Question

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help me with this​

Answer 1

Given:

∠C = 90°

BC = a

CA = b

AB = c

'p' a perpendicular is drawn to AB.

To Prove:

(i) cp = ab

(ii) 1/p² = 1/a² = 1/b²

Proof:

$\sf (i) \ ar(ACB) \ = \dfrac{1}{2} \times bh$

$\sf ar(ACB) \ = \dfrac{1}{2} \times ab$

$\longrightarrow$ 2ar(ACB) = ab → Eq1

$\sf (i) \ ar(CBA) \ = \dfrac{1}{2} \times bh$

$\sf ar(CBA) \ = \dfrac{1}{2} \times cp$

$\longrightarrow$ 2ar(CBA) = cp → Eq2

From Eq1 and Eq2,

a × b = c × p

∴ ab = cp → (Eq3)

Hence Proved.

(ii) From Eq3,

ab = cp

Transposing we get,

$\sf \dfrac{ab}{c} = p$

Taking reciprocal we get,

$\sf \dfrac{c}{ab} = \dfrac{1}{p}$

On Squaring both sides,  

$\sf (\dfrac{c}{ab})^{2}$ = $\sf (\dfrac{1}{p})^{2}$

$\sf \dfrac{c^{2}}{a^{2}b^{2}} = \dfrac{1}{p^{2}}$

By using pythagoras theorem we know that,

c² = a² + b²

Using this in the previous step we get,

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

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

Cancelling a² and b² we get,

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

By adjusting the places,

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

Hence Proved.