Area of triangle ABC, given AB=AC=sqrt{52}, BC=8, DC=BD, altitude is AD

[HaveAQUESTION]

I think it says in the question the hypotenuse is 52, and hypotenuse is suppose to be the sum of two squares (a squared plus b squared equals c squared) I can't think of any squares added together that would equal 52. And 52 isn't a perfect itself. I know the formula formula for area. (B x h x 1/2). But I'm not sure what the length of the legs are either. Someone please clear this up for me.

 

[Deleted member 4993]

I think it says in the question the hypotenuse is 52, and hypotenuse is suppose to be the sum of two squares (a squared plus b squared equals c squared) I can't think of any squares added together that would equal 52. And 52 isn't a perfect itself. I know the formula formula for area. (B x h x 1/2). But I'm not sure what the length of the legs are either. Someone please clear this up for me.

BD = 1/2* BC = 4 = CD

BD^2 = 16

AB^2 = 52

AD^2 = AB^2 - BD^2 = 52 - 16 = 36

AD = ?

Area of the ADC = 1/2 * AD * CD = ?

We do not need "perfect" (I hope you meant rational integers) numbers for geometry (unless otherwise stipulated).

 

[MarkFL]

Consider \(\displaystyle \Delta ABD\). We are given \(\displaystyle \overline{BD}=4\) and \(\displaystyle \overline{AB}=\sqrt{52}\). Using the Pythagorean theorem, what is \(\displaystyle \overline{AD?}\)