mathslover
  • mathslover
Find the area of the pentagon whose vertices are A(1, 1), B(7, 21), C(7, –3), D(12, 2) and E (0, –3).
Mathematics
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SOLVED
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katieb
  • katieb
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mathslover
  • mathslover
@hartnn @.Sam. @amistre64
amistre64
  • amistre64
|dw:1370874673384:dw|
mathslover
  • mathslover
|dw:1370874797801:dw| what is the coordinate of the third point amiste?

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amistre64
  • amistre64
|dw:1370874777508:dw| hmm, it appears that the vertexes are no "in order
mathslover
  • mathslover
Yep.
amistre64
  • amistre64
find the slopes of 2 adjacent line segments; and their midpoints. construct a perp line to each segment to find the center of the pentagon
amistre64
  • amistre64
im thinking of this too geometrically and might be better off using a sum of trapeziods
amistre64
  • amistre64
this might not even be a "regular" pentagon
amistre64
  • amistre64
|dw:1370875038923:dw| move it all up by 3
mathslover
  • mathslover
mathslover
  • mathslover
"Move it all up by 3" , means?
amistre64
  • amistre64
|dw:1370875109068:dw|
amistre64
  • amistre64
it means to move (0,-3) to the origin (0,0) to better view the distances
mathslover
  • mathslover
Oh k.
amistre64
  • amistre64
|dw:1370875252801:dw|
amistre64
  • amistre64
12x24 - 5x5 - 5x21 - 1x4 - 20x1 - 20x6
amistre64
  • amistre64
that 5x21 is off
amistre64
  • amistre64
5x19 is more like it
mathslover
  • mathslover
|dw:1370875389141:dw|
mathslover
  • mathslover
That complete is 24 , and the one length part is 5 , so the other one should be 19. Am I getting something wrong there?
amistre64
  • amistre64
|dw:1370875407273:dw|
mathslover
  • mathslover
Oh k . you wrote it there. 5*19 is more like it.
amistre64
  • amistre64
there are different ways to address it, but moving the lowest leftest point to the origin tends to make life simpler
mathslover
  • mathslover
So it should be : 1*4 + 6*4 + 5*5 + 5*19+ 6*20 , right?
amistre64
  • amistre64
that looks good to me yes
mathslover
  • mathslover
And that is not among the options , I have.
mathslover
  • mathslover
Here are the options : 130/3 135/2 137/3 137/2
amistre64
  • amistre64
check me adding then, and also make sure your post is correctq
amistre64
  • amistre64
A(1, 1), B(7, 21), C(7, –3), D(12, 2) and E (0, –3) +3 +3 +3 +3 +3 ------------------------------------------ 1 4 7 24 7 0 12 5 0 0 7 24 12 5 1 4 0 0 7 0
mathslover
  • mathslover
Adding them gives me, 268 though it is clear that it is a whole number while the options don't have any whole no. . I will say that the question is posted right from my side, not sure of the source from which I got this question, but that is reliable source .
amistre64
  • amistre64
|dw:1370875915752:dw|
ganeshie8
  • ganeshie8
|dw:1370876056936:dw|
ganeshie8
  • ganeshie8
maybe we need to honor the order of vertices
mathslover
  • mathslover
\(\Delta = \cfrac{1}{2} | (x_1 y_2 - x_2 y_1) + (x_2 y_3 - x_3 y_2 ) \\+ (x_3 y_4 - x_4 y_3) + (x_4 y_5 - x_5 y_4 ) + (x_5 y_1 -x_1 y_5) | \)
mathslover
  • mathslover
Using the above formula I get : 137/2 sq. metres. Not sure to prove the formula.
mathslover
  • mathslover
I will get back here after some time. Have to take dinner. Sorry amistre and ganeshi8 to go but I will be back soon. Thanks.
amistre64
  • amistre64
order of verts sounds odd tho :/ but then its worth a try
ganeshie8
  • ganeshie8
giving it to geogebra...
ganeshie8
  • ganeshie8
1 Attachment
amistre64
  • amistre64
\[\int_{0}^{4}4x~dx+\int_{4}^{7}\frac13(10x+2)~dx-\int_{0}^{7}\frac{5}{12}x~dx+\int_{7}^{12}-\frac{7}{12}x+7~dx\]
amistre64
  • amistre64
err, 0 to 1, and 1 to 7\[\int_{0}^{1}4x~dx+\int_{1}^{7}\frac13(10x+2)~dx-\int_{0}^{7}\frac{5}{12}x~dx+\int_{7}^{12}-\frac{7}{12}x+7~dx\] \[148-\int_{0}^{7}\frac{5}{12}(x-7)~dx+\int_{7}^{12}\frac5{12}(x-7)~dx-25\] \[123-\int_{0}^{7}\frac{5}{12}(x-7)~dx+\int_{7}^{12}\frac5{12}(x-7)~dx\]
amistre64
  • amistre64
lol resaw my 12,5 point off that time
amistre64
  • amistre64
\[123-\int_{0}^{7}\frac{5}{12}x~dx+\int_{7}^{12}\frac5{12}x~dx\]
amistre64
  • amistre64
ugh ...i mysterously forgot to divide my traingles by 2
amistre64
  • amistre64
\[86-\int_{0}^{7}\frac{5}{12}x~dx+\int_{7}^{12}\frac5{12}x~dx-\frac{25}{2}\] \[86-\frac{245}{24}+\frac{475}{24}-\frac{25}{2}=\frac{997}{12}\]
mathslover
  • mathslover
Still not right @amistre64 ...
Zarkon
  • Zarkon
if you keep the order of the vertices it looks like it is 137/2
mathslover
  • mathslover
Hmm, well it is not said in the question that the pentagon is ABCDE , it can be anything, right?
mathslover
  • mathslover
@Zarkon , which method you used to get answer as 137/2 ?
Zarkon
  • Zarkon
I'm just going of the answers you provided....keeping the vertices in the original order gives an answer on your list
Zarkon
  • Zarkon
just make 3 triangles
mathslover
  • mathslover
Ok, nice! Well, will that formula also work?
mathslover
  • mathslover
Oh k I think I got it now. Thanks @Zarkon @amistre64 @ganeshie8 .

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