show that in a right triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides
Answers
WITH REFERENCE TO THE DIAGRAM BELOW
Area of the large square = Area of four triangles + Area of small square
Atotal=Afourtriangles+AsmallsquareAtotal=Afourtriangles+Asmallsquare
(a+b)2=4(1/2ab)+c2(a+b)2=4(1/2ab)+c2
a2+2ab+b2=2ab+c2a2+2ab+b2=2ab+c2
a2+b2=c2a2+b2=c2
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 .