Question

In figure 3.99, seg MN is a chord of a circle with centre O. MN=25,L is a point on chord MN such that ML=9 and d(O,L)=5.Find the radius of the circle.

Answer 1

Answer:

Radius of the circle  = 13

Step-by-step explanation:

We have given some data:

MN = 25, ML = 9,

distance from O to L is 5

We have to find the radius of the circle

$\dfrac {25}2 = 9 + 5cos\theta\\\dfrac {25}2 - 9 = 5cos\theta \\\dfrac {25-18}2 = 5cos\theta\\\dfrac 72= 5cos\theta\\cos\theta = \dfrac 7{10}$

We will find tha value of $sin\theta$ by using pythagoras trick

So,

$sin\theta = \dfrac {\sqrt{51}} {10}\\5sin\theta = \dfrac {5\sqrt{51}} {10}$

Now we will find the radius,

$r^2 = \sqrt{(\frac {25}2)^2 +  ({\dfrac {5\sqrt{51}}{10})^2\\\\r^2 = \sqrt{\dfrac {625}{4} +\dfrac{1275}{100}\\\\r^2 = \sqrt{\dfrac {15625+1275}{100}\\\\\\r^2 = \sqrt{\dfrac {16900}{100}\\r^2 = \sqrt {169} \\r = 13$

Radius of the circle  = 13