Tag: maths

Questions Related to maths

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

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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. $16:25$

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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$=4:5$

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:

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

  1. $10$ $cm$

  2. $12$ $cm$

  3. $\cfrac{20}{3}$ $cm$

  4. $8$ $cm$

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Correct Option: A
Explanation:

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

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

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

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

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

  1. $11.2\ cm$

  2. $11.6\ cm$

  3. $11.4\ cm$

  4. $10.8\ cm$

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Correct Option: A
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)
$\Rightarrow \quad \cfrac { 64 }{ 121 } =\cfrac { { BC }^{ 2 } }{ { EF }^{ 2 } } \quad { \left{ \cfrac { BC }{ EF }  \right}  }^{ 2 }={ \left{ \cfrac { 8 }{ 11 }  \right}  }^{ 2 }$
$\Rightarrow \quad \cfrac { BC }{ EF } =\cfrac { 8 }{ 11 } \quad \Rightarrow \quad BC=\cfrac { 8 }{ 11 } \times EF$
$\Rightarrow \quad BC=\cfrac { 8 }{ 11 } \times 15.4cm=11.2cm$

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$^2$)

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

  1. $54$

  2. $36$

  3. $72$

  4. $96$

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Correct Option: D
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 $\displaystyle ar\left ( \triangle DEF \right )=\frac{54\times 16}{9}=96: \text{cm}^{2}$

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:

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

  1. $6.4$ cm

  2. $10$ cm

  3. $8.8$ cm

  4. $3.2$ cm

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Correct Option: C
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.

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:

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

  1. $\dfrac 13$

  2. $\dfrac 14$

  3. $\dfrac 15$

  4. $\dfrac 16$

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Correct Option: A
Explanation:

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

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

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

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