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Area of an Equilateral Triangle

A copper wire...

Question

A copper wire when bent in the form of an equilateral triangle has an area of $121 \sqrt{3} c m^{2}$ . If the same wire is bent into the form of a circle, then the area enclosed by the wire is

A

110.25

c m^{2}

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B

346.5

c m^{2}

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C

121.5

c m^{2}

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D

336.5

c m^{2}

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Solution

The correct option is B 346.5 $c m^{2}$
$G i v e n - A w i r e i s b e n t i n t o a n e q u i l a t e r a l t r i a n g l e . I t s a r e a = 121 \sqrt{3} {c m}^{2} . T h e s a m e w i r e i s b e n t i n t o a c i r c l e . T o f i n d o u t - a r . c i r c l e = ? S o l u t i o n - T h e s a m e w i r e i s b e n t i n t o a n e q u i l a t e r a l t r i a n g l e a n d a c i r c l e . ∴ P e r i m e t e r P o f t h e t r i a n g l e = c i r c u m f e r e n c e C o f t h e c i r c l e . L e t t h e s i d e o f t h e e q u i l a t a r a l t r i a n g l e b e x . i . e P = 3 x = C . . . . . . . . . (i) T h e n a r . t r i a n g l e = \frac{\sqrt{3}}{4} x^{2} {c m}^{2} . = 121 \sqrt{3} {c m}^{2} (g i v e n) ⟹ x = 22 c m . S o P = 3 x = 3 \times 22 c m = 66 c m = C (b y i) . . . . . . . . (i i) A g a i n l e t u s a s s u m e t h a t t h e r a d i u s o f t h e c i r c l e = r . T h e n t h e c i r c u m f e r e n c e = C = 2 π r ⟹ r = \frac{C}{2 π} = \frac{66}{2 π} (f r o m i i) ⟹ r = \frac{66}{2 \times \frac{22}{7}} c m = 10.5 c m . ∴ a r . c i r c l e = π r^{2} = \frac{22}{7} \times {(10.5)}^{2} {c m}^{2} = 346.5 {c m}^{2} . A n s - O p t i o n B .$

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