Question

Triangle FGH is an isosceles right triangle with a hypotenuse that measures 16 units. An altitude, , is drawn from the right angle to the hypotenuse. What is the length of ?

Answer 1

Answer:

8 units

Step-by-step explanation:

mid point of hypotenuse of right angled isosceles triangle is equidistant from three vertices.

length of altitude=16/2=8

Answer 2

The question is incomplete, here is a complete question.

Triangle FGH is an isosceles right triangle with a hypotenuse that measures 16 units. An altitude, GJ, is drawn from the right angle to the hypotenuse.

What is the length of GJ?

A. 2 units

B. 4 units

C. 6 units

D. 8 units

Answer : The correct option is, (D) 8 units

Step-by-step explanation :

Given:

Length FH = 16 unit

As we know that a altitude between the two equal legs of an isosceles triangle creates right angles is a angle and opposite side bisector.

Thus,

Length FJ = Length HJ = $\frac{16}{2}$ = 8 units

As, the triangle is an isosceles. So, length GF = length GH = x unit

First we have to determine the value of 'x'.

Using Pythagoras theorem in ΔFGH :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(FH)^2=(GF)^2+(GH)^2$

Now put all the values in the above expression, we get :

$(16)^2=(x)^2+(x)^2$

$256=2x^2$

$x^2=128$

$x=8\sqrt{2}$

Thus, length GF = length GH = x unit = $8\sqrt{2}$

Now we have to determine the length GJ.

Using Pythagoras theorem in ΔGJH :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(GH)^2=(GJ)^2+(HJ)^2$

Now put all the values in the above expression, we get :

$(8\sqrt{2})^2=(GJ)^2+(8)^2$

$128=(GJ)^2+64$

$(GJ)^2=128-64$

$(GJ)^2=64$

$GJ=\sqrt{64}$

$GJ=8$

Thus, the length of GJ is, 8 units.