Question

Find the area of the following shape. You must show all work to receive credit.

Answer 1

All right, so I know I have to chop this up into a bunch of triangles.

Answer 2

You need to break up this odd-shaped area into shapes whose areas are easy to find: triangles, trapezoids, etc., and then to find the areas of these smaller shapes, and then to sum up your results. Take an educated guess: What's your first sub-area going to look like, and what's its area?

Answer 3

Ignore FG please.

Answer 4

I'm pretty sure I have to cut this up into a bunch of triangles. It looks like AFB is the easiest area to find right now?

Answer 5

Since the height is easy to find for the formula (2 units square).

Answer 6

and the base is 6. So to find the area, it would be 2(6)/2 --> 12/2 --> 6. So the area of that subsection is 6, right?

Answer 7

Here's a case where you'd learn the most from experimentation. So I won't say what I think of your choice of triangle AFB.

Answer 8

I'd prefer you write out the area formula for triangles: A = (1/2) bh, or A=bh/2.

Answer 9

does your first area stem from that formula? Try it.

Answer 10

However, now I'm stuck on the rest of the triangles because they don't have even bases (they're all slanted). xD

Answer 11

a=height, b=base.

Answer 12

If you feel stuck, then now's the time to experiment with other possible sub-areas.

Answer 13

Okay, so the formula for area of the triangle is 1/2 (b)(h). The base is 6 and the height is 2. So, the area would be 1(6)(2)/2, which gives us 12/2. Then we would have to divide 12 by 2, which leaves us with 6.

Answer 14

Maybe I should divide these into trapezoids instead? The formula for that is 1/2 (b1 + b2)h.

Answer 15

The only problem with that is the height is not the same because it has 2 different legs...

Answer 16

You definitely have one or more trap..s here; feel free!

Answer 17

That's not a problem...that's a true trapezoid! ;)

Answer 18

Can I use this as the height?

Answer 19

And this too? Can this be used as a height as well?

Answer 20

Here; this is a little neater. My thought processes tend to be a little jumpy whenever I'm in the middle of something. :)

Answer 21

So, if we use the trapezoid formula, we would get 1/2(b1 + b2)h. We can count the units in the top trapezoid, and then we have to substitute the numbers we get into the formula: 1/2 (6 + 6)(6). Am I doing this correct, professor? :)

Answer 22

Going from there, I would have 12(6)(1)/2 --> 72/2 --> 36. So, the area of the top trapezoid would be equal to 36 units squared?

Answer 23

And then we do the same for the second figure on the bottom, correct?

Answer 24

So in the second trapezoid, B1 is 6, b2 is 6, and the height is 3?

Answer 25

Note that you can use the formula for the area of a trapezoid only if the figure has two parallel sides. Do your trapezoids satisfy that critierion?

Answer 26

No, it definitely doesn't. That was why I was trying to go with the triangles in the first place.

Answer 27

K: Afraid this conversation is getting to be so long that it's hard to follow. Based upon your previous efforts, would you go with trapezoids or triangles or both if you were to start over?

Could you start with the original figure and modify it to show your first and second triangles or trapezoids (or triangle and trapezoid), labeling every vertex that is not already labeled?

Answer 28

Which would be your first triangle, and why?

Answer 29

I would make BCD my first triangle, but want to know your first choice.

Answer 30

Okay, hold on. So, we start with this figure, right?

Answer 31

But the shapes you've created are neither triangles nor trapezoids. Mind explaining your choice, or (better yet) moving on to a different subdivision that creates either triangles or traps or both? Traps MUST have 2 parallel sides.

Answer 32

So, something like this is better?

Answer 33

Consider triange BCD first. How would you subdivide the remaining area?

Answer 34

I would either create a line segment FB or I would add point G at (1, -4) and connect it to point F, creating a rectangle?

Answer 35

I believe your latest subdivision would 'work.' The acid test is whether or not you can easily find the areas of each triangle you've defined. Try it. If you can find all these sep. areas and then their sum, you're done. if not, then you might want to try a different subdiv. YOU have the power already to choose subdivisions that are easiest for you...you do not need my approval.

Answer 36

I don't care for your AFGB shape, because it is neither triangle nor trapezoid. However, your AFGB shape could be subdivided into one triangle and one trap.

Please make your own choice...don't wait for my "ok." Then try to find the areas of each of your interior triangles and / or trapezoids.

Answer 37

Start once more with the original diagram. Construct 1 or 2 triangles and/or traps. YOU yourself must decide how hard it would be for you to find their areas, as well as how hard to construct additional traps. or triangles. Don't ask for my opinion; instead, tell me about your results....were you able to find the areas?

Answer 38

Question, for triangle BCD, I can use the x-axis to measure height, correct?

Answer 39

If you were easily able to find the areas of your triangles/traps, and if the remaining areas are easily subdiv. into triangles and traps, fine. If not, change your subdivision.

Answer 40

Regarding BCD: Yes, vertex D is on the x-axis. The area of BCD is easily found. Want to try finding this area?

Answer 41

Okay, I've found the areas of the triangles so far:

Formula: A=1/2bh
TriangleBCD: 8(2)/2 = 16/2 = 8
TriangleFGB: 4(4)/2 = 16/2 = 8
Triangle AFB = 6(2)/2 = 12/2 = 6.

Answer 42

Trapezoid Area Formula: 1/2(b1 + b2)h
Trapezoid FEDG: 1/2(6+5)(6) = 1(11)(6)/2 = 66/2 = 33.

Answer 43

And this would mean our area is 55 square units total, if the way I separated this was correct.

Answer 44

I agree with your result for Triangle BCD: 8 square units. I have not done the other area calculations.

If you have chosen vertices of your triangles and traps that coincide exactly with intersections of the faint horiz. and vertical lines on the given graph, and have found only integer base and height measurements for each, then it's very likely that your end result is correct. Have you any doubts? If so, what are they?

Again, I wanted you to subdivide the given area in to trap-s and tri-s whose areas would be easiest for YOU to calculate.

Thanks for your persistence.

Answer 45

I'm fairly confident in my answer. The only one I'm not completely certain about is the height of the trapezoid since it's slightly slanted on the bases, but even that I'm pretty confident about. (About 95% certain I did even though correct.)

Thank you for sticking with me, professor! :)

Answer 46

Regarding the trap: Two sides of the figure MUST be parallel. Does your trap satisfy that criterion? Your formula for the area of a trap was correct:\[A _{trap}=\frac{ l _{1}+l ^{_{2}} }{ 2 }*h\]

Answer 47

where l1 and l2 are the (different) lengths of the parallel sides.

Answer 48

Yes, then it definitely satisfies the conditions! Thank you very much! =)

Answer 49

My pleasure. Best to you, Katherine!