Question

if P, Q, R, are the midpoints of side AB, BC, CA respectively of triangle ABC and area of triangle PQR = k (area triangle ABC) then the value of k is?​

Answer 1

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Answer 2

Basic concept used :-

$\large \boxed {\tt \:  ⟼Midpoint \: theorem :- }$

• The midpoint theorem states that “The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.”

$\large \boxed {\tt \:  ⟼ \: Area \: Ratio \: Theorem:- :- }$

•  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.

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

$\tt \:  ⟼Since, \: P \: is \: midpoint \: of \: AB</p><p>$

$\tt \:  ⟼and \: Q \: is \: midpoint \: of \: BC.</p><p>$

$\tt \:  ⟼So, \: by \: midpoint \: theorem,$

$\tt \:  ⟼ \: PQ \: = \: \dfrac{1}{2} AC$

$\bf\implies \:\dfrac{PQ}{AC} = \dfrac{1}{2} - - - (1)$

$\tt \:  ⟼Now, \: R \: is \: midpoint \: of \: AC$

$\tt \:  ⟼and \: Q \: is \: midpoint \: of \: BC.</p><p>$

$\tt \:  ⟼So, \: by \: midpoint \: theorem,$

$\tt \:  ⟼ \: QR \: = \: \dfrac{1}{2} AB$

$\bf\implies \:\dfrac{QR}{AB} = \dfrac{1}{2} - - - (2)$

$\tt \:  ⟼Now, \: R \: is \: midpoint \: of \: AC$

$\tt \:  ⟼ \: and \: \: P \: is \: midpoint \: of \: AB$

$\tt \:  ⟼So, \: by \: midpoint \: theorem,$

$\tt \:  ⟼PR \: = \: \dfrac{1}{2} BC$

$\bf\implies \:\dfrac{PR}{BC} \: = \dfrac{1}{2} - - - (3)$

☆ From (1), (2) and (3), we conclude that

$\tt \:  ⟼\dfrac{PQ}{AC} = \dfrac{QR}{AB} = \dfrac{PR}{BC} = \dfrac{1}{2}$

$\tt \:  ⟼This \: implies \: \triangle \: QRP \: \sim \: \triangle ABC \:$

$\tt \:  ⟼ \: \therefore \: By \: Area \: Ratio \: Theorem$

$\tt \:  ⟼\dfrac{ar(\triangle \: QRP)}{ ar(\triangle ABC)} = \dfrac{ {QR}^{2} }{ {AB}^{2} }$

$\tt \:  ⟼\dfrac{ar(\triangle \: QRP)}{ ar(\triangle ABC)} = \: \bigg( { \dfrac{1}{2} \bigg)}^{2}$

$\tt \:  ⟼\dfrac{ar(\triangle \: QRP)}{ ar(\triangle ABC)} = \: \dfrac{1}{4}$

$\tt \:  ⟼ \: ar(\triangle \: QRP) \: = \: \dfrac{1}{4} ar( \triangle ABC)$

$\large \boxed{\tt \:  ⟼ \: Hence, \: k \: = \: \dfrac{1}{4} }$