Question

Prove the Pythagoras theorem ​

Answer 1

The “Pythagorean theorem,” was inspired by the ancient Greek Mathematician Pythagoras who invented the theorem during 500 B.C. It has been argued that the “Ancient Babylonians” already understood the theorem long before the invention by Pythagoras. They knew the relationship between the sides of the triangle and while solving for the hypotenuse of an isosceles triangle, they came up with the approximate value of up to 5 decimal places.

What is Pythagoras theorem?

In a right-angled triangle, “The sum of squares of the lengths of the two sides is equal to the square of the length of the hypotenuse (or the longest side).”

a² + b² = c²

In the given triangle, side “a” is called as “Perpendicular”, side “b” is called as “Base” and side “c” is called as “Hypotenuse.”

The side opposite to the right angle (90∘) is the longest side known as Hypotenuse, as we know that the side opposite to the greatest angle is longest.

Note- Pythagoras theorem is only applicable to a Right-Angled triangle.

Another way of representation of the Pythagoras Theorem-

Another way of Pythagoras Theorem proof is described here. Consider a right-angled triangle having perpendicular as a, base as b, and hypotenuse as c. Consider three squares of sides a,b,c mounted on the three sides of a triangle having the same sides as shown.

By Pythagoras Theorem –

Area of square A + Area of square B = Area of square C

Example- Prove the Pythagoras Theorem for a right angle triangle having sides to be 3cm, 4cm and 5 cm.Solution –

From Pythagoras Theorem we have,

(Perpendicular)2+(base)2=(Hypotenuse)2

Perpendicular = 3 cm

Base = 4 cm

Hypotenuse = 5 cm

(3)2+(4)2=(5)2

⇒9+16=25

⇒25=25

L.H.S. = R.H.S.

Therefore Pythagoras theorem is proved.

Hope it helps you out. Please mark me as the brainliest.

Answer 2

Step-by-step explanation:

Pythagoras' theorem :-

→ In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

Step-by-step explanation:

It's prove :-

➡ Given :-

→ A △ABC in which ∠ABC = 90° .

➡To prove :-

→ AC² = AB² + BC² .

➡ Construction :-

→ Draw BD ⊥ AC .

➡ Proof :-

In △ADB and △ABC , we have

∠A = ∠A ( common ) .

∠ADB = ∠ABC [ each equal to 90° ] .

∴ △ADB ∼ △ABC [ By AA-similarity ] .

⇒ AD/AB = AB/AC .

⇒ AB² = AD × AC ............(1) .

In △BDC and △ABC , we have

∠C = ∠C ( common ) .

∠BDC = ∠ABC [ each equal to 90° ] .

∴ △BDC ∼ △ABC [ By AA-similarity ] .

⇒ DC/BC = BC/AC .

⇒ BC² = DC × AC. ............(2) .

Add in equation (1) and (2) , we get

⇒ AB² + BC² = AD × AC + DC × AC .

⇒ AB² + BC² = AC( AD + DC ) .

⇒ AB² + BC² = AC × AC .

$\huge \green{ \boxed{ \sf \therefore AC^2 = AB^2 + BC^2 }}$

Hence, it is proved ...... .......... ..............