# Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians

**Solution:**

As we know, if one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. And we know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

In ΔPQR, PM is the median and,

In ΔABC AN is the median

ΔPQR ∼ ΔABC (given)

∠PQR = ∠ABC................ (1)

∠QPR = ∠BAC.........(2)

∠QRP = ∠BCA...........(3)

and-PQAB = QR/BC = RP/CA............ (4)

(If two triangles are similar, then their corresponding angles are equal and corresponding sides are in the same ratio)

Area of ΔPQR / Area of ΔABC = (PQ)^{2} / ( AB)^{2} = (QR)²/(BC)² = (RP)^{2} / (CA)^{2}------ [THEROM 6.6] ……… (5)

Now In ΔPQM and ΔABN

∠PQM = ∠ABN...... (from 1)

And PQ/AB = QM/BN

[Therefore, PQ / AB = QR / BC = 2QM / 2BN; M, N midpoints of QR and BC]

⇒ ΔPQM ∼ ΔABN [SAS similarly]

Area of ΔPQM / Area of ΔABN = (PQ)^{2} / (AB)^{2} = (QM)^{2 }/ (BN)^{2} = (PM)^{2} / (AN)^{2} [by theorem 6.6]....... (6)

From (5) and (6)

Area of ΔPQR / Area of ΔABC = (PM)^{2} / ( AN)^{2}

**Video Solution:**

## Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians

### NCERT Class 10 Maths Solutions - Chapter 6 Exercise 6.4 Question 6:

Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians

Hence proved that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians