Question

Given: AG and CI are common internal tangents of H and B. HG = 7, ED = 7, and ED = EF.

Answer 1

Answer:

EC = 12.12

Step-by-step explanation:

Given H and B are circles of radius 7

ED = 7

CB = 7, so DB is also 7

Therefore, EB = ED + DB

                        = 7 + 7

                        = 14

Since IC is tangent at C we know that it is making a right angle to the radius CB.

Therefore, Triangle EBC is a right angle triangle where CB = 7 and EB = 14

Therefore from Pythagoras theorem,

$(EB)^{2}= (EC)^{2}+(CB)^{2}$

$(EC)^{2}= (EB)^{2}-(CB)^{2}$

$(EC)^{2}= (14)^{2}-(7)^{2}$

$(EC)^{2}= 196-49$

$(EC)^{2}= 147$

Therefore, EC = 12.12

Answer 2

EC = 24

This is because CBE is a right triangle with a right angle at angle ECB. Then we know that BC =7 and EB = 25. Use the Pythagorean Theorem to find EC.