Question

Answer These For Me Please, I’d Really Really Appreciate It So Much :), Appropriate Answers Only!!!

Answer 1

1. Notice that our quadrilateral is EFGH is an isosceles trapezoid, so line segment FE is congruent to line segment GH. Knowing this, we can equate the measures for both sides and solve for $n$:
$FE=GH$
$4n-11=2n+5$
$2n=16$
$n= \frac{16}{2}$
$n=8$
Now wen substituent the value of $n$ in GH:
$GH=2n+5$
$GH=2(8)+5$
$GH=16+5$
$GH=21$
We can conclude that the length of GH is 21 units.

2. To solve this we are going to use the formula: $S=(n-2)180$
where 
$S$ is the sum of the interior angles of the polygon.
$n$ is the number of sides of the polygon.
We know form our problem that the measures of the interior angles of the polygon is 3420°, so the sum if the interior angles of the polygon is 3420°. Lets replace that value in our formula and solve for $n$:
$S=(n-2)180$
$3420=(n-2)180$
$3420=180n-360$
$3060=180n$
$n= \frac{3060}{180}$
$n=17$
We can conclude that the polygon has 17 sides.

3. We can infer from our pictures that both triangles are right triangles, so to find their areas, we are going to find their bases and their heights, and then we are going to use the formula for the area of a triangle: $A= \frac{1}{2}bh$
where
$b$ is the base of the triangle 
$h$ is the height of the triangle 
For triangle ABC:
We can infer from our picture that the base of our triangle is the line segment AC and the height is the line segment AB. Using the distance formula:
$d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
$d_{AC}= \sqrt{(-7--7)^2+(5--1)^2}$
$d_{AC}= \sqrt{(-7+7)^2+(5+1)^2}$
$d_{AC}= \sqrt{6^2}$
$d_{AC}=6$
$b=6$
$d_{AB}= \sqrt{(-4--7)^2+(-1--1)^2}$
$d_{AB}= \sqrt{(-4+7)^2+(-1+1)^2}$
$d_{AB}= \sqrt{3^2}$
$d_{AB}=3$
$h=3$
Now we can find the area of our triangle:
$A= \frac{1}{2} bh$
$A= \frac{1}{2} (6)(3)$
$A=9$ square units 

For triangle DEF:
base=DF and height=FE
Using the distance formula:
$d_{DF}= \sqrt{(2--2)^2+(-3--3)^2}$
$d_{DF}= \sqrt{(2+2)^2(-3+3)^2}$
$d_{DF}= \sqrt{4^2}$
$d_{DF}=4$
$b=4$
$d_{FE}= \sqrt{(2-2)^2+(1--3)^2}$
$d_{FE}= \sqrt{(1+3)^2}$
$d_{FE}= \sqrt{4^2}$
$d_{FE}=4$
$h=4$
Using the are formula:
$A= \frac{1}{2} bh$
$A= \frac{1}{2}(4)(4)$
$A=8$ square units 
We can conclude that the area of triangle ABC is greater than the area of triangle DEF.