Let be △ABD and △BCD equilateral triangles of side 10. Q,R,S and T are middle points of △ABD and triangle BCD. FInd the shaded area.
+781 Notice that the white shapes are 1/6 of a circle. There are 6 of them so in total, their area is equivalent to one circle. We need to find the radius, which is 1/2 of the side of the equilateral triangle. So the radius of the circle is 1/2*10=5. Then the area of the circle is \(\pi{r}^2=\pi(5)^2=25\pi\). The shaded area is then the equilateral triangle x 2 minus 25pi.
To find the area of the equilateral triangle, we already have the base but we need the height. To find this we use the pythagorean theorem. We are using half of the equilateral triangle in this step.
a^2+b^2=c^2
5^2+b^2=10^2
25+b^2=100
b^2=75
b=\(5\sqrt{3}\)
So the area of the equilateral triangle is
\(\frac{1}{2}\cdot5\sqrt{3}\cdot10\\ =5\cdot5\sqrt{3}\\ =25\sqrt{3}\)
Multiply this by 2 to get \(50\sqrt{3}\). Now we subtract 25pi. Therefore the area of the shaded area is \(50\sqrt{3}-25\pi\).
