The area of the inscribed equilateral triangle is 75/4*sqrt(3). Find the perimeter of circumscribing square to the nearest whole number.

Guest Feb 14, 2020

#1**+1 **

I'm assuming that the area of the triangle is (75/4) * sqrt (3)

We can solve for the side length, S, of the triangle as follows

(75/ 4)sqrt (3) = (1/2) S^2 sin (60°)

(75/4)sqrt (3) = (1/2) S^2 * sqrt (3) / 2

(75/4) = (1/4)S^2

75 = S^2

sqrt (75) = S

And using the Law of Cosines we can find the radius, R, of the circle

75 = 2R^2 - 2R^2 cos (120°)

75 = 2R^2 + R^2

75 = 3 R^2

25 = R^2

5 = R

And the side of the square is twice this = 10

So.....the perimeter of the square = 4 * 10 = 40 units

CPhill Feb 14, 2020

#2**+2 **

The area of the inscribed equilateral triangle is 75/4*sqrt(3). Find the perimeter of circumscribing square to the nearest whole number.

The ratio of the area of an equilateral triangle to its circumcircle's area is: **1.299 : pi **or **0.413496**

**Area of the triangle is 75/4*sqrt(3) = 32.47595 u²**

**Area of the circle is 32.47595 / 0.413496 = 78.54 u²**

**Raduis of the circle is r = sqrt( 78.54 / pi ) = 5**

**Perimeter of a square is P = 8r = 40 units _{}**

Dragan Feb 14, 2020