Question

In ∆ PQR and ∆ XYZ, if PQ/XY = QR/YZ = PR/XZ =2/5 ,then find PQR : XYZ​

Answer 1

Correct Statement is

In ∆ PQR and ∆ XYZ, if PQ/XY = QR/YZ = PR/XZ =2/5, then find ar(∆ PQR) : ar(∆ XYZ)

ANSWER

$\begin{gathered}\begin{gathered}\bf \:Given - \begin{cases} &\sf{\rm A \: \triangle \:PQR \: and \: \triangle \: XYZ \: such \: that} \\ &\sf{ \rm \: \dfrac{PQ}{XY} = \dfrac{QR}{YZ} = \dfrac{PR}{XZ} = \dfrac{2}{5} } \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To\:find\:-\begin{cases} &\bf{\dfrac{ar(\triangle \: PQR)}{ar(\triangle \: XYZ)} }  \end{cases}\end{gathered}\end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

$\rm :A \: \triangle \:PQR \: and \: \triangle \: XYZ \: such \: that$

$\rm \: \dfrac{PQ}{XY} = \dfrac{QR}{YZ} = \dfrac{PR}{XZ}$

$\rm :\implies\: \triangle \: PQR \: \sim \: \triangle \: XYZ$

So,

• By area ratio theorem,

we have

• Theorem: If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

So,

$\rm :\implies\:\dfrac{ar(\triangle \: PQR)}{ar(\triangle \: XYZ)} = \dfrac{ {PQ}^{2} }{ {XY}^{2} }$

$\rm :\implies\:\dfrac{ar(\triangle \: PQR)}{ar(\triangle \: XYZ)} = \dfrac{ {2}^{2} }{ {5}^{2} }$

$\rm :\implies\: \boxed{ \pink{\dfrac{ \bf \: ar(\triangle \: PQR)}{ \bf \: ar(\triangle \: XYZ)} = \dfrac{ 4 }{ 25 } }}$

Additional Information :-

Basic Proportionality theorem

• Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

Pythagoras Theorem

• Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“. 

Converse of Pythagoras Theorem

• The converse of Pythagoras theorem states that “If the square of a side is equal to the sum of the square of the other two sides, then triangle must be right angle triangle”.

Answer 2

Answer:

4/25

Step-by-step explanation:

Correct Statement is

In ∆ PQR and ∆ XYZ, if PQ/XY = QR/YZ = PR/XZ =2/5, then find ar(∆ PQR) : ar(∆ XYZ)

ANSWER

\begin{gathered}\begin{gathered}\begin{gathered}\bf \:Given - \begin{cases} &\sf{\rm A \: \triangle \:PQR \: and \: \triangle \: XYZ \: such \: that} \\ &\sf{ \rm \: \dfrac{PQ}{XY} = \dfrac{QR}{YZ} = \dfrac{PR}{XZ} = \dfrac{2}{5} } \end{cases}\end{gathered}\end{gathered}\end{gathered}

Given−

⎩

⎪

⎨

⎪

⎧

A△PQRand△XYZsuchthat

XY

PQ

=

YZ

QR

=

XZ

PR

=

5

2

\begin{gathered}\begin{gathered}\bf \: To\:find\:-\begin{cases} &\bf{\dfrac{ar(\triangle \: PQR)}{ar(\triangle \: XYZ)} } \end{cases}\end{gathered}\end{gathered}

Tofind−{

ar(△XYZ)

ar(△PQR)

\large\underline\purple{\bold{Solution :- }}

Solution:−

\rm :A \: \triangle \:PQR \: and \: \triangle \: XYZ \: such \: that:A△PQRand△XYZsuchthat

\rm \: \dfrac{PQ}{XY} = \dfrac{QR}{YZ} = \dfrac{PR}{XZ}

XY

PQ

=

YZ

QR

=

XZ

PR

\rm :\implies\: \triangle \: PQR \: \sim \: \triangle \: XYZ:⟹△PQR∼△XYZ

So,

By area ratio theorem,

we have

Theorem: If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

So,

\rm :\implies\:\dfrac{ar(\triangle \: PQR)}{ar(\triangle \: XYZ)} = \dfrac{ {PQ}^{2} }{ {XY}^{2} }:⟹

ar(△XYZ)

ar(△PQR)

=

XY

2

PQ

2

\rm :\implies\:\dfrac{ar(\triangle \: PQR)}{ar(\triangle \: XYZ)} = \dfrac{ {2}^{2} }{ {5}^{2} }:⟹

ar(△XYZ)

ar(△PQR)

=

5

2

2

2

\rm :\implies\: \boxed{ \pink{\dfrac{ \bf \: ar(\triangle \: PQR)}{ \bf \: ar(\triangle \: XYZ)} = \dfrac{ 4 }{ 25 } }}:⟹

ar(△XYZ)

ar(△PQR)

=

25

4