Please help with partial circle areas...
- Thread starter adevine
- Start date
Hello, Anthony!
I must assume you know some Trigonometry . . . and some basic Geomtry.
\(\displaystyle \text{In No.5, we have a quarter-circle with area: }\;\frac{1}{4}\pi(4^2) \:=\:4\pi\)
. . \(\displaystyle \text{and a right triangle with area: }\:\frac{1}{2}(4)(4) \:=\:8\)
\(\displaystyle \text{The shaded area is: }\,4\pi - 8\)
\(\displaystyle \text{In No.6, we have a sixth of a circle with area: }\;\frac{1}{6}\pi(6^2) \:=\:6\pi\)
. . \(\displaystyle \text{and an equilateral triangle with side 6; its area is: }\;\frac{\sqrt{3}}{4}(6^2) \:=\:9\sqrt{3}\)
\(\displaystyle \text{The shaded area is: }\,6\pi - 9\sqrt{3}\)
\(\displaystyle \text{In No.7, we have a third of a circle.}\)
\(\displaystyle \text{The two small triangles are 30-60 right triangles.}\)
\(\displaystyle \text{Since the shortest side is 3, the hypotenuse (radius) is 6.}\)
. . \(\displaystyle \text{The third side is: }\:3\sqrt{3}\)
\(\displaystyle \text{The area of the sector is: }\;\frac{1}{3}\pi(6^2) \:=\:12\pi\)
\(\displaystyle \text{The area of the triangle is: }\;\frac{1}{2}(6\sqrt{3})(3) \:=\:9\sqrt{3}\)
\(\displaystyle \text{The shaded area is: }\,12\pi - 9\sqrt{3}\)
I must assume you know some Trigonometry . . . and some basic Geomtry.
\(\displaystyle \text{In No.5, we have a quarter-circle with area: }\;\frac{1}{4}\pi(4^2) \:=\:4\pi\)
. . \(\displaystyle \text{and a right triangle with area: }\:\frac{1}{2}(4)(4) \:=\:8\)
\(\displaystyle \text{The shaded area is: }\,4\pi - 8\)
\(\displaystyle \text{In No.6, we have a sixth of a circle with area: }\;\frac{1}{6}\pi(6^2) \:=\:6\pi\)
. . \(\displaystyle \text{and an equilateral triangle with side 6; its area is: }\;\frac{\sqrt{3}}{4}(6^2) \:=\:9\sqrt{3}\)
\(\displaystyle \text{The shaded area is: }\,6\pi - 9\sqrt{3}\)
\(\displaystyle \text{In No.7, we have a third of a circle.}\)
\(\displaystyle \text{The two small triangles are 30-60 right triangles.}\)
\(\displaystyle \text{Since the shortest side is 3, the hypotenuse (radius) is 6.}\)
. . \(\displaystyle \text{The third side is: }\:3\sqrt{3}\)
\(\displaystyle \text{The area of the sector is: }\;\frac{1}{3}\pi(6^2) \:=\:12\pi\)
\(\displaystyle \text{The area of the triangle is: }\;\frac{1}{2}(6\sqrt{3})(3) \:=\:9\sqrt{3}\)
\(\displaystyle \text{The shaded area is: }\,12\pi - 9\sqrt{3}\)