BF is the angle bisector of $∠ A B C$ constructed using the radius of the arc as 3 cm. If DF = 5 cm. Find the value of 2 EF + BE.

10 cm

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13 cm

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5 cm

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20 cm

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Solution

The correct option is B 13 cm
Given
Radius of the arc = 3 cm.
BD = BE = 3 cm (radii of same arc)
DF = EF (Arcs of equal radii) [ Construction of angle bisector]
$\Rightarrow$ EF = DF = 5 cm.
$∴$ 2EF + BE = ( 2 $\times$ 5) + 3 = 10 + 3 = 13 cm.

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