Question

In the given figure ΔABC ~ ΔPQR, PM is median of ΔPQR. If ar ΔABC = 289 cm², BC = 17 cm, MR = 6.5 cm then the area of ΔPQM is​

Answer 1

Answer:

ar ∆PQM = 169 cm²

Step-by-step explanation:

since PM is median therefore it divides QR into QM and MR where QM = MR

since given MR = 6.5 therefore QM = 6.5 and we know

QR = MR + QM = 6.6 +6.5 = 13

since ∆ABC ~ ∆PQR

therefore (AB/PQ) = (BC/QR)=(CA/RP)= √(ar∆ABC/ar∆PQR)

that is (BC/QR) =✓ (ar∆ABC/ar∆PQR)

17/13 = ✓(289/ ar∆PQR)

squaring both side

289/169 = 289/ ar ∆PQR

ar ∆PQR = 289 × 169/289

= 169

Answer 2

Given:

ΔABC ≈ ΔPQR

PM is a median of ΔPQR

ar(ΔABC) = 289 cm²

BC = 17cm

MR = 6.5 cm

To find:

Area of ΔPQM

Solution:

Since, ΔABC ≈ ΔPQR then their corresponding sides are proportional.

BC and QR are the corresponding sides and thus their lengths will be proportional,i.e.,  $\frac{BC}{QR}$

PM is a median of ΔPQR. A median is a line segment drawn from one vertex of a triangle to the midpoint of the opposite side of triangle.

Here, M is the midpoint of side QR. That means,

$QR=2QM=2MR$

$QR=2*6.5=13$ cm

If two triangles are similar, then the ratio of the square of their corresponding sides will be equal to the ratio of the areas of the triangles.

$\frac{(BC)^{2}}{(QR)^{2}}=\frac{ar(ABC)}{ar(PQR)}$

$\frac{(17)^{2}}{(13)^{2}}=\frac{289}{ar(PQR)}$

$ar(PQR)=289*\frac{(13)^{2}}{(17)^{2}}$

$ar(PQR)=13^{2}=169$ cm²

Area of a triangle = $\frac{1}{2}$ × base × height

Area of ΔPQR = $\frac{1}{2}*QR*PM$

$169=\frac{1}{2}*13*PM$

$PM=\frac{169*2}{13}$

∴ $PM=13*2=26$ cm

Area of ΔPQM = $\frac{1}{2}*QM*PM$

$ar(PQM)=\frac{1}{2}*6.5*26$           (∵ QM = MR)

$ar(PQM)=84.5$ cm²

Area of ΔPQM is 84.5 cm²