Tag: maths

Questions Related to maths

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

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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. $20cm^{2}$

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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)$
$\Rightarrow \text{Ar}(\triangle EBC)=15 \ \ \text{cm}^2$

Through a point $P$ inside the triangle $ABC$ a line is drawn parallel to the base $AB$, dividing the triangle into two equal area. If the altitude to $AB$ has a length of $1$, then the distance from $P$ to $AB$ is

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

  1. $\dfrac {1}{2}$

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

  3. $2 - \sqrt {2}$

  4. $\dfrac {2 - \sqrt {2}}{2}$

  5. $\dfrac {2 + \sqrt {2}}{8}$

Show answer
Correct Option: D
Explanation:

Let $x$ be the distance from $P$ to $AB$. By similar triangles.
$\dfrac {1}{2} = \dfrac {(1 - x)^{2}}{1^{2}}; \therefore 1 - x = \pm \dfrac {1}{\sqrt {2}}; \therefore x = \dfrac {2 - \sqrt {2}}{2}$
(negative sq. root rejected).

Triangles ABC and DEF are similar. If their areas are 64 $cm^2$ and 49 $cm^2$ and if AB is 7 cm, then find the value of DE.

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

  1. 8 cm

  2. $\dfrac{49}{8}$ cm

  3. $\dfrac{8}{49}$ cm

  4. $\dfrac{64}{7}$cm

Show answer
Correct Option: B
Explanation:

$\Delta ABC \Delta DEF$
$\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF}$.
We know that,
$\dfrac{Area of \Delta  ABC}{ Area of \Delta  DEF} = $ $\Rightarrow \dfrac{64}{49} = \left ( \dfrac{7}{DE} \right )^2$
$\Rightarrow \left ( \dfrac{8}{7} \right )^2 = \left ( \frac{7}{DE} \right )^2 \Rightarrow \left ( \dfrac{8}{7} \right )^2 = \left ( \dfrac{7}{DE} \right )^2 = \dfrac{49}{8}$cm 

If $\triangle ABC\sim \triangle QRP,\dfrac{Ar(ABC)}{Ar(QRP)}=\dfrac{9}{4}$,$AB=18\ cm$ 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. $20\ cm$

  4. $8\ cm$

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Correct Option: A
Explanation:

Given $ \triangle  ABC \sim  \triangle  QRP $
Therefore, $ \frac { Area\triangle ABC\quad  }{ Area\triangle QRP\quad }=\frac { { BC }^{ 2 } }{ { PR }^{ 2 } }  $
or $ \frac { 9 }{ 4 }  = \frac { { 15 }^{ 2 } }{ { PR }^{ 2 } }  $
or $ { PR }^{ 2 }\quad =\quad \frac { { 15 }^{ 2 }\quad \times \quad 4 }{ 9 }  $ cm.
Therefore, $ { PR }=\frac { { 15 }\times \quad 2 }{ 3 }  = $ 10 cm.

Which among the following is/are correct?
(I) If the altitudes of two similar triangles are in the ratio $2:1$, then the ratio of their areas is $4 : 1$.
(II) $PQ \parallel BC$ and $AP : PB=1:2$. Then, $\dfrac{A(\triangle APQ)}{A(\triangle ABC)}=\dfrac{1}{4}$

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

  1. $(I)$

  2. $(II)$

  3. Both $(I)$ and $(II)$

  4. None of the above

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Correct Option: A
Explanation:

Option A: This statement is correct. The ratio of the altitudes of the similar triangles  is  $2:1$

Ratio of the areas of the similar triangles $=$ Square of the ratio of  the  altitudes.

$\therefore$ Ratio  of  the  areas $ =  { \left( \dfrac { 2 }{ 1 }  \right)  }^{ 2 }=  4:1$


Option B: If  $PQ\parallel BC$,  then $\triangle APQ \sim \triangle ABC$ by AA test of similarity.

Hence, $\dfrac {A( \triangle APQ)}{A(\triangle ABC)}=\dfrac { AP^2 }{ AB^2 }$

If $AP = x$ and $BP = 2x$, then $AB = 3x$.

$\therefore \dfrac {A( \triangle APQ)}{A(\triangle ABC)}=\dfrac 19$

So, the given statement is false

Two triangles ABC and PQR  are similar, if $BC : CA : AB = $1: 2 : 3, then $\dfrac{QR}{PR}$ is

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

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

  2. $\dfrac{1}{2}$

  3. $\dfrac{1}{{\sqrt{2}}}$

  4. $\dfrac{2}{3}$

Show answer
Correct Option: B
Explanation:

Two triangle ABC and PQR are similar then,
$AB: BC:CA :: PQ: QR: PR$
Given $BC: CA = 1:2$ 
Hence, $QR : PR = BC : CA = 1:2$

Let $\displaystyle \Delta XYZ$ be right angle triangle with right angle at Z. Let $\displaystyle A _{X}$ denotes the area of the circle with diameter YZ. Let $\displaystyle A _{Y}$ denote the area of the circle with diameter XZ and let $\displaystyle A _{Z}$ denotes the area of the circle diameter XY. Which of the following relations is true?

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

  1. $\displaystyle A _{Z}=A _{X}+A _{Y}$

  2. $\displaystyle A _{Z}=A^{2} _{X}+A^{2} _{Y}$

  3. $\displaystyle A^{2} _{Z}=A^{2} _{X}+A^{2} _{Y}$

  4. $\displaystyle A^{2} _{Z}=A^{2} _{X}-A^{2} _{Y}$

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Correct Option: A
Explanation:

In $\triangle XYZ$, using Pythagoras theorem,
$XY^2 = XZ^2 + YZ^2$
$\pi XY^2 = \pi XZ^2 + \pi YZ^2$ (Multiply by $\pi$)
$A _z = A _x + A _y$