A circle has a radius of 6 in. The circumscribed equilateral triangle will have an area of:
do you have a figure? i mean is the triangle inside or outside the circle?
No it just asking this
I think the circle is inside the tringle
Circumscribed means outside. Inside would be inscribed.
The height of the equilateral triangle will be \[\frac 3 2 \times 6 = 9\] Length of the side \( = 9 \times \frac 2 {\sqrt{3}}=6\sqrt{3} \)
Area \(= \frac 12 6\sqrt{3} \times 9 =27\sqrt{3} \; in^2\) It's still early in the morning so make sure that my algebra is correct.
But the process is right.
My answer is suppose to be \[XXX \sqrt{X}\]
i have to fill in the x
X=3
\[3\times 3\times 3\times \sqrt{3} \]
Do you understand?
The three x is a big number
I have computer work and it grades it show half correct
|dw:1338514431053:dw|
I get 12sqrt(3) for the side-length: |dw:1338515519577:dw| The height would then be the longer leg of a 30-60-90 triangle, 6sqrt(3)*sqrt(3) = 6*3 = 18 A = 1/2 b h = 1/2 18 * 12sqrt(3) = 9 * 12sqrt(3) = 108sqrt(3)
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