Question

in the figure given below PA QB RC and SD are all perpendiculars to a line AB=6 BC=9 CD=12 SP=36 find PQ QR and RS

Answer 1

$ANSWER:$

GIVEN:

Seg PA perpendicular to line l

Seg QB perpendicular to line l

Seg RC perpendicular to line l

Seg SD perpendicular to line l.

$seg \: PA || seg \: QB || seg \: RC \: || seg \: SD....1.....(lines \: perpendicular \: to \: same \: lines \: are\: parallel \: )$

Now,

$seg \: PA \: || seg \: QB \: || seg \: RC \\ \\ \: \: \: \\ \: \: \: \: \: \frac{PQ}{QR} = \frac{AB}{AC} .......(property \: of \: three \: parallel \: lines \: and \: their \: transversals) \\ \\ \: \: \\ \: \: \: \: \frac{PQ}{QR} = \frac{6}{9} .......given \: AB \: = 6 \: and \: BC = 9cm \\ \\ \: \: \\ \: \: \: \frac{PQ}{QR} = \frac{2}{3}.......2$

Also,

$seg \: QB || seg \: RC \: || seg \: SD......(from \: 1) \\ \\ \frac{QR}{BC} = \frac{BC}{CD} \\ \\ \frac{QR}{RS} = \frac{9}{12}......(given \: BC = 9 \: and \: CD = 12 \\ \\ \frac{QR}{RS} = \frac{3}{4} .......(3)$

Therefore,

PQ:QR:RS=2:3:4.......(from 2 & 3)

Let the common multiple be x.

Therefore,

PQ=2x

QR=3x

&

RS=4x.

Now,

PQ+QR+RS=PS.....(P-Q-R, Q-R-S)

Therefore,

2x+3x+4x=36

9x=36

$\\ x = \frac{36}{9}$

x=4.

PQ=2x=2×4

PQ=8 units.

QR=3x=3×4

QR=12 units

RS=4x=4×4

RS=16 units.

Therefore,

PQ=8 units, QR=12 units and RS=16 units.