Tag: similarity

Questions Related to similarity

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

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

  1. $5:7$

  2. $5:6$

  3. $6:5$

  4. $6:7$

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Correct Option: B
Explanation:

The areas and sides of similar triangles are related as 

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

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

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

  1. $10\ cm$

  2. $12\ cm$

  3. $\displaystyle \frac{20}{3}\ cm$

  4. $8\ cm$

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Correct Option: A
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

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

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

  1. $\displaystyle {{25} \over 4}$

  2. $\displaystyle {4 \over {25}}$

  3. $\displaystyle {5 \over 2}$

  4. $\displaystyle {{25} \over 2}$

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Correct Option: B
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

$\dfrac{ar(ABC)}{ar(PQR)}=\dfrac{AB^2}{PQ^2}=\dfrac{4}{25}$.

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 

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

  1. 8 cm

  2. 10 cm

  3. 12 cm

  4. 15 cm

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Correct Option: A
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)
$\therefore AB=kDE, BC=kEF, AC=kDF$
$AB+BC+AC=k(DE+EF+DF)$
$\Rightarrow 30=k\times 20$
$\Rightarrow k=\dfrac{3}{2}$
$\Rightarrow \dfrac{AB}{DE}=\dfrac{3}{2}$
$\Rightarrow \dfrac{12}{DE}=\dfrac{3}{2}$
$\Rightarrow DE=8$.

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 

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

  1. 4 : 5

  2. 5 : 4

  3. 25 : 16

  4. 16 : 25

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Correct Option: D
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 

$\Rightarrow \dfrac{Area\, of\, 1^{st}}{Area\, of\, 2^{nd}}=\left(\dfrac{(side \,of\, 1^{st})^2}{(side\, of\, 2^{nd})^2}\right)=\dfrac{16}{25}$.

If the vector $a=2i+3j+6k$ and $b$ are collinear and $|b|=21$, then $b=$

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

  1. $\pm(2i+3j+6k)$

  2. $\pm3(2i+3j+6k)$

  3. $(2i+j+k)$

  4. $\pm21(2i+3j+6k)$

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Correct Option: A

 The area of the ratio of two similar triangles is equal to the ratio of the square of their corresponding sides.

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

  1. True

  2. False

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Correct Option: A

The areas of two similar triangles are $49 \ {cm}^{2}$ and $64 \ {cm}^{2}$ respectively. The ratio of their corresponding sides is:

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

  1. $49:64$

  2. $7:8$

  3. $64:49$

  4. none of these

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Correct Option: B
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, $\dfrac{A _1}{A _2} = \dfrac{(s _1)^2}{(s _2)^2}$
$\Longrightarrow \dfrac{49}{64} = \dfrac{(s _1)^2}{(s _2)^2}$
$\Longrightarrow\dfrac{s _1}{s _2} = \dfrac{7}{8}$

$\Delta ABC \sim  \Delta PQR$ 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:

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

  1. $9$ cm, $24$ cm

  2. $24$ cm, $9$ cm

  3. $32$ cm, $6.75$ cm

  4. $13.5$ cm, $16$ cm

Show answer
Correct Option: B
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.

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

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

  1. $4:5$

  2. $25:16$

  3. $5:4$

  4. $16:25$

Show answer
Correct Option: A
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$=4:5$