Question

in a right angle triangle the square of an a hypotenuse is equal to the sum of square the other to sides ​

Answer 1

Answer:

Given :

A right triangle ABC right angled at B.

To prove :

AC² = AB² + BC²

Construction :

Draw BD ⊥ AC

Proof :

In Δ ADB and Δ ABC

∠ A = ∠ A    [ Common angle ]

∠ ADB = ∠ ABC   [ Both are 90° ]

∴  Δ  ADB  Similar to Δ ABC   [ By AA similarity ]

So , AD / AB = AB / AC   [ Sides are proportional ]

= > AB² = AD . AC  ... ( i )

Now in Δ BDC and Δ ABC

∠ C = ∠ C    [ Common angle ]

∠ BDC = ∠ ABC   [ Both are 90° ]

∴  Δ  BDC Similar to Δ ABC   [ By AA similarity ]

So , CD / BC = BC / AC

= > BC² = CD . AC   ... ( ii )

Now adding both equation :

AB² + BC² = CD . AC +  AD . AC

AB² + BC² = AC ( CD + AD )

AB² + BC² = AC² .

AC² = AB² + BC² .

Hence proved .

Answer 2

$\bf{\underline{\underline{Solution:-}}}$

➡ Given :-

→ A △ABC in which ∠ABC = 90° .

➡To prove :-

→ AC² = AB² + BC² .

➡ Construction :-

→ Draw BD ⊥ AC .

➡ Proof :-

In △ADB and △ABC , we have

∠A = ∠A ( common ) .

∠ADB = ∠ABC [ each equal to 90° ]

∴ △ADB ∼ △ABC [ By AA-similarity ]

⇒ AD/AB = AB/AC .

⇒ AB² = AD × AC ............(1) .

In △BDC and △ABC , we have

∠C = ∠C ( common ) .

∠BDC = ∠ABC [ each equal to 90° ] .

∴ △BDC ∼ △ABC [ By AA-similarity ] .

⇒ DC/BC = BC/AC .

⇒ BC² = DC × AC. ............(2) .

Add in equation (1) and (2) , we get

⇒ AB² + BC² = AD × AC + DC × AC .

⇒ AB² + BC² = AC( AD + DC ) .

⇒ AB² + BC² = AC × AC

$\large{\boxed{\underline{\red{\mathcal{AB^2 + BC^2=AC^2}}}}}$

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