In the figure,PA,QB,RC,and SD are all perpend | vedclass

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(N/A) Given: $AB = 6 \, cm$,$BC = 9 \, cm$,$CD = 12 \, cm$,and $PS = 36 \, cm$.Also,$PA$,$QB$,$RC$,and $SD$ are all perpendiculars to line $l$,which i

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(N/A) Given: $AB = 6 \, cm$,$BC = 9 \, cm$,$CD = 12 \, cm$,and $PS = 36 \, cm$.
Also,$PA$,$QB$,$RC$,and $SD$ are all perpendiculars to line $l$,which implies $PA \parallel QB \parallel RC \parallel SD$.
By the intercept theorem (or Thales' theorem application for parallel lines),the ratio of intercepts on one transversal is equal to the ratio of intercepts on another transversal:
$PQ : QR : RS = AB : BC : CD$
$PQ : QR : RS = 6 : 9 : 12$
Let $PQ = 6x$,$QR = 9x$,and $RS = 12x$.
Since the total length $PS = 36 \, cm$:
$PQ + QR + RS = 36$
$6x + 9x + 12x = 36$
$27x = 36$
$x = \frac{36}{27} = \frac{4}{3}$
Now,calculating the lengths:
$PQ = 6x = 6 \times \frac{4}{3} = 8 \, cm$
$QR = 9x = 9 \times \frac{4}{3} = 12 \, cm$
$RS = 12x = 12 \times \frac{4}{3} = 16 \, cm$

In the figure,$PA$,$QB$,$RC$,and $SD$ are all perpendiculars to a line $l$. Given $AB = 6 \, cm$,$BC = 9 \, cm$,$CD = 12 \, cm$,and $PS = 36 \, cm$. Find $PQ$,$QR$,and $RS$.

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