Question

Prove that :-

In a right the square of hypotenuse is equal to the sum of the square of the other two sides .

$<Marquee>$

Class - 10th
CHAP - Triangle.

❤️❤️ THANKS ❤️❤️

Answer 1

$<b >$ Given: ∆ABC right angle at B

To Prove: AC^2= AB^2+ BC^2

Construction: Draw BD ⊥ AC

Proof: Since BD ⊥ AC

We know that if a perpendicular is drawn from the vertex of the right angle of the a right triangle to the hypotenuse then triangle on both side of the perpendicular are similar to whole triangle and to each other.

IN Δ ADB ∼ Δ ABC
Since, sides of similar triangles are in the same ratio,

⇒ AD/AB = AB/AC

⇒AD . AC= AB^2..... (1)

Since, sides of similar triangles are in the same ratio in BDC ∼ Δ ABC

⇒ CD/BC = BC/AC
⇒CD . AC= BC ^2 .......(2)

Adding (1) and (2)

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

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

AC × AC = AB^2 + BC^2

AC^2 = AB^2 + BC^2

Hence Proved

Answer 2

Given: ∆ABC right angle at B 

To Prove: AC^2= AB^2+ BC^2 

Construction: Draw BD ⊥ AC 

Proof: Since BD ⊥ AC 

We know that if a perpendicular is drawn from the vertex of the right angle of the a right triangle to the hypotenuse then triangle on both side of the perpendicular are similar to whole triangle and to each other. 

IN Δ ADB ∼ Δ ABC 
Since, sides of similar triangles are in the same ratio, 

⇒ AD/AB = AB/AC

⇒AD . AC= AB^2..... (1)

Since, sides of similar triangles are in the same ratio in BDC ∼ Δ ABC

⇒ CD/BC = BC/AC
⇒CD . AC= BC ^2 .......(2)

Adding (1) and (2) 

AD . AC + CD . AC = AB^2 + BC ^2 

AC (AD + CD) = AB^2 + BC ^2 

AC × AC = AB^2 + BC^2 

AC^2 = AB^2 + BC^2

Hence Proved