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  • In Fig., PA, QB, RC and SD are all perpendiculars to a line l, AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm. Find PQ, QR and RS.

In Fig., PA, QB, RC and SD are all perpendiculars to a line l, AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm. Find PQ, QR and RS.

 

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Answer: \left [ PQ=8_{cm},QR=12\; cm,RS=16\; cm\right ]

Given :- PA, QB, RC and SD are all perpendiculars to a line l, AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm

We know that 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.

\therefore PA \parallel QB \parallel RC \parallel SD     (parallel lines)

Then according to basic proportionality theorem

PQ : QR : RS = AB : BC : CD = 6 : 9 : 12

         \text {Let}\\PQ = 6x\\ QR = 9x\\ RS = 12x

length of PS = 36        (given)

\therefore \; \; \; PQ + QR + RS = 36\; \; \; \Rightarrow \; \; \; 6x + 9x + 12x = 36 \; \; \; \; \; \Rightarrow \; \; \; \; \; 27x = 36

x=\frac{36}{27}=\frac{4}{3}

PQ=6x=6 \times\frac{4}{3}=8\; cm

QR=9x=\frac{9\times 4}{3}=12\; cm

RS=12x=\frac{12\times 4}{3}=16\; cm

Posted by

infoexpert23

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