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<h3>Refer to the diagram below</h3>
- Draw one smaller circle inside another larger circle. Make sure the circle's edges do not touch in any way. Based on this diagram, you can see that any tangent of the smaller circle cannot possibly intersect the larger circle at exactly one location (hence that inner circle tangent cannot be a tangent to the larger circle). So that's why there are no common tangents in this situation.
- Start with the drawing made in problem 1. Move the smaller circle so that it's now touching the larger circle at exactly one point. Make sure the smaller circle is completely inside the larger one. They both share a common point of tangency and therefore share a common single tangent line.
- Start with the drawing made for problem 2. Move the smaller circle so that it's partially outside the larger circle. This will allow for two different common tangents to form.
- Start with the drawing made for problem 3. Move the smaller circle so that it's completely outside the larger circle, but have the circles touch at exactly one point. This will allow for an internal common tangent plus two extra external common tangents.
- Pull the two circles completely apart. Make sure they don't touch at all. This will allow us to have four different common tangents. Two of those tangents are internal, while the others are external. An internal tangent cuts through the line that directly connects the centers of the circles.
Refer to the diagram below for examples of what I mean.
<h3>Given</h3>
trapezoid PSTK with ∠P=90°, KS = 13, KP = 12, ST = 8
<h3>Find</h3>the area of PSTK
<h3>Solution</h3>It helps to draw a diagram.
∆ KPS is a right triangle with hypotenuse 13 and leg 12. Then the other leg (PS) is given by the Pythagorean theorem as
... KS² = PS² + KP²
... 13² = PS² + 12²
... PS = √(169 -144) = 5
This is the height of the trapezoid, which has bases 12 and 8. Then the area of the trapezoid is
... A = (1/2)(b1 +b2)h
... A = (1/2)(12 +8)·5
... A = 50
The area of trapezoid PSTK is 50 square units.