Ans.

it is given that area of circle = πr^{2}

πr^{2} = 154

⇒ (22/7) × r^{2} = 154

⇒ r^{2} = 154 × (7/22) = 49

∴ r = 7 cm

‘r’ can not be –ve , So r=7cm.

OD=7cm

⇒AO=14cm

⇒AO=OC=OB=14cm

In right triangle ODB, By using the formula

⇒BD=12.12

⇒BC=2BD=2×12.12=24.24cm

So, Perimeter of the equilateral triangle

⇒3×length of one side

⇒3×24.24

⇒72.74cm

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If sin theta = 4/5 , find the value of 4tan theta -5 cos theta/ sec theta plus 4 cot theta.

sec squared theta minus sin squared theta minus two sin power four/ 2cos power 4 – cos squared = 1

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