In PA, QB, RC and SD are all perpendiculars to a line l, AB = 6cm, BC = 9cm, CD = 12cm and SP = 36cm. Find PQ, QR and RS.
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Given: PA, QB, RC and SD are perpendiculars on line l.
AB = 6cm, BC = 9cm, CD = 12cm
To find: PQ, QR and RS
Construction: Produce SP and I to meet each other at E.
Proof: In $\triangle\text{EDS},$
AP || BQ || DS || CR [Given]
⇒ PQ : QR : RS = AB : BC : CD
⇒ PQ : QR : RS = 6 : 9 : 12
Let PQ = 6x
Then = 9x
and RS = 12x
⇒ PQ + QR + RS = 36cm
⇒ 6x + 9x + 12x = 36
⇒ 27x = 36
$\Rightarrow\text{x}=\frac{36}{27}=\frac{4}{3}$
$\therefore\text{PQ}=6\times\frac{4}{3}=8\text{cm}$
$\text{QR}=9\times\frac{4}{3}=12\text{cm}$
$\text{RS}=12\times\frac{4}{3}=16\text{cm}$
AB = 6cm, BC = 9cm, CD = 12cm
To find: PQ, QR and RS
Construction: Produce SP and I to meet each other at E.
Proof: In $\triangle\text{EDS},$
AP || BQ || DS || CR [Given]
⇒ PQ : QR : RS = AB : BC : CD
⇒ PQ : QR : RS = 6 : 9 : 12
Let PQ = 6x
Then = 9x
and RS = 12x
⇒ PQ + QR + RS = 36cm
⇒ 6x + 9x + 12x = 36
⇒ 27x = 36
$\Rightarrow\text{x}=\frac{36}{27}=\frac{4}{3}$
$\therefore\text{PQ}=6\times\frac{4}{3}=8\text{cm}$
$\text{QR}=9\times\frac{4}{3}=12\text{cm}$
$\text{RS}=12\times\frac{4}{3}=16\text{cm}$
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