Question

state and Prove the Pythagoras theorem of triangle​

Answer 1

Pythagoras Theorem is an important topic in Maths, which explains the relation between the sides of a right-angled triangle. It is also sometimes called Pythagorean Theorem. The formula and proof of this theorem are explained here. This theorem is basically used for the right-angled triangle and by which we can derive base, perpendicular and hypotenuse formula. Let us learn this theorem in detail here.Where

“a” is the perpendicular side,

“b” is the base,

“c” is the hypotenuse side.

According to the definition, the Pythagoras Theorem formula is given as:

Hypotenuse2 = Perpendicular2 + Base2

c2 = a2 + b2

The side opposite to the right angle (90°) is the longest side (known as Hypotenuse) because the side opposite to the greatest angle is the longest.pythagoras theorem proveGiven: A right-angled triangle ABC.

To Prove- AC2 = AB2 + BC2

Pythagoras Theorem Proof

Proof: First, we have to drop a perpendicular BD onto the side AC

We know, △ADB ~ △ABC

Therefore, ADAB=ABAC (Condition for similarity)Or, AB2 = AD × AC ……………………………..……..(1)

Also, △BDC ~△ABC

Therefore, CDBC=BCAC (Condition for similarity)

Or, BC2= CD × AC ……………………………………..(2)

Adding the equations (1) and (2) we get,

AB2 + BC2 = AD × AC + CD × AC

AB2 + BC2 = AC (AD + CD)

Since, AD + CD = AC

Therefore, AC2 = AB2 + BC2

Hence, the Pythagorean thoerem is proved.

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.