In the figure, seg PS 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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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.
Hope it helps:)
khubanivijay:
Hey! It is said there clearly to prove but you have not proved it . You have given the solution which is totally wrong.
Please see it in the last 2nd line .I have proved it
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