MATH SOLVE

8 months ago

Q:
# 1. What is the area of a triangle whose vertices are D(3, 3) , E(3, −1) , and F(−2, −5) ?2.The coordinates of the vertices of a polygon are (−2,−2), (3,−3), (4,−6), (1,−6), and (−2,−4).What is the perimeter of the polygon to the nearest tenth of a unit?

Accepted Solution

A:

1. 10 square units.
2. 16.9 units
1. The area of a triangle is 1/2bh where b is the base and h is the height of the triangle. Any of the 3 sides of the triangle may be the base. So looking at the 3 points, I'll consider the line segment DE to be the base since both D and E have an X value of 3, so the length of the base is 3 - (-1) = 4. Since the base of the triangle is a vertical line with X = 3, the height of the triangle will be the absolute value of the X value of vertex F minus 3. So abs(-2 - 3) = abs(-5) = 5. We now have a base of 4 and height of 5 and using the 0.5bh formula, that gives us 0.5 * 4 * 5 = 10.
2. I'll call the points A(-2,-2), B(3,-3), C(4,-6), D(1,-6), and E(-2,-4). Using the pythagorean theorem, we can calculate the length of each side. SO
length AB = sqrt((-2 - 3)^2 + (-2 - (-3))^2) = sqrt(-5^2 + 1^2) = sqrt(25 + 1) = sqrt(26) = 5.099
length BC = sqrt((3 - 4)^2 + (-3 - (-6))^2) = sqrt(-1^2 + 3^2) = sqrt(1 + 9) = sqrt(10) = 3.162
Do the same for the lengths of CD, DE, and EA getting 3.000, 3.606, and 2 respectively.
Now just add them together. So
5.099 + 3.162 + 3.000 + 3.606 + 2.000 = 16.867, which rounds to 16.9