In ∆ PQR, PD ⊥ QR such that D lies on QR . If PQ = a, PR = b, QD = c and DR = d, prove that (a + b) (a - b) = (c + d) (c – d)

Solution:

Given, in triangle PQR, PD ⊥ QR such that D lies on QR.

Also, PQ = a

PR = b

QD = c

DR = d

We have to prove that (a + b)(a - b) = (c + d)(c - d)

In △PDQ,

D = 90°

By pythagoras theorem,

PQ² = PD² + DQ²

a² = PD² + c²

PD² = a² - c² ------------- (1)

In △PDR,

D = 90°

By pythagoras theorem,

PR² = PD² + DR²

b² = PD² + d²

PD² = b² - d² -------------- (2)

From (1) and (2),

a² - c² = b² - d²

On rearranging.

a² - b² = c² - d²

By using algebraic identity,

(x² - y²) = (x + y)(x - y)

So, (a² - b²) = (a + b)(a - b)

c² - d² = (c + d)(c - d)

(a + b)(a - b) = (c + d)(c - d)

Therefore, it is proved that (a + b)(a - b) = (c + d)(c - d)

✦ Try This: D is a point on side QR of ∆PQR such that PD ⊥ QR. Will it be correct to say that ∆PQD ∼ ∆RPD? Why?

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 6

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NCERT Exemplar Class 10 Maths Exercise 6.4 Problem 11

In ∆ PQR, PD ⊥ QR such that D lies on QR . If PQ = a, PR = b, QD = c and DR = d, prove that (a + b) (a - b) = (c + d) (c – d)

Summary:

In ∆ PQR, PD ⊥ QR such that D lies on QR . If PQ = a, PR = b, QD = c and DR = d, it is proven that (a + b) (a - b) = (c + d) (c – d)

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