Question

1. What is the length of the 2nd base of a trapezoid if the length of one base is 24 and the length of the midsegment is 19? Show your work.
2. What is the sum of the measures of the exterior angles in a heptagon? Explain.



3. Find the measure of each interior angle and each exterior angles of the following regular polygons. Show your work.
a. Decagon

b. Pentagon

c. Dodecagon

d. 16-gon

e. 25-gon

Answer 1

3. Check the first picture.

let

"a" be the number of angles (or sides)
"t" be the number of triangular regions, that the diagonals drawn from one angle divide the regular polygon into.

it is clear that for a polygon of n-sides, there are n-2 triangular regions, each of which has a total measure of angles equal to 180°.

so the sum of the interior angles of a n-gon is (n-2)180°.

In a regular polygon, the measure of each interior angle is equal, so:

the measure of each interior angle of a regular n-gon is:  $\frac{(n-2)180}{n}$

apply the formula to each angle.

a. n=10,   $\frac{(n-2)180}{n}=\frac{(10-2)180}{10}$=(8*180)/10=144
b. n=5,   $\frac{(n-2)180}{n}=\frac{(5-2)180}{5}$=(3*180)/5=108
c. n=12,   $\frac{(n-2)180}{n}=\frac{(12-2)180}{12}$=(10*180)/12=150
d. n=16,   $\frac{(n-2)180}{n}=\frac{(16-2)180}{16}$=(14*180)/16=157.5
e. n=25,   $\frac{(n-2)180}{n}=\frac{(25-2)180}{25}$=(23*180)/25=165.5

2. 

the sum of the interior angles of a heptagon is (7-2)180°=5*180°=900°

let the measures of the interior angles be a,b,c,d,e,f and g.

so a+b+c+d+e+f+g=900°

the exterior angle measures are 180-a, 180-b, 180-c, 180-d, 180-e, 180-f, 180-g, 

so the sum of the measure angles of the exterior angles is 

(180°-a)+(180°-b) +(180°-c)+(180°-d)+(180°-e)+(180°-f)+(180°-g)
= 180°*7-(a+b+c+d+e+f+g)=1260°-900°=360°.

In fact, the sum of the measures of the exterior angles of any polygon is 360°.

1. Let the length of the second base be x

then 
$\frac{x+24}{2}=19$
x+24=38
x=38-24=14 units