Answer:
Part a) The radii are segments AC and AD and the tangents are the segments CE and DE
Part b) [tex]DE=4\sqrt{10}\ cm[/tex]
Step-by-step explanation:
Part a)
we know that
A radius is a line from any point on the circumference to the center of the circle
A tangent to a circle is a straight line which touches the circle at only one point. The tangent to a circle is perpendicular to the radius at the point of tangency.
In this problem
The radii are the segments AC and AD
The tangents are the segments CE and DE
Part b)
we know that
radius AC is perpendicular to the tangent CE
radius AD is perpendicular to the tangent DE
CE=DE
Triangle ACE is congruent with triangle ADE
Applying the Pythagoras Theorem
[tex]AE^{2}=AC^{2}+CE^{2}[/tex]
substitute the values and solve for CE
[tex]14^{2}=6^{2}+CE^{2}[/tex]
[tex]CE^{2}=14^{2}-6^{2}[/tex]
[tex]CE^{2}=160[/tex]
[tex]CE=\sqrt{160}\ cm[/tex]
[tex]CE=4\sqrt{10}\ cm[/tex]
remember that
CE=DE
so
[tex]DE=4\sqrt{10}\ cm[/tex]
