Question

prove that∆abc is right angled at a if ab=2n+1,ac=2n(n+1)and bc=2n(n+1)​

Answer 1

Consider the right triangle ABC, right angled at C.

We have to prove:

Consider triangle ABC,

by Pythagoras theorem, we get

Consider triangle ACD,

by Pythagoras theorem, we get

(Equation 1)

Since, D is the midpoint of BC.

CD=DB and

Substituting the value of CD in equation 1, we get

Hence, proved.

Answer 2

Answer:

being A as 90 degree

Hypotenuse BC = 2n(n+1) + 1

Sides AB = 2n + 1

And AC = 2n (n+1)

For the triangle to be right angled we have to prove

BC^2 = AB^2 + AC^2

Plugging on R H S

(2n+1)^2 + [2n (n+1)]^2

==> [2n (n+1)]^2 + 4n^2 + 4n + 1

==> [2n (n+1)]^2 + 2* 2n (n + 1) + 1

==> A^2 + 2 * A * B + B^2 {here A = 2n(n+1), B = 1}

Thus we have (A+B)^2

That is [2n(n+1) + 1}^2 = L H S

Thus by converse of Pythagoras it is confirmed that BAC is a right angled triangle