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1.) Three circles are drawn, so that each circle is externally tangent to the other two circles. Each circle has a radius of 2. A triangle is then constructed, such that each side of the triangle is tangent to two circles, as shown below. Find the perimeter of the triangle.
2.) A circle centered at A with radius 10 is externally tangent to a circle centered at B with radius 7. A line that is externally tangent to both circles is drawn, where both circles lie on the same side of the line. This line intersects line AB at C. Find the length BC.
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hi Firestar134360
Welcome to the forum.
Q1. It's tempting to assume that triangle is equilateral, but, as I warn in another geometry post, you shouldn't make such assumptions; instead make sure you can prove it. Here's how:
Let the centres be at A, B and C. That triangle is equilateral because all the sides are length 4. Let the circles touch at D,E and F. Let the large triangle have a top vertex at H and left bottom at I. Let G be the point on the 'C' circle where HI touches and similarly, J on the 'A' circle touching HI at J
As CGJ = AJG = 90 ACGJ is a rectangle, so GJ is parallel to AC. A similar analysis will show that the other two sides of the large triangle are parallel to the sides of the small triangle. Thus the large triangle is also equilateral. You can use trig to calculate HG. GJ = AC and by symmetry JI = HG. So you can calculate the length of a side.
Q2. Let DE be the external tangent with D on circle 'A' and E on circle 'B'. In the triangles ADC and BEC, one angle is 90 and angle ACD is common so these triangles are similar. If you call BC length x, you can use the similar triangles to say AD:BE = AC:BC. Form an equation in x from this and solve.
Hope that helps,
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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