on a circular table cover of radius 42cm , a design is formed by a girl leaving an equilateral triangle ABC in rhe middle,

Asked by akshay6098 | 9th Aug, 2020, 07:32: PM

Expert Answer:

Let a be the length of each side of the equilateral triangle ABC in the circle.
Let space AD space be space the space altitude space of space ABC thin space from space straight A space on space BC
Altitude space of space an space equilateral space triangle equals fraction numerator square root of 3 straight a over denominator 2 end fraction
AD thin space is space also space the space median space of space triangle space ABC thin space from space straight A space on space BC space and space straight O space is space the space centroid space of space the space triangle space ABC
rightwards double arrow straight O space divides space AD space in space the space ratio space 2 colon 1
rightwards double arrow OA equals 2 over 3 AD
rightwards double arrow 42 equals 2 over 3 AD
rightwards double arrow AD equals 63 space cm
rightwards double arrow fraction numerator square root of 3 straight a over denominator 2 end fraction equals 63
rightwards double arrow straight a equals 42 square root of 3 space cm
Area space of space equilateral space triangle equals fraction numerator square root of 3 over denominator 4 end fraction straight a squared equals fraction numerator square root of 3 over denominator 4 end fraction open parentheses 42 square root of 3 close parentheses squared equals fraction numerator square root of 3 over denominator 4 end fraction cross times 42 cross times 42 cross times 3 equals 1323 square root of 3 space cm squared
Area space of space circle equals πr squared equals 22 over 7 cross times 42 cross times 42 equals 5544 space cm squared
Area space of space design space equals space 5544 space minus space 1323 square root of 3 space equals space 5544 space minus space 2288.79 space equals space 3255.21 space cm squared

Answered by Renu Varma | 11th Aug, 2020, 02:00: PM