Question

In rectangle abcd, points p and q lie on side AB and DC respectively. Angle PMQ is a right angle, M is the midpoint of side BC and PB=4/3 BC. What is the ration PM: MQ

Answer 1

Answer:

PM:MQ = 8:3.

Step-by-step explanation:

$\rm \angle B\hat{M}P + 90^{\circ} + \angle C\hat{M}Q = 180^{\circ}$;

$\implies \rm \angle B\hat{M}P + \angle C\hat{M}Q = 90^{\circ}$;

$\implies \rm 90^{\circ} - \angle B\hat{M}P = \angle C\hat{M}Q$.

Also,

$\rm \angle B\hat{P}M = 90^{\circ} - \angle B\hat{M}P$ in right triangle PBM.

Thus $\rm \angle{P\hat{B}M} = \angle C\hat{M}Q$.

Additionally $\rm \angle \hat{B} = 90^{\circ} = \angle \hat{C}$.

Therefore $\rm \triangle PBM \sim \triangle MCQ$.

$\rm \displaystyle BC = 2\;MC$ for M is the midpoint of segment BC.

$\rm \displaystyle PB = \frac{4}{3}BC = \frac{8}{3}MC$.

$\rm \triangle PBM \sim \triangle MCQ$ implies that

$\displaystyle \rm PM:MQ = PB:MC = 1:\frac{8}{3} = 8:3$.