Question

1. What is the area of a triangle whose vertices are D(3, 3) , E(3, −1) , and F(−2, −5) ?

Answer 1

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

Answer 2

Answer: Area of triangle is 10 square units   and perimeter of polygon is 16.87

Explanation:

We have formula for area of triangle when three vertices are given

 $\frac{1}{2}[ (x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

we have values $x_1=3 ,x_2=3 ,x_3=-2\\y_1=3 ,y_2=-1 ,y_3=-5$

substituting the values in the formula we will get

$\frac{1}{2} [3(-1+5)+3(-5-3)+(-2)(3+1)]=\frac{1}{2}[3(4)+3(-8)-2(4)] =\frac{1}{2} [12-24-8]=-10$

Hence, area of triangle ABC is 10 square units we will neglect the negative sign since area can not be negative.

To find the perimeter we will add up all the sides

A(-2,-2) , B(-2,-4), C(1,-6),D(4,-6) ,E(3,-3)

First we need to find  the sides  for that we will be using distance formula below

$\sqrt{(y_2-y_1)^{2}+ (x_2-x_1)^{2}$

Now to find AB $x_1 =-2, x_2=-2\\y_1=-2,y_2=-4$

we will get $\sqrt{(-4+2)^2+(-2+2)^2}= \sqrt{4}=2$

Similarly, we will find BC $x_1=-2,x_2=1\\y_1=-4,y_2=-6$

$\sqrt{(-6+4)^2+(1+2)^2}= \sqrt{13}$

In same manner CD $x_1=1,x_2=4\\y_1=-6,y_2=-6$

 $\sqrt{(-6+6)^2+(4-1)^2}= \sqrt{9}=3$

For DE$x_1=4,x_2=3\\y_1=-6,y_2=-3$

$\sqrt{(-3+6)^2+(3-4)^2}= \sqrt{10}$

For AE$x_=3,x_2=-2\\y_1=-3,y_2=-2$

$\sqrt{(-2+3)^2+(-2-3)^2}=\sqrt{26}$

we will add all sides will give$2+3+\sqrt{13}+\sqrt{10}+ \sqrt{26} =2+3+3.6055+3.1622+5.0990=16.8667$$\approx16.87$