Question

In the given figure, area of PQR = 44.8 cm2, PL = LR and QM = MR. Find the area of LMR.

Answer 1

Given:

Area of ΔPQR = 44.8 cm²

PL = LR

QM = MR

To find:

The area of ΔLMR

Solution:

Since PL = LR and QM = MR (given)

∴ L is the midpoint of the side PR and M is the midpoint of the side of QR ....... (i)

We know,

$\boxed{\bold{Midpoint\:Theorem}}:$ The line segment joining the midpoints of any two sides of a triangle is parallel to the third side and is equal to half of the third side.

Based on the above theorem and from (i), we can say,

LM // PQ ..... (ii)

and

LM = $\frac{1}{2}$ PQ ..... (iii)

Considering ΔLMR and ΔPQR, ∵ LM // PQ, we have

∠LRM = ∠PRQ ..... [common angle]

∠RML = ∠RQP ...... [corresponding angles]

∴ ΔLMR ~ ΔPQR ..... [By AA similarity]

We know that → the ratio of the areas of the two similar triangles is equal to the ratio of the squares of their corresponding sides.

So, based on this, we can say that,

$\frac{Area (\triangle LMR)}{Area (\triangle PQR)} = \frac{LM^2}{PQ^2}$

substituting from (iii), we get

$\implies \frac{Area (\triangle LMR)}{Area (\triangle PQR)} = \frac{[\frac{1}{2} PQ]^2}{PQ^2}$

$\implies \frac{Area (\triangle LMR)}{Area (\triangle PQR)} = \frac{\frac{1}{4} PQ^2}{PQ^2}$

$\implies \frac{Area (\triangle LMR)}{Area (\triangle PQR)} = \frac{1}{4}$

substituting Area (Δ PQR) = 44.8 cm²

$\implies \frac{Area (\triangle LMR)}{44.8} = \frac{1}{4}$

$\implies Area (\triangle LMR) = \frac{44.8}{4}$

$\implies\bold{ Area (\triangle LMR) = 11.2\: cm^2}$

Thus, the area of Δ LMR is → 11.2 cm².

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Answer 2

Thus, the area of Δ Lmr is → 11.2 cm².

Step-by-step explanation:

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