Challenge Problem - Looks Incorrcet to me. Pls help

[howidid]

I gave the answer as D, but the answer is B.

Can someone explain pls.

Regds,
Deep

[shaji]

I gave the answer as D, but the answer is B.

Can someone explain pls.

Regds,
Deep

The maximum area a right angled triangle with hypotenuse 6 is 25. Therefore B indeed is the correct answer. A clever trap by the question setter.

[shaji]

I gave the answer as D, but the answer is B.

Can someone explain pls.

Regds,
Deep

The maximum area a right angled triangle with hypotenuse 6 is 25. Therefore B indeed is the correct answer. A clever trap by the question setter.

"6" should read 10 in the above post. Regret inconvenience.

[geometrically challenged]

I gave the answer as D, but the answer is B.

Can someone explain pls.

Regds,
Deep

The maximum area a right angled triangle with hypotenuse 6 is 25. Therefore B indeed is the correct answer. A clever trap by the question setter.

"6" should read 10 in the above post. Regret inconvenience.

Does this mean that a right triangle with a hypotenuse with a 5, 10, or any multiple of 5 is a 3:4:5 triangle?

[Guest]

I gave the answer as D, but the answer is B.

Can someone explain pls.

Regds,
Deep

The maximum area a right angled triangle with hypotenuse 6 is 25. Therefore B indeed is the correct answer. A clever trap by the question setter.

"6" should read 10 in the above post. Regret inconvenience.

3,4 & 5 is indeed a pythagorian triplet. This isn't implied by what I have stated above. The problem is evaluate the maximum possible area for a rightangled triangle of hypotenuse 10, which is 10^2/4=25.

Does this mean that a right triangle with a hypotenuse with a 5, 10, or any multiple of 5 is a 3:4:5 triangle?

[rfernandez]

Good work, shaji.

In fact, the right triangle in question (with hypotenuse = 10) would not have its maximum possible area if the side lengths were 6-8-10. The right triangle with the largest area is isosceles, with each leg equaling 5*root(2).