Question

In ∆ PQR, seg MN || seg OR such that P - M - Q and P - N - R. If A(∆PMN) = 20 sq.units, 3PM = 2MQ, then finf A([] MQRN)

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Answer 1

Step-by-step explanation:

By applying contradiction, we can prove that NM is parallel to RQ.

Let's assume, NM || RQ

Then,

ΔPRQ ≈ ΔPNM, as

∠P is common to both the triangles

∠PNM = ∠PRQ (as corresponding angle of parallel lines)

∠PMN= ∠PQR (as corresponding angle of parallel lines)

Applying similar triangle properties,

$\frac{pn}{pr \: \: } = \frac{pm}{pq}$

$\frac{pr - nr}{pr} = \frac{pm}{pq}$

$\frac{20 - 8}{20} = \frac{15}{25}$

$\frac{12}{20} = \frac{15}{25}$

$\frac{3}{5} = \frac{3}{5}$

As the ratios came out to be same, so what we had assumed was correct.

Therefore, NM || RQ.(Proved)

I hope it will help you