Question

Two chords of lengths 24cm and 18cm, are on the same side of the center, and parallel to each
other. If the distance between the chords is 3cm, then the radius of the circle equals ______ cm.

Answer 1

S O L U T I O N :

$\underline{\bf{Given\::}}$

Two chords of lengths are 24 cm & 18 cm are on the same side of the center, & parallel to each other. If the distance between the chords is 3 cm.

$\underline{\bf{Explanation\::}}$

Firstly, attachment a figure of circle with two chords according to the given question.

Let the radius of circle be M.

A/q

→ OS + OR = 3

→ OR = 3 - x

→ OS = x

Now, we can see that;

In Δ OAR :

• AR = 12 cm
• OR = 3 - OS
• OA = M

$\underline{\underline{\tt{Using\:by\:Pythagoras\:theorem\::}}}$

$\mapsto\sf{(Hypotenuse)^{2} = (Base)^{2} + (Perpendicular)^{2} }$

$\mapsto\sf{(OA)^{2} = (AR)^{2} + (OR)^{2} }$

$\mapsto\sf{(M)^{2} = (12)^{2} + (3-x)^{2}}$

$\mapsto\sf{(M)^{2} = 144 + (3)^{2} + (x)^{2} -2\times 3 \times x}$

$\mapsto\sf{(M)^{2} = 144 + 9 + x^{2} -6x}$

$\mapsto\sf{(M)^{2} = 153 + x^{2} -6x...........(1)}$

&

In Δ OCS :

• OC = M
• CS = 9 cm
• OS = x

$\underline{\underline{\tt{Using\:by\:Pythagoras\:theorem\::}}}$

$\mapsto\sf{(Hypotenuse)^{2} = (Base)^{2} + (Perpendicular)^{2} }$

$\mapsto\sf{(OC)^{2} = (CS)^{2} + (OS)^{2} }$

$\mapsto\sf{(M)^{2} = (9)^{2} + (x)^{2} }$

$\mapsto\sf{(M)^{2} = 81 + x^{2} ...............(2)}$

So, comparing both equation because both radius equal.

$\mapsto\tt{153 + x^{2} - 6x = 81 + x^{2}}$

$\mapsto\tt{153 \cancel{+ x^{2}} - 81 =\cancel{x^{2}} + 6x}$

$\mapsto\tt{153 - 81 = 6x}$

$\mapsto\tt{72 = 6x}$

$\mapsto\tt{6x = 72}$

$\mapsto\tt{x = \cancel{72/6}}$

$\mapsto\bf{x = 12\:cm}$

∴ Putting the value of x in equation (2),we get;

$\mapsto\tt{(M)^{2} = 81 + (12)^{2}}$

$\mapsto\tt{(M)^{2} = 81 + 144}$

$\mapsto\tt{(M)^{2} = 225}$

$\mapsto\tt{M = \sqrt{225} }$

$\mapsto\bf{M = 15\:cm}$

Thus,

The radius of the circle will be M = 15 cm .