Question

2) Prove the following statement In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of remaining two sides.​

Answer 1

Pythagoras Theorem

Pythagoras theorem states that In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of remaining two sides.

[Please the attachment for the diagram]

We are given a right triangle $ABC$ at angled $B$. And we need to prove that $AC^2 = AB^2 + BC^2$.

Let us draw $BD$ perpendicular to $AC$, i.e. $BD \perp AC$. Then,

$\triangle ADB \sim \triangle ABC \quad$ [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 perpendicular are similar to the whole triangle and to each other]

So,

$\implies \dfrac{AD}{AB} = \dfrac{AB}{AC} \quad$ [Sides are proportional]

$\implies AD.AC = AB^2 \quad \bf{.... .(2)}$

Also,

$\triangle BDC \sim \triangle ABC \quad$ [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 perpendicular are similar to the whole triangle and to each other]

So,

$\implies \dfrac{CD}{BC} = \dfrac{BC}{AC} \quad$ [Sides are proportional]

$\implies CD.AC = BC^2 \quad \bf{.... .(2)}$

Now, adding equation (1) and equation (2), we get:

$\implies AD.AC + CD.AC = AB^2 + BC^2 \\ \\ \implies AC(AD + CD) = AB^2 + BC^2 \\ \\ \implies AC.AC = AB^2 + BC^2 \\ \\ \implies \boxed{AC^2 = AB^2 + BC^2}$

Hence, proved that in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of remaining two sides.

Answer 2

Answer:

Question :-

$\leadsto$ Prove that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of remaining two sides.

Given :-

$\leadsto$ A right triangle ABC right angled at B.

To Prove :-

$\leadsto$ AC² = AB² + BC²

Construction :-

$\leadsto$ Draw BD $\perp$ AC.

Proof :-

$\implies$ In ∆ABC and ∆ABD

We can write as,

$\implies \sf \angle{ABC} =\: \angle{ADB}$

$\implies \sf \angle{A} =\: \angle{A}\: (Taking\: as\: common)$

Hence,

$\mapsto \bf \triangle ABC \sim \triangle ABD\: (By\: Angle-Angle(AA)$

$\implies \sf \dfrac{AB}{AC} =\: \dfrac{AD}{AB}\: (Sides\: of\: triangles\: are\: proportional)$

By doing cross multiplication we get,

$\implies \sf AB .\: AB =\: AD .\: AC$

$\implies \sf\bold{\purple{AB^2 =\: AD .\: AC\: ------\: (Equation\: No\: 1)}}$

Again similarly,

$\implies \sf In\: \triangle ABC\: and\: \triangle BCD$

We can write as,

$\implies \sf \angle{ABC} =\: \angle{BDC}$

$\implies \sf \angle{C} =\: \angle{C}\: (Taking\: as\: common)$

Hence,

$\mapsto \bf \triangle ABC \sim \triangle BCD (By\: Angle-Angle(AA)$

Now,

$\implies \sf \dfrac{DC}{BC} =\: \dfrac{BC}{AC}\: (Sides\: of\: triangles\: are\: proportional)$

By doing cross multiplication we get,

$\implies \sf BC .\: BC =\: DC .\: AC$

$\implies \sf \bold{\purple{BC^2 =\: DC .\: AC\: ------\: (Equation\: No\: 2)}}$

By adding both equation no 1 and equation no 2 we get,

$\implies \sf AB^2 + BC^2 =\: AD .\: AC + DC .\: AC$

$\implies \sf AB^2 + BC^2 =\: AC(AD + DC)$

$\implies \sf AB^2 + BC^2 =\: AC(AC)$

$\implies \sf\bold{\red{AC^2 =\: AB^2 + BC^2}}$

$\dashrightarrow \: \: \: \: \: \: \: \: \sf\boxed{\bold{\green{HENCE\: PROVED}}}$

[Note:- Please refer that attachment for the diagram. ]

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