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

ABC is an isosceles triangle right angled at B. Similar triangles ACD and ABE are constructed in sides AC and AB. Find the ratio between the areas of $\triangle ABE$ and $\triangle ACD$.

  1. $2:1$
  2. $1:1$
  3. $1:2$

  4. none
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In $\triangle ABC$

$\implies { AB }^{ 2 }+{ BC }^{ 2 }={ AC }^{ 2 }$
$\implies\quad { AB }^{ 2 }+{ AB }^{ 2 }={ AC }^{ 2 }$
$\implies\quad { AC }^{ 2 }={ 2AB }^{ 2 }\quad -(1)$
Ratio of areas of similar triangle is equal to ratio of squares of their corresponding sides.
$\implies\quad \cfrac { Area\quad (\triangle ABE) }{ Area\quad (\triangle ACD) } =\cfrac { { AB }^{ 2 } }{ { AC }^{ 2 } } $
 using(1)
$\implies\quad \cfrac { Area\quad (\triangle ABE) }{ Area\quad (\triangle ACD) } =\cfrac { { AB }^{ 2 } }{ { 2AB }^{ 2 } } =\cfrac { 1 }{ 2 } $

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

Area of similar triangles are in the ratio $25:36$ then ratio of their similar sides is _________?

  1. $5:7$
  2. $5:6$
  3. $6:5$
  4. $6:7$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The areas and sides of similar triangles are related as 

$\dfrac{Ar(\Delta ABC)}{Ar(\Delta PQR)}=\left(\dfrac {AB}{PQ}\right)^2\\dfrac {25}{36}=\left(\dfrac {AB}{PQ}\right)^2\\dfrac{AB}{PQ}=\sqrt {\dfrac {25}{36}}=\dfrac 56$

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

If $\Delta ABC \sim \Delta QRP, \displaystyle \frac{ar (ABC)}{ar (PQR)} = \frac{9}{4}, AB = 18 cm$ and $BC=15 cm$; then PR is equal to

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

Area of ∆ ABCArea of ∆QRP = 94


AB = 18 cm , BC = 15 cm So PR = ?

We know when two triangles are similar then " The areas of two similar triangles are proportional to the squares of their corresponding sides.

Area of ∆ ABCArea of ∆ QRP = $AB^2QR^2$ = $Bc^2PR^2$ = $AC^2QP^2$
So, we take 
Area of ∆ ABC Area of ∆ QRP = $BC^2PR^2$

Now substitute all given values and get

94 = 152PR2

Taking square root on both hand side, we get

32 = 15PR

PR = 10 cm

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

If $\Delta ABC \sim \Delta PQR$ and $\displaystyle {{PQ} \over {AB}} = {5 \over 2}$ then area $(\Delta ABC):$ area $(\Delta PQR) = ?$

  1. $\displaystyle {{25} \over 4}$
  2. $\displaystyle {4 \over {25}}$
  3. $\displaystyle {5 \over 2}$
  4. $\displaystyle {{25} \over 2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\Delta ABC\sim \Delta PQR$

Also $\dfrac{PQ}{AB}=\dfrac{5}{2}$

If triangles are similar then the ratio of their is equal to ratio of square of the sides

$\dfrac{ar(ABC)}{ar(PQR)}=\dfrac{AB^2}{PQ^2}=\dfrac{4}{25}$.
Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

The perimeter of two similar triangles is 30 cm and 20 cm. If one altitude of the former triangle is 12 cm, then length of the corresponding altitude of the latter triangle is 

  1. 8 cm

  2. 10 cm

  3. 12 cm

  4. 15 cm
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\Delta$ABC and $\Delta$DEF be two similar triangle. Perimeter of first and second triangles are $30$cm and $20$cm respectively.
Then $\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{AC}{DF}=k$ (say)
$\therefore AB=kDE, BC=kEF, AC=kDF$
$AB+BC+AC=k(DE+EF+DF)$
$\Rightarrow 30=k\times 20$
$\Rightarrow k=\dfrac{3}{2}$
$\Rightarrow \dfrac{AB}{DE}=\dfrac{3}{2}$
$\Rightarrow \dfrac{12}{DE}=\dfrac{3}{2}$
$\Rightarrow DE=8$.
Multiple choice maths similarity areas of similar figures areas of similar triangles relations between the areas of triangles

The perimeter of two similar triangles is 40 cm and 50 cm. Then the ratio of the areas of the first and second triangles is 

  1. 4 : 5

  2. 5 : 4

  3. 25 : 16

  4. 16 : 25
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
We know the ratio of perimeters of $2$ similar triangles are equal to ratio of corresponding sides

i.e., $\dfrac{perimeter \,of \,1^{st}}{perimeter\, of \,2^{nd}}=\dfrac{side\, of\, 1^{st}}{side\, of\, 2^{nd}}$

$\Rightarrow \dfrac{40}{50}=\dfrac{side\, of\, 1^{st}}{side\, of \,2^{nd}}=\dfrac{4}{5}$

As both the triangles are similar 

$\Rightarrow \dfrac{Area\, of\, 1^{st}}{Area\, of\, 2^{nd}}=\left(\dfrac{(side \,of\, 1^{st})^2}{(side\, of\, 2^{nd})^2}\right)=\dfrac{16}{25}$.
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 $49 \ {cm}^{2}$ and $64 \ {cm}^{2}$ respectively. The ratio of their corresponding sides is:

  1. $49:64$
  2. $7:8$
  3. $64:49$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Areas of two similar triangles are $49 $ cm $^2$ and $64$ cm $^2.$
For similar triangles the ratio of areas is equal to the ratio of square of corresponding sides.
Hence, $\dfrac{A _1}{A _2} = \dfrac{(s _1)^2}{(s _2)^2}$
$\Longrightarrow \dfrac{49}{64} = \dfrac{(s _1)^2}{(s _2)^2}$
$\Longrightarrow\dfrac{s _1}{s _2} = \dfrac{7}{8}$

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

$\Delta ABC \sim  \Delta PQR$ and $\displaystyle\frac{A( \Delta ABC)}{A( \Delta PQR)}=\dfrac{16}{9}$. If $PQ=18$ cm and $BC=12$ cm, then $AB$ and $QR$ are respectively:

  1. $9$ cm, $24$ cm
  2. $24$ cm, $9$ cm
  3. $32$ cm, $6.75$ cm
  4. $13.5$ cm, $16$ cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle\frac{16}{9}=\left[\frac{AB}{PQ}\right]^2=\left[\frac{BC}{QR}\right]^2$

$\displaystyle\Rightarrow \frac{16}{9}=\left[\frac{AB}{18}\right]^2$ and $\displaystyle\frac{16}{9}=\left[\frac{12}{QR}\right]^2$

$\displaystyle \Rightarrow \frac{4}{3}=\frac{AB}{18}$ and $\displaystyle \frac{4}{3}=\frac{12}{QR}$

$\Rightarrow AB=24$ cm, $QR=9$ cm.