Geometry book, Hard Question Set, #8

[DanC]

In the geometry book, in the hard practice questions, number 8, why does the 1:SR3:2 ratio not work for that problem?

I analyzed the problem properly, discovered that triangle ABC was a 30-60-90, and then attempted to apply that rule to it, and it failed. Why? The side opposite the angle of 60 degrees is 6. That is given. Shouldn't that mean the multiplier for the ratio is 6/SR3? I got that by doing x times SR3=6. Which would make BC = 6/SR3 and AC = 12/SR3.

That obviously is wrong. But why? Did I apply the 30-60-90 rule incorrectly or does that rule not work on some problems?

Note: SR = square root.

[DanC]

Similar problem on question 15. After figuring the measurements of the sides for triangle PQT, how do we know that RT must = 1/4? Am I to assume that because all three are 30-60-90 that their sides are all in direct proportion to each other? Even so, that would seem to make ST = 1/4, not RT, because the 1 and 1/2 measurements are for the hypotenuses of the other triangles and TS is the hypotenuse of the RST triangle. RT is the long side, which corresponds with PR in the triangle PTQ and with QR in QRT.

[DanC]

Can someone explain the last line in the answer for question 15, starting with x= and ending with SR3/12?

Prior to that we have established that x*SR3=1/4, which would mean that x=(1/4)/SR3. But how does that get multiplied by SR3/SR3? Shouldn't it be multiplied by 1/SR3 (the reciprocal of what it is being divided by)? I mean, SR3/SR3=1, so how do you go from dividing one quarter by SR3 to multiplying one quarter by 1?

[tommywallach]

Hey Guys,

First thing, about posting. Always put the question/book/page that you're having trouble with as the subject of the post. That way other people with a similar question can locate it (I've changed the name of this post). Also, please limit every post to the subject of a single question. So Dan, would you mind posting your question as a separate thread? Thanks!

Now, to the question:

Everything you did was perfect Dan, you just didn't finish the question!

The base is indeed 6/rt. 3. So area of triangle =

1/2 * b * h = 1/2 * 6 * 6/rt. 3 = 18/rt.3

Multiply top and bottom by rt. 3 -->

(18 * rt. 3) / 3 = 6 * rt. 3

Tada!

-t

[DanC]

Thanks for clarifying that I was correct in finding 6/rt3 as the multiplier in the 1:rt3:2 ratio. I did the area like you said and got 18/rt3, but how does that become 6rt3? Why do you do that last multiplication? My answer (18/rt3) and the official answer equal the same amount, so I am confused as to why 18/rt3 is not considered simplified enough to stand as the answer.

[tommywallach]

Hey Dan,

Sorry for the slow response. It's considered incorrect to ever leave radicals in the denominator. So in this case, we multiply top and bottom by rt. 3.

Hope that helps!

-t