BB Math p.684 #15

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1

<p>
[quote]
15 on 684</p>

<p>To find the answer, you'll have to construct or visualize two right triangles that include the midpoints.</p>

<p>The first one will create a hypothenuse that joins the corner below B and the point A. It is quite easy to find the value since you have two sides of 2 and 1. Now, you simple create another right triangle with a hypo that joins A and B. You have the value for the right sides as the first calculated value (it is sqrt 5) and 1.</p>

<p>Leave the values in root form (based on a quick look at the proposed solutions). The answer is sqrt 6.

[/quote]
<-- xiggi</p>

<p>There was an explanation in the consolidated list, but I still don't get # 15 (I can't really write the problem, because there's a diagram)</p>

<p>I thought it was (E), which is route 10 (10^0.5) but I was wrong. I don't get it, because I thought that line AB meant that the line went ON the edge which is in between A & B, which would result in one big right-triangle with a '3' as a base and '1' as height, therefore sqrt(3^2 + 1^2)=sqrt(10)</p>

<p>Why is my explanation wrong? Thank you for your help (this problem is driving me up the wall!)</p>

2

<p>AB is going through the cube, not around the outside. It is still a right triangle. The height is 1, but as the BB explained it, the base is the diagonal segment from point A to the corner below Point B. So, you have to do 2 right triangles.
The 1st step is to find the distance from A to the corner 1^2 + 2^2 = 5. the hypotenuse is therefore sqrt5.
Then you use sqrt 5 as the base for your 2nd right triangle, which AB is the hypotenuse. sqrt5^2 + 1^2 = 6
AB is then sqrt 6.</p>

<p>hope it helps</p>