Answer:
Solution given:
<J=90°[tangent to the circle is perpendicular to radius]
In right angled triangle ∆KJL
base[b]=11
hypotenuse [h]=11+7=18
perpendicular [p]=x
by using Pythagoras law
p²+b²=h²
x²+11²=18²
x²=18²-11²
x=√203
x=14.25
