Question

in right angled triangle,the square of the hypotenuse is equal to the sum of the squares of other two sides​

Answer 1

Step-by-step explanation:

Given:

• A right angled triangle.

To Prove:

• Square of hypotenuse is equal to the sum of the squares of other two sides.

Construction:

• Draw BD ⟂ AC

Proof: Let in a right angled (∠ABC is 90°) triangle ABC.

• AC = Hypotenuse
• AB and BC = Other sides

We have to prove AC² = AB² + BC²

Now, in ∆ADB and ∆ABC we have

• ∠A = ∠A {common}
• ∠ADB = ∠ABC {90° each}

∴ ∆ADB ~ ∆ABC by AA similarity.

$\implies{\rm }$ AD/AB = AB/AC

$\implies{\rm }$ AD(AC) = AB(AB)

$\implies{\rm }$ AD(AC) = AB²......(1)

Now, in ∆BDC and ∆ABC we have

• ∠C = ∠C {common}
• ∠BDC = ∠ABC {90° each}

∴ ∆BDC ~ ∆ABC by AA similarity.

$\implies{\rm }$ DC/BC = BC/AC

$\implies{\rm }$ DC(AC) = BC(BC)

$\implies{\rm }$ DC(AC) = BC².....(2)

Adding equation (1) and (2) we got

➮ AD(AC) + DC(AC) = AB² + BC²

➮ (AD + DC)AC = AB² + BC²

➮ AC(AC) = AB² + BC²

➮ AC² = AB² + BC²

$\large\bold{\texttt {Proved }}$

Answer 2

Given:-

∆ ABC is a right triangle right angled at B and BD perpendicular to AC .

To prove:-

(AC)² = (AB)² + (BC)²

Construction:-

BD perpendicular to AC was drawn.

Proof:-

∆ ABC ~ ∆ ADB (°.° if a perpendicular drawn from the right vertex of a Triangles then Triangles form either side to the perpendicular are similar to each other and similar to the whole Triangle)

$so \: \frac{AB}{AD} = \frac{AC}{AB}$

➬AB × AB = AC × AD.

➬(AB)² = AC × AD. _____ Eq 1

∆ CBA ~ ∆ CDB (°.° if a perpendicular drawn from the right vertex of a Triangles then Triangles form either side to the perpendicular are similar to each other and similar to the whole Triangle)

$so, \: \frac{CB}{CD} = \frac{CA}{CB}$

➬CB × CB = CA × CD

➬(CB)² = CA. CD _____ Eq 2.

adding equation 1 and 2.

AB² + CB² = ( AC × AD ) + ( AC × CD )

➬ AB² + CB² = AC( AD + CD )

➬AB² + CB² = AC × AC .

➬AB² + CB² = AC²

➬AC² = AB² + CB².

so, according to the question the square of the hypotenuse is equal to the sum of the square of other two sides.

HENCE PROVED