Kaplan geometry

resilient · Posted: Wed Apr 02, 2008 3:08 pm Post subject: Kaplan geometry

please check the attachment that I have uploaded. thanks

Stuart Kovinsky · Posted: Wed Apr 02, 2008 3:17 pm Post subject:

To what "unneeded" information are you referring on the first question?

Just knowing that we have a hypotenuse of 26 does NOT necessarily mean that we have a 5x:12x:13x triangle. All that we know is that:

bc^2 + cd^2 = 26^2,

an equation with infinite solutions for bc and cd.

We definitely need both triangles and the ratios on the base to determine the length of bc.

On the second question, we have a 30/60/90 triangle with an area of 8root3. We're asked to solve for the length of the hypotenuse.

We know that the sides of a 30/60/90 triangle are in the ratio of x:xroot3:2x.

So, our triangle has:

base = x
height = xroot3
area = 8root3

We also know that area = 1/2(base)(height)

So, we know that:

8root3 = (1/2)(x)(xroot3)
8root3 = (1/2)(x^2)(root3)

dividing through by root3, we get:

8 = (1/2)(x^2)

multiplying both sides by 2, we get:

16 = x^2

and solving for x we get:

x = 4

(remember: in geometry, we can ignore non-positive solutions, which is why we don't worry about x = -4).

Going back to our ratio, we know that the hypotenuse is 2x. Since x = 4, we know that the hypotenuse (and the answer to the question) is 8.