A 45° -45° - 90° right triangle has hypotenuse of length h. What is th

Question

A 45° -45° - 90° right triangle has hypotenuse of length h. What is the area of the triangle in terms of h?

A) h/ √2
B) h/2
C) h/4
D) (h)^2
E) (h)^2/4

Note: Answer is E, but I need some help with a step-by-step process.

Thanks.

ENGRTOMBA2018 · Answer

Make sure to put your own analyses/question within "spoiler".

As for your question, look below:

As the triangle is a 45-45-90 triangle , it is called isosceles right triangle with the 2 sides (perpendicular and the base) equal in length.

Let 'a' be the length of the perpendicular and the base. h is given to be height of the triangle.

Apply pythagoras theorem to get,

a^2+a^2=h^2  --->  2a^2=h^2 --->  a^2=\frac{h^2}{2}  ....(1)

Now, the area of the triangle = A = 0.5*perpendicular*base = 0.5*a*a= 0.5*a^2  .....(2)

Now substitute the value of  a^2  from (1) in (2), we get

A =  0.5*\frac{h^2}{2}  =  \frac{h^2}{4}  .

E is the correct answer.

Hope this helps.

EMPOWERgmatRichC · Answer

Hi jbabaganoosh,

This question can be solved by TESTing VALUES.

IF...each of the two shorter sides = 3, then the hypotenuse = H = 3(root2).

The AREA = (1/2)(Base)(Height) = (1/2)(3)(3) = 9/2.

So we're looking for an answer that = 9/2 when H = 3(root2). There's only one answer that matches...

Final Answer:  E

GMAT assassins aren't born, they're made,
Rich

jbabaganoosh · Answer

Thank you.

My solution:
The triangle is a x, x, x√2 triangle

Therefore, divide everything by √2 to get h = 1....

Now you have...
h/√2, h/√2, and 1

(*legs are now in terms of h)

Solve: 
base*height/2 
= (h√2*h√2) / 2
= (h√2)²/2
= h²/4

chetan2u · Answer

Hi,
the sides will be  h/\sqrt{2} , h/\sqrt{2}  and h....

jbabaganoosh · Answer

My apologies! You are correct.

ScottTargetTestPrep · Answer

The side ratio of a 45-45-90 right triangle is x : x : x√2, thus:

h = x√2

h/√2 = x

So, the area is (h/√2)^2 * 1/2 = h^2/2 * 1/2 = h^2/4.

Answer: E

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A 45° -45° - 90° right triangle has hypotenuse of length h. What is th

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