4 years ago

Can someone help me with this?

Mathematics

1 answer:

rewona [7]4 years ago

8 0

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

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