equation solution (Question:277965.0)

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Anil Kumar

· started a discussion

· 1 Months ago

equation solution (Question:277965.0)

equation solution not understand

Question:

ABC is a quarter circle and a smaller circle is inscribed in it. If AB = 1 cm., then find the radius of smaller circle.

Options:

( $\sqrt{2} -1)$ cm.

( $2+ \sqrt{2}$ ) cm.

( $2 \sqrt{2} -1)$ cm.

1- $2 \sqrt{2}$ cm

Solution:

Let, the radius of the inner smaller circle be ‘r’ and that of the outer quarter circle be R

$∴$ OB = R-r and BD = OD =r

In ∆BDO,

By Pythagoras theorem

${(R-r)}^{2} = r^{2} + r^{2}$ . . . .. (i)

$∵ AB=R=1 cm$

$1-2r+ r^{2} = 2 r^{2} or r^{2} +2r-1=0$

By solving quadratic equation, we get

$r=(\sqrt{2} -1)$

$∵ The radius of smaller circle = \sqrt{2} -1 cm$

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equation solution (Question:277965.0)

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