Question

In a triangle if the saqure of on side is equal to the sum of the squares of the other to sides then prove that the angle opposite to the frist side is a right angle ​

Answer 1

Question

in a triangle if the square on one side is equal to the sum of the squares of the other two sides then prove that the angle opposite to the first side is a right angle

Given

In ∆ ABC, Square of one side is equal to the sum of square of other two sides

$\sf {AB}^{2} + {BC}^{2} = {AC}^{}$

To Prove

The angle opposite to the first step de is a right angle.

Solution

∠B = 90°

∆PQR is a right triangle

so, by using Pythagoras theorem we have,

$\sf {PR}^{2} = {PQ}^{2} + {QR}^{2}$

$\sf \implies \: {PR}^{2} = {AB}^{2} + {BC}^{2} (pq = ab \: and \: qr = bc)$

$\sf {PR}^{2} = {AC}^{2} ( {AB}^{2} + {BC }^{2} = {AC}^{2} )$

$\sf \: PR = AC$

Thus, ∆ ABC = ∆ PQR =∠ B = 90°

Answer 2

Answer:

Question

• in a triangle if the square on one side is equal to the sum of the squares of the other two sides then prove that the angle opposite to the first side is a right angle

Solution

Given

• In ∆ ABC, Square of one side is equal to the sum of square of other two sides

$\boxed{ \rm{ {AB}^{2} + {BC}^{2} = {AC}^{2} }}$

To Prove

• The angle opposite to the first step de is a right angle.

Let us solve

∠B = 90°

∆PQR is a right triangle

so, by using Pythagoras theorem we have,

PR² = PQ² + QR²

PR² = AB² + BC²

PR² = AC²7 [AB² + BC² = AC²]

PR = AC

[We can cancel square in both sides]

Therefore,

∆ ABC = ∆ PQR =∠ B = 90°