Twenty gremlins and fifteen imps are at the Annual Mischief Convention. The imps have had a lot of in-fighting lately and refuse
to shake hands with each other, but they readily shake hands with all of the gremlins. Meanwhile, all the gremlins are quite friendly and shake hands with all of the other gremlins as well as imps. Each pair of creatures shakes hands at most once. How many handshakes were at the convention?
Let us first count the handshakes between two gremlins. There are $20$ gremlins, so there must be $\dfrac{20 \cdot 19}{2} = 190$ handshakes between two gremlins, making sure to divide by two to avoid over-counting.
Meanwhile, there are $15$ imps that shake hands with each of the $20$ gremlins, which makes $15 \cdot 20 = 300$ handshakes between imps and gremlins.
Adding them up, we have $300 + 190 = \boxed{490}$ total handshakes.
Let us first count the handshakes between two gremlins. There are 20 gremlins, so there must be 20*19/2=190 handshakes between two gremlins, making sure to divide by two to avoid over-counting.
There are 15 imps that shake hands with each of the 20 gremlins, which makes 15* 20 = 300 handshakes between imps and gremlins.
Adding these up, you have $300 + 190 = 490 total handshakes.
The circumference of the circle, 2πr, is the measure completely around the circle. When the pieces of the circle are rearranged, half of this circumference will be on the top of the parallelogram and half will be on the bottom. This means the base will be 1/2(2πr).
The approximate height of the parallelogram is the radius of the circle; this makes the area
