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Similaire à CPM Geometry Unit 10 Day 6 Cs50 Common Tangent
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CPM Geometry Unit 10 Day 6 Cs50 Common Tangent
- 1. Unit 10 Day 6
- 3. CS50 a IN = 12” IL = 9” Find SA
- 4. CS50 a IN = 12” IL = 9” Find SA Hint #1 Draw IS and IA
- 5. CS50 a IN = 12” IL = 9” Find SA Hint #2 Draw IT Answer: SA =
- 6. CS50 b Find RE
- 7. CS50 b 8’ 8’ 14’ MA = 14 + 17 + 8 = 39 MB = 14 – 8 = 6 So since triangle MBA is a right triangle, 39 2 – 6 2 = BA 2 , therefore BA = 38.54’, which is the same as RE. 6’
- 8. How to solve common tangent problems. If the radius of the large circle is 14, the radius of the smaller circle is 10 and the circles are 10’ apart, find the length of the common tangent.
- 9. Step #1 – Draw the segment joining the centers of the circles If the radius of the large circle is 14, the radius of the smaller circle is 10 and the circles are 10’ apart, find the length of the common tangent.
- 10. Step #2 – Draw the radii to the points of contact. If the radius of the large circle is 14, the radius of the smaller circle is 10 and the circles are 10’ apart, find the length of the common tangent.
- 11. Step #3 – From the center of the smaller circle , draw a segment parallel to the tangent line. If the radius of the large circle is 14, the radius of the smaller circle is 10 and the circles are 10’ apart, find the length of the common tangent.
- 12. Step #3 – From the center of the smaller circle , draw a segment parallel to the tangent line. If the radius of the large circle is 14, the radius of the smaller circle is 10 and the circles are 10’ apart, find the length of the common tangent. 10 10 14 10 10 4
- 13. Step # 4 – Use the Pythagorean theorem and properties of rectangles to find the length of the common external tangent. If the radius of the large circle is 14, the radius of the smaller circle is 10 and the circles are 10’ apart, find the length of the common tangent. 10 10 14 10 10 4
- 14. Step # 4 – Use the Pythagorean theorem and properties of rectangles to find the length of the common external tangent. 10 10 14 10 10 4 34 2 – 4 2 = 1140, so √1140 = 33.76, therefore the length of the tangent is 33.76
- 17. Finish CS52 – CS58 for tomorrow!