Question

State and prove .. Pythagoras theorem

Answer 1

Answer:

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).”

Right Angled Triangle

In the given triangle, side “a” is the “Perpendicular”, side “b” is the “Base” and side “c” is the “Hypotenuse.” Then according to the definition;

Hypotenuse2 = Perpendicular2 + Base2

c2 = a2 + b2

The side opposite to the right angle (90°) is the longest side (known as Hypotenuse) because side opposite to the greatest angle is the longest.

Note- Pythagorean theorem is only applicable to Right-Angled triangle.

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.