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Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

 

 
 
 
 
 

Answers (1)

Given in the question,

    A right-angled triangle ABC in which \mathrm{\angle B = 90\degree}

To prove:- (hypotenuse)2 = (base)2 + (perpendicular)2

                           \mathrm{AC^2 = AB^2 + BC^2}

Construction    \mathrm{BD \perp AC}

Proof:    In \DeltaADB and \DeltaABC

               \mathrm{\angle ADB = \angle ABC \quad (each \ = 90\degree)}

               \mathrm{\angle A = \angle A}

So, By AA-similarity criterion

                \mathrm{\Delta ADB \sim \Delta ABC}

        \mathrm{\Rightarrow\frac{AD}{AB} = \frac{AB}{AC} \quad [\text{In similar tringles corresponding sides are equal}]}

\mathrm{\Rightarrow AB^2 = AD\cdot AC\quad -(i)}

In \mathrm{\Delta BDC} and \mathrm{\Delta ABC}

        \mathrm{\angle CDB = \angle ABC \quad[\text{each} = 90\degree]}

                \mathrm{\angle C = \angle C}

So, by AA-similarity 

            \mathrm{\Delta BDC \sim \Delta ABC}

\mathrm{\Rightarrow \frac{DC}{BC} = \frac{BC}{AC}}

\mathrm{\Rightarrow BC^2 = AC\times DC \quad -(ii)}

Adding eq (i) and (ii)

            \mathrm{AB^2 +BC^2 = AC(AD + DC)}

      \mathrm{\Rightarrow AB^2 +BC^2 = AC\cdot AC}

    \mathrm{\Rightarrow AC^2 = AB^2 +BC^2}

Hence \mathrm{AC^2 = AB^2 + BC^2}

 

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Safeer PP

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