Answer:
Part a) The radii are AC and AD and the tangents are CE and DE
Part b) [tex]DE=4\sqrt{10}\ cm[/tex]
Step-by-step explanation:
Part a) we know that
The radius of a circle is the distance from the center of the circle to any point on the circle
A tangent to a circle is a straight line which touches the circle at only one point
so
In this problem
The radii are AC and AD
The tangents are CE and DE
Part b) If given that AC is 6 cm and AE is 14, what is DE?
we know that
The triangle ADE is a right triangle , because the radius AD is perpendicular to the tangent DE
Applying the Pythagoras Theorem
[tex]AE^{2}= AD^{2}+DE^{2}[/tex]
we have
[tex]AD=AC=6\ cm[/tex] -----> is the radius of the circle
[tex]AE=14\ cm[/tex]
substitute and solve for DE
[tex]14^{2}= 6^{2}+DE^{2}[/tex]
[tex]DE^{2}=196-36[/tex]
[tex]DE^{2}=160[/tex]
[tex]DE=\sqrt{160}=4\sqrt{10}\ cm[/tex]
