Question

In the fig, from a point Q, the tangent to a circle is 24cm and the distance of

Q from the centre is 25 cm, find the radius of the circle



please answer me fast please ​

Answer 1

Answer:

In right ∆POQ,

By Pythagoras Theorem,

${h}^{2} = {b}^{2} + {p}^{2} \\ { 25 }^{2} = {24}^{2} + {p}^{2} \\ 625 = 576 + {p}^{2} \\ 625 - 576 = {p}^{2} \\ 49 = {p}^{2} \\ p = \sqrt{49} \\ p = 7 \: cm$

Therefore, the radius of the circle is 7 cm.

Step-by-step explanation:

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Answer 2

By Pythagorean theorem

OQ^2 =PQ^2+ OP^2

=> 25^2=24^2+ OP^2

=> 625=576+OP^2

=> OP^2=625-576=49

=>OP= 7

hence the radius of the circle is 7cm

Hope it will help u