Question

prove that in a right triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides​

Answer 1

Answer:

fkjgzxvghk

ggjkkggkj

tyikogf

Step-by-step explanation:

ccgucgjchgchmg bvcnh b

xjhxgmxgghxkgxkhffkzlgzfyflfzfytlfzfjfzfyzfyzfhzmfhzfhhggzhmfzfusyrsryosorysotspsoysoyzlyzlhlzhflfjzlfuzlitzlitghhcyxgjxfg ggzhzgf yfkzgljgdyfiidylustpst7suptsp85pd58dupfcngzhxjkhDulzrlsutldtilditlsitlsitlsitlsitldyidptxcgxfhmc

• fzdyzdzreiptdilsgyhdrarktislitlitsdpidyiyididpttdi^'$

Answer 2

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 .