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The problem didn't give me any angles. I was approaching the angle of 640 and 650 as a right angle to calculate the hypotenuse and thus have an edge, but I don't think it is the most correctenter image description here

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    $\begingroup$ without any angles, I doubt that there is an unique solution. A pentagon is not uniquely determined by its side lengths. $\endgroup$ – person Nov 3 '20 at 17:13
  • $\begingroup$ The angle between 640 and 650 is about $95°$ and even if you knew it , you needed another one, let's say between 132 and 1140. $\endgroup$ – Raffaele Nov 3 '20 at 19:28
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    $\begingroup$ Did your instructor gave you indications for solving this problem ? Did he say that there is an infinite number of solutions ? $\endgroup$ – Jean Marie Nov 4 '20 at 8:48
  • $\begingroup$ See this similar question with the interesting comment that, if two angles are given, then there is a unique solution. $\endgroup$ – Jean Marie Nov 4 '20 at 10:00
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$Area\approx 344,430\text{m}^2$

I used GeoGebra. Imported the picture, drew a polygon over the given one, made a proportion and got the result. There is no other way to solve without angles.

enter image description here

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  • $\begingroup$ I am sorry but I don't agree with this method and the affirmation that "there is no other way to solve (it) without angles..."... $\endgroup$ – Jean Marie Nov 3 '20 at 21:31
  • $\begingroup$ @JeanMarie Suggest something more effective $\endgroup$ – Raffaele Nov 3 '20 at 21:41
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    $\begingroup$ The picture is not drawn to scale. I measure the side lengths of polygon in GeoGebra and they are not porprotional (or even close) to the given side lengths. So even this last resort of way of estimating the area won't work. $\endgroup$ – achille hui Nov 4 '20 at 1:16
  • $\begingroup$ See the method I propose. $\endgroup$ – Jean Marie Nov 4 '20 at 1:39
  • $\begingroup$ @achillehui I added a picture to show my method with GeoGebra. $\endgroup$ – Raffaele Nov 4 '20 at 8:19
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Here is a systematic method, taking into account the fact that, fundamentaly, this issue has an infinite range of solutions.

enter image description here

Let $C(M,R)$ denote the circle with center $M$ and radius $R$.

Consider line $OA$ with length $1140$, which will be the $x$ axis with origin $0$.

Take a point $D$ on (green) circle $C(O,132)$ and a point $B$ on (red) circle $C(A,650)$.

If distance $BD \le 958$, these circles $C(D,318)$ and $C(B,640)$ have a common point that we will take for point $C$.

In this way, all the constraints are fulfilled.

Knowing the coordinates of the vertices of the pentagon, its area can be computed by the shoelace formula. In the case displayed on the figure, one obtains:

$$A=404,473 \ \text{m}^2$$

Remark: I have tried to stick to the appearance of the figure given in the question, but as @Achille Hui rightly remarks, this figure isn't faithful.

Taking different positions for $D$ and especially $B$ would lead to different values of the area: see for example Fig. 2.

Appendix: Points $O,A,B,C,D$ of Fig. 1 have the following (rounded) coordinates:

$$\begin{pmatrix} 0 & 1140 & \ \ \ 812 & \ \ \ 199 & -4 \\ 0& 0 &-561 &-377 &-132\end{pmatrix}$$

Here is another very different "solution", in this case with area $298,151 \ \text{m}^2$:

enter image description here

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    $\begingroup$ I added a picture to show my method with GeoGebra. $\endgroup$ – Raffaele Nov 4 '20 at 8:18

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