Question

Q5 plz tell the correct way ​

Answer 1

(i) ar of ∆ABC = 1/2 × AC × BC

=. 1/2 × b×a = ab/2 -------------(1)

ALSO ar. ∆ABC= 1/2 × AB × CD

= 1/2 × c×p = cp/2 --------------(2)

equating 1 and 2, we get

ab=cp

therefore proved.

(ii) Now , using Pythagoras theorm we get

${a}^{2} + {b}^{2} = {c}^{2}$

putting c=ab/p. from (i)

a^2 + b^2 = (ab)^2/p^2

a^2 + b^2/ (ab)^2=1/p^2

$\frac{1}{ {b}^{2} } + \frac{1}{ {a}^{2} } = \frac{1}{ {p}^{2} }$

therefore proved

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