Question

State and prove converse of Pythagoras
theorem from similar triangles.

​

Answer 1

✭ Question : -

• State and prove the converse of Pythagoras theorem !

✭ Statement : -

• In a triangle, if the square of one side is equal to the sum of the squares of other two sides ,then show that the angle opposite to the first side is a right angle.

✭ Given : -

• In ∆ABC,
• AC² = AB² + BC²

✭ Required to prove : -

• ∠B = 90°

✭ Construction : -

• Construction an imaginary right angle ∆PQR such that ∠Q = 90° and AB = PQ, BC = QR.

✭ Proof : -

• In ∆ABC,
• It is given that;
• AC² = AB² + BC² ..... (1)

✭ Now,

• In ∆PQR,∠Q = 90°
• Using the Pythagoras theorem;
• (hypotenuse)² = (side)² + (side)²

✭ This implies;

• PR² = PQ² + QR²

✭ Since,

• AB = PQ
• BC = QR

[ By construction ]

✭ So,

• PR² = (AB)² + (BC)²
• PR² = AB² + BC²

✭ From equation - 1

• PR² = AC²
• Taking square root on both sides
• √(PR²) = √(AC²)
• ★ PR = AC

✭ Now,

Consider ∆ABC & ∆PQR

• In ∆ABC & ∆PQR
• AB = PQ [ By construction ]
• BC = QR [ " " ]
• AC = PR [ Proved above ]
• Using the SSS congruency rule !
• ∆ABC ≅ ∆PQR

✭ Here,

• ∠B = ∠Q [ corresponding parts of congruent triangles ]
• This implies;
• ∠B = 90°

Hence Proved !

✭ Note ✭

• Diagram is in attachment!

Answer 2

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➪State and Prove converse of Pythagoras theorem from the similar Triangles.

$\huge\red{\mid{\underline{\overline{\textbf{Solution\:࿐}}}\mid}}$

● Statement : -

In a triangle, if the square of one side is equal to the sum of the squares of other two sides ,then show that the angle opposite to the first side is a right angle.

■ note ■ - Diagram in the attachment.

● Given : -

In ∆ABC,

AC² = AB² + BC²

● Required to prove : -

∠B = 90°

● Construction : -

Construction an imaginary right angle ∆PQR such that ∠Q = 90° and AB = PQ, BC = QR.

● Proof : -

In ∆ABC,

It is given that;

AC² = AB² + BC² .. (1)

● Now,

In ∆PQR,∠Q = 90°

Using the Pythagoras theorem;

(hypotenuse)² = (side)² + (side)²

● This implies;

PR² = PQ² + QR²

● Since,

AB = PQ

BC = QR

[ By construction ]

● So,

PR² = (AB)² + (BC)²

PR² = AB² + BC²

● From equation - 1

PR² = AC²

Taking square root on both sides

√(PR²) = √(AC²)

● PR = AC

● Now,

Consider ∆ABC & ∆PQR

In ∆ABC & ∆PQR

AB = PQ [ By construction ]

BC = QR [ " " ]

AC = PR [ Proved above ]

Using the SSS congruency rule !

∆ABC ≅ ∆PQR

● Here,

∠B = ∠Q [ corresponding parts of congruent triangles ]

This implies;

∠B = 90°

Hence Proved !

$\sf\red{hope\:it\:helps\:you}$