I'm studying for the SAT, and I came upon a website that gives random problems every time you refresh. I was working on them, and I found one that I have no clue how to approach. It includes a circle tangent to the insides of an equilateral triangle, with a radius of 8. I'm supposed to find the perimeter of the triangle based on this information- but how?!
Circle in Equilateral Triangle
- Thread starter sphelps17
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Draw a line segment from the center of the circle to one of the tangent points, then draw a line segment from the center of the circle to one of the vertices adjacent to the chosen tangent point.
You now have a 30°-60°-90° triangle, and the shorter leg is the radius of the circle. The longer leg is 1/6 the perimeter of the equilateral triangle.
Can you proceed from here?
You now have a 30°-60°-90° triangle, and the shorter leg is the radius of the circle. The longer leg is 1/6 the perimeter of the equilateral triangle.
Can you proceed from here?
The only way to solve this problem is to draw it out. Work with your lines and imagination. Think angles and triangles, specifically a right triangle (90 degree triangle).
From the center of the circle, draw two lines - one joining a vertex of the triangle, and the other line joining the center of the circle and the midpoint of the side of the triangle next to the vertex.
If a is the side of the triangle, r is the radius of the circle, and hopefully if you are familiar with trignometry, can you see that
tan(60/2)=r/(a/2)=Opposite side/Adjacent side
Note that tan 30 degrees=1/sqrt(3). Thus, you can find 'a', which is twice of 'a/2'.
Thus 1/sqrt(3)=2r/a =>a=2*sqrt(3)*r or r=a/(2*sqrt(3)). Thus, if you know 'r', you can solve for 'a' or vice versa.
Cheers,
Sai.
From the center of the circle, draw two lines - one joining a vertex of the triangle, and the other line joining the center of the circle and the midpoint of the side of the triangle next to the vertex.
If a is the side of the triangle, r is the radius of the circle, and hopefully if you are familiar with trignometry, can you see that
tan(60/2)=r/(a/2)=Opposite side/Adjacent side
Note that tan 30 degrees=1/sqrt(3). Thus, you can find 'a', which is twice of 'a/2'.
Thus 1/sqrt(3)=2r/a =>a=2*sqrt(3)*r or r=a/(2*sqrt(3)). Thus, if you know 'r', you can solve for 'a' or vice versa.
Cheers,
Sai.
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