Question

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

Answer 1

GIVEN :-

• ∆ ABC is right angled at B

TO PROVE :-

• AC² = AB² + BC²

CONSTRUCTION :-

• draw BD ⊥ AC

FIGURE :-

• reffered to the attachment

STATEMENT :-

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

SOLUTION :-

using Theorem 6.7 :

( If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other )

(I) In ∆ABD ~ ∆ABC

since , sides of similar triangles are in same ratio ,

$\implies \rm{ \dfrac{ad}{ab} = \dfrac{ab}{ac} }$

$\implies \rm{ \bold{ad \times ac = {ab}^{2}} - - - eq1 }$

(I) In ∆BDC ~ ∆ABC

since , sides of similar triangles are in same ratio

$\implies \rm{ \dfrac{cd}{bc} = \dfrac{bc}{ac} }$

$\implies \rm{ \bold{ac \times cd = {bc}^{2}} - - - eq2 }$

now adding eq 1 and eq 2

$\implies \rm{(ad \times ac) + (ac \times cd) = {ab}^{2} + {bc}^{2} }$

$\implies \rm{ad \: . \: ac + ac \: . \: cd = {ab}^{2} + {bc}^{2} }$

$\implies \rm{\: ac (ad+ cd) = {ab}^{2} + {bc}^{2} }$

$\implies \rm{\: ac \times ac = {ab}^{2} + {bc}^{2} \: \: \: (ad + cd = ac) }$

$\implies \boxed {\boxed{ \rm{\: ac {}^{2} = {ab}^{2} + {bc}^{2} }} }$

hence proved