If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR prove that \( \dfrac{AB}{PQ}\) = \( \dfrac{AD}{PM}\).

Asked by Abhisek | 1 year ago |  84

1 Answer

Solution :-

Given, ΔABC ~ ΔPQR

Ncert solutions class 10 chapter 6-27

We know that the corresponding sides of similar triangles are in proportion.

\( \dfrac{AB}{PQ}=\dfrac{AC}{PR}=\dfrac{BC}{QR}\)……(i)

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ……..(ii)

Since AD and PM are medians, they will divide their opposite sides.

∴ BD = \( \dfrac{BC}{2}\) and QM =\( \dfrac{QR}{2}\)  ……….(iii)

From equations (i) and (iii), we get

\( \dfrac{AB}{PQ}=\dfrac{BD}{QM}\) ………….(iv)

In ΔABD and ΔPQM,

From equation (ii), we have

∠B = ∠Q

From equation (iv), we have,

\( \dfrac{AB}{PQ}=\dfrac{BD}{QM}\)

ΔABD ~ ΔPQM (SAS similarity criterion)

⇒ \( \dfrac{AB}{PQ}=\dfrac{BD}{QM}\)\( \dfrac{AD}{PM}\)

Answered by Pragya Singh | 1 year ago

Related Questions

In an trapezium ABCD, it is given that AB ∥ CD and AB = 2CD. Its diagonals AC and BD intersect at the point O such that ar(∆AOB) = 84 cm2. Find ar(∆COD)

Class 10 Maths Triangles View Answer

Find the length of the altitude of an equilateral triangle of side 2a cm.

Class 10 Maths Triangles View Answer

A ladder 10 m long reaches the window of a house 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.

Class 10 Maths Triangles View Answer

In the given figure, DE ∥ BC such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm. Find the value of x.

Class 10 Maths Triangles View Answer

If ∆ABC ∼ ∆DEF such that 2 AB = DE and BC = 6 cm, find EF.

Class 10 Maths Triangles View Answer