Question

write the theorem of Pythagoras and prove it

Answer 1

Pythagoras theorem is applied in right angled triangle.
Its formula is = Base square + Perpendicular square = Hypotenuse square
It means that= a sq.+b sq.= c sq.

Suppose in the above triangle ABC;

Let BC=3cm and AB=4cm and AC= 5cm
Square of 3=9
Square of 4=16
Square of 5=25
According to pythagoras theorem;
Base sq.+ Perpendicular sq.= Hypotenuse sq.
9+ 16=25
25=25

Hope it will help you.

PLEASE MARK IT AS 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.