For in GOD we live, and move, and have our being.
- Acts 17:28
The Joy of the Teacher is the Success of the Students.
-
Samuel Dominic Chukwuemeka
Welcome to our site.
I greet you this day,
The polygons discussed in this site are regular polygons only.
First: read the notes. Second: view the videos. Third: solve the questions/solved
examples.
Fourth: check your solutions with my thoroughly-explained solutions. Fifth: check
your answers with the calculators as applicable.
I wrote the codes for the calculators using JavaScript, a client-side scripting language. Please use the
latest Internet browsers. The calculators should work.
Only integers and decimals are allowed. Fractions are not allowed.
Comments, ideas, areas of improvement, questions, and constructive criticisms are welcome. You may
contact me.
If you are my student, please do not contact me here. Contact me via the school's system.
Thank you for visiting!!!
Samuel Dominic Chukwuemeka (Samdom For Peace) B.Eng., A.A.T, M.Ed., M.S
Equilateral Triangle inscribed in a Circle OR a Circle circumscribed about an Equilateral Triangle
| $ h = \dfrac{3r}{2} \\[5ex] b = r\sqrt{3} \\[3ex] a = \dfrac{r}{2} \\[5ex] r = 2a \\[3ex] A_T = \dfrac{bh}{2} \\[5ex] A_C = \pi r^2 \\[3ex] \dfrac{A_T}{A_C} = \dfrac{3\sqrt{3}}{4\pi} \\[5ex] \dfrac{A_C}{A_T} = \dfrac{4\pi \sqrt{3}}{9} \\[5ex] A_r = A_C - A_T \\[3ex] A_{ep} = \dfrac{A_r}{3} $ | $ h = \dfrac{b\sqrt{3}}{2} \\[5ex] b = \dfrac{2h\sqrt{3}}{3} \\[5ex] a = \dfrac{b\sqrt{3}}{6} \\[5ex] r = \dfrac{2h}{3} \\[5ex] A_T = \dfrac{3r^2\sqrt{3}}{4} \\[5ex] $ | $ h = 3a \\[3ex] b = 2a\sqrt{3} \\[3ex] a = \dfrac{h}{3} \\[5ex] r = \dfrac{b\sqrt{3}}{3} \\[5ex] $ |
Circle inscribed in an Equilateral Triangle OR an Equilateral Triangle circumscribed about a Circle
| $ h = 3r \\[3ex] b = 2r\sqrt{3} \\[3ex] a = r \\[3ex] r = a \\[3ex] A_T = \dfrac{bh}{2} \\[5ex] A_C = \pi r^2 \\[3ex] \dfrac{A_T}{A_C} = \dfrac{3\sqrt{3}}{\pi} \\[5ex] \dfrac{A_C}{A_T} = \dfrac{\pi \sqrt{3}}{9} \\[5ex] A_r = A_T - A_C \\[3ex] A_{ep} = \dfrac{A_r}{3} $ | $ h = \dfrac{b\sqrt{3}}{2} \\[5ex] b = \dfrac{2h\sqrt{3}}{3} \\[5ex] a = \dfrac{b\sqrt{3}}{6} \\[5ex] r = \dfrac{h}{3} \\[5ex] A_T = 3r^2\sqrt{3} \\[3ex] $ | $ h = 3a \\[3ex] b = 2a\sqrt{3} \\[3ex] a = \dfrac{h}{3} \\[5ex] r = \dfrac{b\sqrt{3}}{6} \\[5ex] $ |
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Square inscribed in a Circle OR
a Circle circumscribed about a Square
The diameter of the circle is equal to the diagonal of the square
| $ d = d_S \\[3ex] l = r\sqrt{2} \\[3ex] a = \dfrac{l}{2} \\[5ex] r = \dfrac{l\sqrt{2}}{2} \\[5ex] A_S = l^2 \\[3ex] A_C = \pi r^2 \\[3ex] \dfrac{A_S}{A_C} = \dfrac{2}{\pi} \\[5ex] \dfrac{A_C}{A_S} = \dfrac{\pi}{2} \\[5ex] A_r = A_C - A_S \\[3ex] A_{ep} = \dfrac{A_r}{4} $ | $ l = 2a \\[3ex] a = \dfrac{r\sqrt{2}}{2} \\[5ex] r = a\sqrt{2} \\[3ex] A_S = 2r^2 \\[3ex] $ |
Circle inscribed in a Square OR a Square circumscribed about a Circle