Question

The length of tangent from a point A at distance 41 cm from the centre of the circle is
Find the radius of the circle​

Answer 1

Correct Question:-

The length of tangent from a point A at distance 5 cm from the centre of the circle is 5 cm. Find the radius of the circle

Required Solution:-

$\sf\purple{By \:Pythagoras \:theorem}$

$\implies$ $\sf OA² = OB²+AB²$

$\implies$ $\sf 5² = OB²+4²$

$\implies$ $\sf 25-16 = OB²$

$\implies$ $\sf 9 = OB²$

$\implies$ $\sf OB = √9 = 3$

hence, radius is 3 cm.

Answer 2

Corrected Question :

• The length of tangent from a point A at distance 5 cm from the centre of the circle is 5 cm. Find the radius of the circle ?

A N S W E R :

• Radius of the circle is 3 cm.

$~~~~~$

Given :

• The length of tangent from a point A at distance 5 cm from the centre of the circle is 5 cm.

To find :

• Find the radius of the circle ?

$\large\star$ As we know that,

$\large\dag$By Using Pythagoras theorem :

• $\boxed{\bf{(Hypotenuse)^2 \:= \; (Height)^2 \:+ \: (Base)^2}}$

Solution :

• Let the circle be with center O

• Since AB is tangent

• Hence OB ⊥ AB

• $\angle$OBA = 90°

• So, ∆OAB is a right triangle

★ Given Substituting the values,

:$\implies$${\sf{AO^2 \:=\:OB^2\:+ \; AB^2}}$

$~~~~~$:$\implies$${\sf{5^2 \:=\:OB^2\:+ \; 4^2}}$

$~~~~~~~~~~$:$\implies$${\sf{OB^2 \:=\:25\:-\; 16}}$

$~~~~~~~~~~~~~~~$:$\implies$${\sf{OB^2 \:=\;9}}$

$~~~~~~~~~~~~~~~~~~~~$:$\implies$${\sf{OB\:=\; \sqrt{9}}}$

$~~~~~~~~~~~~~~~~~~~~~~~~~$:$\implies$${\sf{OB\:=\; \sqrt{3^2}}}$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$:$\implies$${\underline{\boxed{\frak{\pink{OB\:=\; 3}}}}}$★

$~~~~~$

Hence,

• ${\underline{\sf{Radius\: of\: the\; circle\:is\; \bf{OB\:=\:3\:cm}.}}}$

$~~~~$$\qquad\quad\therefore{\underline{\textsf{\textbf{Hence, Proved!}}}}$

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