+2448 I did try to do this but seriously don't know if what i'm doing is right. Do I have to do pythagorean theorem?
+129852 The perimeter is all solved using the distance formula, RP
The formula for the distance between two points is given by
√ [ (subtract the x coordinates of the two points)^2 + (subtract the y coordinates)^2 ]
So....for instance....the distance between A and B is
√ [ ( -2 - 2)^2 + ( -1 -1)^2 ] = √ [ (-4)^2 + (-2)^2] = √ [16 + 4] = √20 = 2√5
Now..using this example..compute the distances between
A and D
D and C
C and B
Add these four distances....this will be the perimeter
Do this part first....then....I will show you how to calculate the area
Let me know if you run into trouble......
+2448 I got:
2sqrt(5) for A and D
sqrt(3i) for D and C
C and B is sqrt(9) sqrt(5)
+129852 AD is correct !!!
And you got BC correct.... we can simplify it as 3√5
Let's look at DC.....
√ [ ( -4 - (-1) )^2 + ( -5 - (-5) )^2 ] = √ [ ( -3)^2 + ( 0 )^2] = √9 = 3
So.... the perimeter is
AB + AD + DC + BC =
2√5 + 2√5 + 3 + 3√5 =
7√5 + 3 units ≈ 18.65 units
You did pretty well....!!!!
The area is a little tricky, RP.....I'll get back to it in just a bit.....I have a few more questions to get to....just hold on...!!!
+129852 Area....
We have a quadrilateral with two parallel sides AD and BC...so....this is a trapezoid
The area is
(1/2) height of trapezoid * ( sum of the base lengths)
The bases are DC and AB.....we know these
The height is the tricky part....it is found as:
[ The y coordinate of B - The y coordinate of C ] = [ 1 - (-5) ] = [ 1 + 5] = 6 units
So.....the area is
(1/2) (6) ( 3 + 2√5 ) ≈ 22.42 units^2
And that's it....!!!
