<p>I still need an answer for this one:
Circle, with radius 1, is inscribed in a triangle, perimeter 16. Three of the radii are each perpendicular to one side of a triangle. What is the area of the triangle? </p>
<p>Figure Not Drawn To Scale</p>
<p>Answer Choices:
8
9
12
15
16</p>
<p>Did anyone at least have this problem?</p>
<p>I need to know!</p>
<p>answer is 8... its some obscure geometry formula that "area = semiperimenter * inradius" ..theres no other way at all to do it</p>
<p>"theres no other way at all to do it"</p>
<p>Actually, there is a way to do it that does not require to know the "obscure" formula. </p>
<p>As usual you need to pay attention to ALL the clues. One of them says that the radii are perpendicular. </p>
<p>Now, visualize the removal of the circle and draw three lines from the three angles to the center of the circle. This forms three smaller triangles with the radius as height. What do you have? The height of the three triangles is always the radius or 1. What it is the formula? bh/2. The sums of the three b is given as 16. This means that the sum of the 3 triangles is 16*1/2 or 8. </p>
<p>This said, I think this is a very hard problem, short of knowing the "obscure" formula or some knowledge of the incenter properties.</p>
<p>
</p>
<p>But the original problem says the radii are perpendicular to a side. How can you draw lines from the angles? It changes the problem, doesn't it?</p>
<p>OA, I do not have the original figure, but it should like this:</p>
<p>Draw a circle. Add three radii. Now draw 3 tangents that are perpendicular to the 3 radii. The lines should cross at three points and form the triangle. Let's call those points ABC. The center of the circle could be called D. </p>
<p>Now, you can proceed by drawing the ADDITIONAL lines from the three points A, B, and C to the center D. The three triangles are ADB, BDC, and ADC. The three bases are AB, BC, and AC. The height of the 3 triangles is equal to the radius, in this case 1.</p>
<p>In "general" math questions, and especially SAT ones, this approach often works:
in geometry - assume the easiest configuration, in algebra - assign the most "convenient" numeric values to variables.
Good algebraic example - Blue Book / 657 / 18 /.</p>
<p>For a circle in a triangle question.
Since this question does not give any specifics on a triangle, the answer will be the same for any kind of a triangle.</p>
<p>Let's then assume that our triangle is equilateral.
We still have to make the most difficult in geometry solutions leap - draw additional lines.
Since xiggi took care of it, we can just enjoy the view: three equal triangles, each having the incenter of the big triangle as a "top" vertex and a side of the big triangle as a base.
Solution.
One side of the big triangle = 16/3,
area of one small triangle - (1/2)x(16/3)x1 = 8/3,
total area = (8/3)x3 = 8.</p>
<p>This is a classic question from a higher level elementary geometry, and I think it belongs to the SAT II math.
Oh well, time's a changin'.</p>
<p>This question was in the experimental section.</p>
<p>An interesting twist is to look at the problem of a circle with radius 1 that is inscribed in a square. Such problem would be trivial for most SAT takers because the diameter also represents the value of one side. Hence, the area of the square is 4. What is the perimeter? 2 times 4 or 8. Hmm, the value for the area is one half of the value for the perimeter?</p>
<p>Could it be that the principles are the same, but that students are more used to see the relations between a square and a circle? </p>
<p>A further interesting twist would be to see what happens if the triangle was given and we needed to find the relationship with a circle that circumscribes the triangle. :)</p>
<p>Just out of curiosity Xiggi, what did you make on the SAT?</p>