# Sprint 17600 and solutions

May 1, 2019

### Click here to see the Monday April 30 sprint.

##### Problem 1:
The shaded region is the area left over after the area of the circle is subtracted from the area of the square. That can be expressed as
Area_(shade) = Area_S - Area_C
Substitute formulas:
Area_(shade) = S^2 - \pi r^2
Plug in values:
Area_(shade) = 17^2 - \pi (17/2)^2
and compute the answer to the nearest tenth. 62.0
##### Problem 2:
The shaded region is the area left over after the area of the circle is subtracted from the area of the square. That can be expressed as
Area_(shade) = Area_S - Area_C
Substitute formulas:
Area_(shade) = S^2 - \pi r^2
Plug in values:
Area_(shade) = 16^2 - \pi (8)^2
and compute the answer to the nearest tenth. 54.9
##### Problem 3:
To solve the problem you need to know how many degrees the interior angles of a Pentagon add up to. One way to figure that out is to draw all a pentagon and then draw all the diagonals from one vertex as shown:

Since the pentagon can be built from 3 triangles as illustrated, you know the interior angles of the pentagon must be 3 * 180 or 540 degrees. Remember that the 3 angles of a triangle always add to 180 degrees.

We don't know the 2 missing angles but we do know they're equal so we can express the sum of the 5 interior angles as:

540 = 35 + 11 + 44 + x + x

where x is the measure of one the two missing angles. Simplify the equation:

 540 = 90 + 2x
540 - 90 = 100 - 90 +2x
450 = 2x
450/2 = (2x)/2
225 = x
##### Problem 4:
The hexagon can be built from 4 triangles as shown:

4 * 180 = 720 so the interior angles of a hexagon add to 720 degrees.

This time we're given 4 angles and asked to find the remaining 2 equal sized angles. The formula is:

720 = 33 + 61 + 20 + 40 + x + x
720 = 154 + 2x
720 - 154 = 154 - 154 +2x
566 = 2x
566/2 = (2x)/2
283 = x
##### Problem 5:
This problem was a mistaken duplicate. It's solved the same way problem 4 is except the numbers are different.
720 = 33 + 60 + 23 + 30 + x + x
720 = 146 + 2x
720 - 146 = 146 - 146 + 2x
574 = 2x
574/2 = (2x)/2
287 = x
##### Problem 6:
The octagon can be built from 6 triangles as shown:

6 * 180 = 1080 so the interior angles of a hexagon add to 1080 degrees.

This time the problem supplies 3 angles and asks for the value of one of the equal sized remaining angles. An octagon has 8 interior angles so 8-3 = 5 meaning there are 5 equal sized angles plus the three given that add to 1080 degrees.

1080 = 25 + 50 + 5 + x + x + x + x + x
1080 = 80 + 5x
1080 - 80 = 80 - 80 + 2x
1000 = 5x
1000/5 = (5x)/5
200 = x