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20 Using Figure 5.33, find the missing piece of information in parts (a)-(d). B BE = 12, AE = 9, DE = 1, CE = ? (b)* BC = 5, CE = 4, DE = 9, AD = ? (c)* BC = 6, AD = 4, DE = 2, CE = ? (d) CE = 4, DE = 3, AD = 4, BC ? = A C D Figure 5.33 E

20 Using Figure 5.33, find the missing piece of information in parts (a)-(d). B BE = 12, AE = 9, DE = 1, CE = ? (b)* BC = 5, CE = 4, DE = 9, AD = ? (c)* BC = 6, AD = 4, DE = 2, CE = ? (d) CE = 4, DE = 3, AD = 4, BC ? = A C D Figure 5.33 E

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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**Problem 20: Using Figure 5.33, find the missing piece of information in parts (a)–(d).** **Figure 5.33:** The diagram shows a circle with a secant and a tangent originating from point E. The secant intersects the circle creating segments BE and AE. The tangent is segment CE, and DE is a segment inside the circle, intersecting the tangent and secant at points C and D, respectively. **Given Dimensions and Required Solutions:** (a) \( BE = 12 \), \( AE = 9 \), \( DE = 1 \), \( CE = ? \) (b) \( BC = 5 \), \( CE = 4 \), \( DE = 9 \), \( AD = ? \) (c) \( BC = 6 \), \( AD = 4 \), \( DE = 2 \), \( CE = ? \) (d) \( CE = 4 \), \( AE = 3 \), \( AD = 4 \), \( BC = ? \) **Explanation for Approach:** To solve this problem, use the Power of a Point theorem, which states that for a point outside a circle, the product of the whole secant and its external segment is equal to the square of the tangent segment. In mathematical terms: - For part (a), use \( BE \times AE = CE^2 \). - For part (b), use \( (BC + CE) \times CE = AD \times DE \). - For part (c), use \( BC \times (BC + DE) = CE^2 \). - For part (d), use \( (BC + CE) \times CE = AD \times AE \).
Transcribed Image Text:**Problem 20: Using Figure 5.33, find the missing piece of information in parts (a)–(d).** **Figure 5.33:** The diagram shows a circle with a secant and a tangent originating from point E. The secant intersects the circle creating segments BE and AE. The tangent is segment CE, and DE is a segment inside the circle, intersecting the tangent and secant at points C and D, respectively. **Given Dimensions and Required Solutions:** (a) \( BE = 12 \), \( AE = 9 \), \( DE = 1 \), \( CE = ? \) (b) \( BC = 5 \), \( CE = 4 \), \( DE = 9 \), \( AD = ? \) (c) \( BC = 6 \), \( AD = 4 \), \( DE = 2 \), \( CE = ? \) (d) \( CE = 4 \), \( AE = 3 \), \( AD = 4 \), \( BC = ? \) **Explanation for Approach:** To solve this problem, use the Power of a Point theorem, which states that for a point outside a circle, the product of the whole secant and its external segment is equal to the square of the tangent segment. In mathematical terms: - For part (a), use \( BE \times AE = CE^2 \). - For part (b), use \( (BC + CE) \times CE = AD \times DE \). - For part (c), use \( BC \times (BC + DE) = CE^2 \). - For part (d), use \( (BC + CE) \times CE = AD \times AE \).
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