x^{2}}=\dfrac{25}{16}$

$\Rightarrow 25x^{2}=169\times 16$

$\Rightarrow x^{2}=\dfrac{2704}{25}=108.16$

$\Rightarrow x=10cm$

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 $2 : 3$, then their areas are in the ratio:

  1. $9 : 4$
  2. $4 : 9$
  3. $2 : 3$
  4. $3 : 2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, sides of two similar triangles are in the ratio $2:3$

Thus, the ratio of their areas are $($ side $)^2$
$=\left (\dfrac {2}{3}\right)^2=\dfrac {4}{9}$ 
Therefore, the areas are in the raatio $4:9$.

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

In $\Delta ABC$, $D$ is a point on $BC$ such that $3BD = BC$. If each side of the triangle is $12 cm$, then $AD$ equals:

  1. $4\sqrt { 5 } cm$
  2. $4\sqrt { 6 } cm$
  3. $4\sqrt { 7 } cm$
  4. $4\sqrt { 11 } cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given $\triangle ABC$ with $D$ a point on $BC$ such that $3BD=BC$

$\therefore$ $BD=\dfrac{BC}{3}=4cm$
$AD\  \bot\ BC$
Let's take a point $E$ on $BC$ which makes a right angle triangle $ADE$ and $AEB$ at $E$ such that $BE=\dfrac{1}{2}BC=6cm$
$\therefore\ DE=BE-BD=2cm$.
$\therefore\ AE^2=AB^2-BE^2=144-36=108$
$\because\ AED=90^{o}$
$\therefore\ AD^2=AE^2+DE^2=108+4=112$
$\therefore\ AD=4\sqrt{7}cm$.

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

In $\Delta ABC \sim  \Delta PQR$, $M$ is the midpoint of $BC$ and $N$ is the midpoint of $QR$. If the area of $\Delta ABC =$ $100$ sq. cm and the area of $\Delta PQR =$ $144$ sq. cm. If $AM = 4$ cm, then $PN$ is:

  1. $4.8$ cm
  2. $12$ cm
  3. $4$ cm
  4. $5.6$ cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\dfrac { ar(\triangle ABC) }{ ar(\triangle PQR) } =\dfrac { 100 }{ 144 } $

If triangles are similar then ratio of their areas is equal to ratio of square of their corresponding sides
$\dfrac { AB^{ 2 } }{ PQ^{ 2 } } =\dfrac { 100 }{ 144 } \ \dfrac { AB }{ PQ } =\dfrac { 10 }{ 12 } $
$AM$ and $PN$ are medians
Therefore, $ \dfrac { AM }{ PN } =\dfrac { AB }{ PQ } $
$\Rightarrow  \dfrac { 4 }{ PN } =\dfrac { 10 }{ 12 } \ \Rightarrow PM=4.8$

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

D and E are the points on the sides AB and AC respectively of triangle ABC such that $ DE||BC$. If area of $ \triangle DBC =15 cm^2$, then area of $\triangle EBC $ is:

  1. $30cm^{2}$
  2. $7.5cm^{2}$
  3. $15cm^{2}$
  4. $20cm^{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, $DE||BC$

Therefore, the altitudes of $\triangle EBC$ and $\triangle DBC$ are equal.
Also, they have a common base $BC$.
Thus, $ \text{Ar}(\triangle EBC)=\text{Ar}(\triangle DBC)$
$\Rightarrow \text{Ar}(\triangle EBC)=15 \ \ \text{cm}^2$