Given:
CD = 18 cm
CE = 8 cm
To find:
The length of the radius of the circle A.
Solution:
Let the radius of the circle be x.
AD = AE = x
CA = CE + AE
CA = 8 + x
The angle between a tangent and radius is always right angle.
Therefore triangle ADC is a right triangle.
Using Pythagoras theorem:
[tex]AD^2+CD^2=CA^2[/tex]
[tex]x^2+18^2=(8+x)^2[/tex]
Using algebraic identity: [tex](a+b)^2=a^2+2ab+b^2[/tex]
[tex]x^2+324=8^2+16x+x^2[/tex]
[tex]x^2+324=64+16x+x^2[/tex]
Subtract x² from both sides.
[tex]324=64+16x[/tex]
Subtract 64 from both sides.
[tex]260=16x[/tex]
Divide by 16 on both sides, we get
[tex]16.25=x[/tex]
The length of the radius of the circle is 16.25 cm.
