Question

Seg PS is perpendicular to side QR. If PQ=a PR=b QS=c and RS=d then prove that (a+b) (a-b) =(c+d) (c-d)

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Answer 1

Solution:

In Right Δ P SQ

P Q²= PS² + SQ²→→[By Pythagoras theorem]

a²= PS² + c²

→PS² = a² - c²-------(1)

In Right Δ P SR

PR²= PS² + SR²→→[By Pythagoras theorem]

b²= PS² + d²

PS²= b² - d² ------(2)

From (1) and (2)

b²- d²= a² - c²

a² - b²= c² - d²

(a -b)(a+b)= (c-d)(c+d)→→Using the identity, A²-B²= (A-B)(A+B)

Hence proved.

Answer 2

Hope it will help you....