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Quadrilaterals — Maths STD 9 — Question

Maharashtra BoardEnglish MediumSTD 9MathsQuadrilaterals5 Marks
Question
In the adjoining figure, $\triangle\text{ABC}$ is a triangle and through A, B, C, lines are drawn, parallel respectively to BC, CA and AB, intersecting at P, Q and R. Prove that the perimeter of $\triangle\text{PQR}$ is double the perimeter of $\triangle\text{ABC}.$
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Answer

Given: A $\triangle\text{ABC},$ in which through points A, B and C, lines QR, QP and RP have been draw parrallel to BC, AC and AB of$\triangle\text{ABC}$ respectively.

To Prove: Perimeter of $\triangle\text{PQR}=2(\text{Perimeter of} \triangle\text{ABC})$ Proof: Since AR || BC and AB || RC [Given] So, ABCR is a parallelogram. Therefore AR = BC ...(i) Also, AQ || BC and QB || AC So, AQBC is a parallelogram, Therefore QA = BC...(ii) Adding both side of (i) and (ii), we get AR + QA = BC + BC ⇒ QR = 2BC$\Rightarrow\text{BC}=\frac{\text{QR}}{2}$
$\therefore\text{BC}=\frac{1}{2}\text{QR}$
Similarly we can prove $\text{AB}=\frac{1}{2}\text{RP}$ and $\text{AC}=\frac{1}{2}\text{PQ}$ So, Perimeter of $\triangle\text{PQR}$ = PQ + QR + RP = 2AC + 2BC + 2AB = 2(AC + BC + AB) =2(Perimeter of $\triangle\text{ABC}$)

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