Question

please answer this question​

Answer 1

ANSWER:-

Given:

∆ABC is right angled at C.If p is the length of the perpendicular from C to AB & a,b,c are the length of the sides opposite angle A, B & C.

To prove:

1/p² = 1/a² + 1/b²

Proof:

∆ABC is a right angled at C.

Let,

⚫BC=a,

⚫CA=b,

⚫AB=c.

We know that, area of ∆= 1/2 × b ×h

=) 1/2 × BC × AC

=) 1/2 ab

Again,

=) 1/2 × AB × CD

=) 1/2 cp

Comparing both the areas of the ∆;

=) 1/2ab = 1/2cp

=) ab = cp........(1)

Now,

In right angled ∆ABC,

Using Pythagoras Theorem:

(Hypotenuse)²=(base)²+(perpendicular)²

=) AB² = BC² + AC²

$= > {c}^{2} = {a}^{2} + {b}^{2} \\ \\ = > ( \frac{ab}{p} ) {}^{2} = {a}^{2} + {b}^{2} \\ \\ = > \frac{ {a}^{2} {b}^{2} }{ {p}^{2} } = {a}^{2} + {b}^{2} ............[From \: \: (1)] \\ \\ = > \frac{1}{ {p}^{2} } = \frac{( {a}^{2} + {b}^{2}) }{ {a}^{2} {b}^{2} } \\ \\ = > \frac{1}{ {p}^{2} } = ( \frac{ {a}^{2} }{ {a}^{2} {b}^{2} } + \frac{ {b}^{2} }{ {a}^{2} {b}^{2} } ) \\ \\ = > \frac{1}{ {p}^{2} } = ( \frac{1}{ {b}^{2} } + \frac{1}{ {a}^{2} } ) \\ \\ = > \frac{1}{ {p}^{2} } = ( \frac{1}{ {a}^{2} } + \frac{1}{ {b}^{2} } ) \: \: \: \: \: \: \: <strong>[proved</strong>]$

Hope it helps ☺️