Tag: relations between the areas of triangles

Questions Related to 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

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. $6 : 5$

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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

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 $\Delta ABC$?

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

  1. 144 cm$^2$

  2. 121 cm$^2$

  3. 100 cm$^2$

  4. 81 cm$^2$

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Correct Option: C
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 ]

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

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

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

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.

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

  1. $3 cm$

  2. $2.5 cm$

  3. $4 cm$

  4. $3.5 cm$

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Correct Option: D
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 } $
$100{x}^{2}=25\times 49$
${x}^{2}=\cfrac{25\times 49}{100}$
${x}^{2}=\cfrac{49}{4}$
$x=\cfrac{7}{2}$
$x=3.5cm$

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

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

  1. $9$

  2. $3$

  3. $\dfrac {1}{3}$

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

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Correct Option: A
Explanation:

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

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

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

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

  1. $\sqrt{15} : \sqrt{19}$

  2. $15 : 19$

  3. $225 : 361$

  4. $125 : 144$

Show answer
Correct Option: C
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.

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.

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

  1. 20 cm

  2. 15 cm

  3. 10 cm

  4. 16 cm

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Correct Option: C
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
$\therefore \dfrac{(13)^{2}}{(x)^{2}}=\dfrac{25}{16}$

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

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

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

$\Rightarrow x=10cm$

If the sides of two similar triangles are in the ratio $2 : 3$, then their areas are in the ratio:

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

  1. $9 : 4$

  2. $4 : 9$

  3. $2 : 3$

  4. $3 : 2$

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Correct Option: B
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 $4:9$.

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:

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

  1. $4\sqrt { 5 } cm$

  2. $4\sqrt { 6 } cm$

  3. $4\sqrt { 7 } cm$

  4. $4\sqrt { 11 } cm$

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Correct Option: C
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 $BE=\dfrac{1}{2}BC=6cm$
$\therefore\ DE=BE-BD=2cm$.
$\therefore\ AE^2=AB^2-BE^2=144-36=108$
$\because\ AED=90^{o}$
$\therefore\ AD^2=AE^2+DE^2=108+4=112$
$\therefore\ AD=4\sqrt{7}cm$.

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:

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

  1. $4.8$ cm

  2. $12$ cm

  3. $4$ cm

  4. $5.6$ cm

Show answer
Correct Option: A
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 } $
$\Rightarrow  \dfrac { 4 }{ PN } =\dfrac { 10 }{ 12 } \ \Rightarrow PM=4.8$

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:

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

  1. $30cm^{2}$

  2. $7.5cm^{2}$

  3. $15cm^{2}$

  4. $20cm^{2}$

Show answer
Correct Option: C
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)$
$\Rightarrow \text{Ar}(\triangle EBC)=15 \ \ \text{cm}^2$