Dec 14, 2009
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Two sides of an isoceles triangle have length 10 with the third side shorter than these two. If the area is 48, what is the longest possible third side?

correct answer is 16
 

UndergradGuy7

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Jun 23, 2007
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I'm not sure but...

if you have a triangle like this
http://mathworld.wolfram.com/images/eps-gif/IsoscelesTriangle_800.gif

Then the height is (0.5a)^2 + h^2 = b^2

But also area = 1/2 * a * h
so 48 = 0.5 a * h
if we solve for h = 96/a

then plug it back into
(0.5a)^2 + h^2 = b^2
so
0.25a^2 + (96^2)/a^2 = b^2

and solve for a. b = 10. The problem gives us this.

This is my guess.

Edit: I tried solving it. I don't think you can solve it using what I wrote above. So I don't know.
 
Last edited:

Streetwolf

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The triangle with base 16 makes sense...

Except it's supposed to have a base SHORTER than the other two sides.
 
Apr 1, 2010
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this is actually really easy....

that theory pthagorian? the ratio of a right triangle is 3:4:5 right?...

let's multiply this by 2 so that it becomes 6:8:10 and I'm going to put 10 as

one of the side length...

h = height b/2 = base of a half triangle:

this means that h and b/2 and 10 has to equal to 6:8:10

so just plug in... h=6 b/2=8 then you will get entire base of 16 area of 48