Question

State and prove Pythagorous theorem​

Answer 1

Pythagoras theorem states that “ In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

The sides of the right-angled triangle are called base, perpendicular and hypotenuse .

According to Pythagoras theorem ,

(AC)2=(AB)2 + (BC)2

Proof:

Given, a triangle ABC in which ∠ABC is 900.

Construction: Draw a perpendicular BD on AC i.e. BD ⊥ AC.

In ΔABD and ΔABC we have,

∠BAD = ∠BAC i.e. ∠A is common in both triangles.

∠ABC = ∠ADB = 900

Therefore ΔABC∼ΔABD ( By AA similarity i.e. angle-angle similarity)

So,⇒ADAB=ABAC⇒AB2 = AD×AC ...(1)

In ΔBDC and ΔABC we have,

∠BCD = ∠BCA i.e. ∠C is common in both triangles.

∠ABC = ∠ADC = 900

Therefore ΔABC∼ΔBDC ( By AA similarity i.e. angle-angle similarity)

So,⇒DCBC=BCAC⇒BC2 = AC×DC ...(2)

Adding equation (1) and (2) , we get

⇒AB2 + BC2 = AD×AC + AC× DC⇒AB2 + BC2 = AC(AD + DC)⇒AB2 + BC2 = AC(AC)⇒AB2 + BC2 = AC2

Hence, proved.

Answer 2

Answer:

Pythagoras theorem states that “ In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”. The sides of the right-angled triangle are called base, perpendicular and hypotenuse . Proof: ... ∠BAD = ∠BAC i.e. ∠A is common in both triangles.

Step-by-step explanation:

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