Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • In a right-angled triangle, prove that the square of the hypotenuse is equal to the sum of the squares of the other two sides.       </di

In a right-angled triangle, prove that the square of the hypotenuse is equal to the sum of the squares of the other two sides. 

 

 
 
 
 
 

Answers (1)

In \triangleADB and \triangleABC

\angle BDA = \angle CBA\: \: \left \{ \because 90^{\circ} angles \right \}

\angle BAD = \angle BAC\: \: \left \{ \because common \right \}

\therefore \triangle ADB \sim \triangle ABC\: \: \left \{ \because AA \right \}

\therefore \frac{AD}{AB}= \frac{AB}{AC}   (from the theorem)

AB^2=AD.AC ........(1)

In \triangleBDC and \triangleABC,

\angle BDA = \angle ABC\: \: \left \{ \because 90^{\circ} angles \right \}

\angle BCD = \angle BCA\: \: \left \{ \because common \right \}

\therefore \triangle BDC \sim \triangle ABC\: \: \left \{ \because AA \right \}

\therefore \frac{CD}{BC}= \frac{BC}{AD}   (from the theorem)

BC^2=AC.CD ........(2)

 

Adding eqn(1) and eqn(2)

AB^2+BC^2=AD.AC+AC.CD

AB^2+BC^2=AC(AD+CD)

AB^2+BC^2=AC(AC)\: \: \: \left \{ \because AD+CD=AC \right \}

AB^2+BC^2=AC^2

Hence Proved.

Posted by

Safeer PP

View full answer

Similar Questions

Trending Articles/News

anam-khan

Delhi Board scrapped, SoSE schools set for shift to CBSE board, ‘CM SHRI’ tag
anam-khan

CBSE 10th Result 2025: Evaluation criteria, marking scheme, supplementary exams; what we know
anam-khan

Holding CBSE Class 10 twice can lead to ‘paper leaks, irregularities’, warns parliament panel

Latest Question