Question

In the given figure, ∆ABC ~ ∆PQR, PM is median of ∆PQR. If ar(∆ABC) = 289 cm2 , BC = 17 cm, MR= 6.5 cm, then the area of ∆PQM is

Answer 1

Step-by-step explanation:

W.K.T medians bisect the base....

Therefore $QM = MR = 6.5 \: cm$

=> $QR = 13 \: cm$

The ratio of areas of two similar triangles are equal to the square of the corresponding sides....

Therefore,

$\frac{ar(∆ABC)}{ar(∆PQR)} = \frac{ {BC}^{2} }{ {QR}^{2} } \\ \\ By \: Substituting \: The \: Values \\ \\ \frac{289}{ar(∆PQR)} = \frac{ {17}^{2} }{ {13}^{2} }$

By Cross Multiplication,

$ar(∆PQR) = \frac{289 \times 169}{289} = 169 \: {cm}^{2}$

Median bisects the triangle into two Triangles of equal areas...

Therefore $ar(∆PQM) = ar(∆PQR)/2 = 84.5 \: sq. cm$

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Answer 2

Solution :-

given that, PM is the median of ∆PQR .

So,

→ QM = MR = 6.5 cm { Median bisects the base. }

→ QR = QM + MR = 6.5 * 2 = 13 cm

now, given that,

→ ∆ABC ~ ∆PQR

So,

→ Ar(∆ABC)/Ar(∆PQR) = BC²/QR² { when two ∆'s are similar, ratio of their area = ratio of square of their corresponding sides . }

→ 289/Ar(∆PQR) = 17²/13²

→ 289/Ar(∆PQR) = 289/169

→ Ar(∆PQR) = 169 cm²

now, since QM = MR

→ Ar(∆PQM) = (1/2) • Ar(∆PQR) { Median bisect are ∆ into two equal areas. }

→ Ar(∆PQM) = (1/2) • 169

→ Ar(∆PQM) = 84.5 cm² (Ans.)

Learn more :-

In the figure along side, BP and CP are the angular bisectors of the exterior angles BCD and CBE of triangle ABC. Prove ∠BOC = 90° - (1/2)∠A .

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