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 Post subject: Questions from Geometry diagnosticPosted: Sat Jun 12, 2010 1:51 pm

Joined: Sat Jun 12, 2010 12:36 pm
Posts: 1
Hi there,

I'm curious on the two questions below - why is is that in the first one we can assume 30-60-90 but in the second one when we are told the diagonal is 13, we can’t assume 30-60-90 in order to answer? (I had answered D for the first one, and D for the second one)

1. Circle O is inscribed in equilateral triangle ABC, which is itself inscribed in circle P. What is the area of circle P?
(1) The area of circle O is 4π.
(2) The area of triangle ABC is 12√2
Answer D

2. What is the perimeter of rectangle R?
(1) The area of rectangle R is 60.
(2) The length of a diagonal of rectangle R is 13.
Answer C

Thanks

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 Post subject: Re: Questions from Geometry diagnosticPosted: Thu Jun 17, 2010 1:13 pm

Joined: Thu May 06, 2010 3:17 pm
Posts: 28
Location: Everywhere
For #2, we need the dimensions of rectangle R in order to find the perimeter.

Statement 1 tells us that the area of R is 60, but there are many possible pairs for length and width that would give us an area of 60: 20x3, 15x4, 10x6, and so on. Since we don't know which pair it actually is, we can't find a definite perimeter. We can eliminate A and D at this point.

Statement 2 gives us the length of the diagonal, which is 13. Careful! This statement is designed to lead you into believing that it must be a 5-12-13 triangle when the diagonal is drawn. Could it be 5-12-13? Certainly, but it doesn't have to be. Using the Pythagorean theorem, we can find several values to plug in for A and B that would give us 169 for the square of the hypotenuse (13). We can eliminate choice B, leaving C and E as possible answers.

When we combine the statements, the only pair of values for length and width that will give us an area of 60 and a diagonal of 13 are 5 and 12, giving us a perimeter of 34. Of course, we don't have to find the actual perimeter; once we've got definite values for R's dimensions, there is only one possible value for its perimeter.

C is correct.

I will take a look at #1 in a bit and hopefully have a good explanation for you.

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 Post subject: Re: Questions from Geometry diagnosticPosted: Thu Jun 17, 2010 1:36 pm

Joined: Thu May 06, 2010 3:17 pm
Posts: 28
Location: Everywhere
[quote="Lynnut"]Hi there,

I'm curious on the two questions below - why is is that in the first one we can assume 30-60-90 but in the second one when we are told the diagonal is 13, we can’t assume 30-60-90 in order to answer? (I had answered D for the first one, and D for the second one)

1. Circle O is inscribed in equilateral triangle ABC, which is itself inscribed in circle P. What is the area of circle P?
(1) The area of circle O is 4π.
(2) The area of triangle ABC is 12√2
Answer D
[/quote]

We're not truly assuming that it's 30-60-90. Since it's an equilateral triangle, we know that drawing a height is guaranteed to give us a 30-60-90 triangle.

1) With the area of O, we can find a radius of 2. With our knowledge of equilateral triangles, we can create a 30-60-90 triangle where the radius of 2 is the side opposite the 30-degree angle. The hypotenuse then becomes 4; conveniently for us, the hypotenuse that we drew is also the radius of circle P, so we can find the area. Statement 1 is sufficient on its own, so we can eliminate B, C, and E.

2) I'm going to guess that you meant the area of ABC is 12√3? If so, we can use the equilateral area formula to find that each side of the triangle is 4√3; from there, we can derive the same 30-60-90 triangle as in Statement 1, and once we find the hypotenuse of the triangle, the area of Circle R is pretty easy to find. Statement 2 is also sufficient, so our answer is D.

Hopefully my diagram makes sense; I drew it with my finger on my iPad.

[attachment=0]idea-1.png[/attachment]

 Attachments: idea-1.png [ 42.85 KB | Viewed 1682 times ]
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