Question

In triangle PQR , point M is on the side PQ and S is on the side PR such that QRSM is trapezium. If MS:QR = 3:5 , then find the area (PMS) : area (QRSM).

Answer 1

Since MS ∥QR and PQ is a transversal, then∠PMS = ∠PQR   (corresponding angles)Since MS ∥QR and PR is a transversal, then∠PSM = ∠PRQ   (corresponding angles)In ∆PMS and ∆PQR, ∠PMS = ∠PQR∠PSM = ∠PRQ  ∆PMS ~ ∆PQR  [AA]⇒PMMQ = MSQR = PSPR   (Correponding sides of similar ∆'s are proportional)Now, MSQR = 35Now, we know that ratio of areas of 2 similar ∆'s is equal to ratio of squares of their corresponding sides.ar(∆PMS)ar(∆PQR) = (MSQR)2 = 925⇒ar(∆PMS)ar(∆PMS) − ar(trap.QRSM) = 925⇒25 ar(∆PMS) = 9ar(∆PMS) − 9ar(trap.QRSM)⇒16 ar(∆PMS) = 9ar(trap.QRSM)⇒ar(∆PMS)ar(trap.QRSM) = 9/16

Answer 2

first in ∆PMS and PQR
BY AA they are similar
angle P = angle P (common)
angle PMS=angle PQR ( corresponding angles)
now according to th:6.6
at(PMS)/at(PQR)= MS^2/QR^2
9/25
therefore ar(PMS)= 9
and ar (PQR)=25
ar(QMSR)=ar(PQR)-ar(PMS)
= 25-9. =16
therefore
ar(PMS)/ar(QMSR)= 9/16
Hope it might be right ans and helps u☺