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`