[Question] What is the area of this shape? (geometry question)

[bhamv3]

I saw this question in an advertisement on the train home, and I can't solve it. According to the advert, this can be solved by a fifth grader.

Take a regular octagon, all sides and all angles are equal. Each side of the octagon is 30cm long. Since it's a regular octagon, each of the eight angles is 135°. What is the area of the blue triangle?

This advert made me feel really stupid.

 

[drifter]

~1098 cm^2, I hope.

 

[Azurephoenix]

The answer is 1086.36 cm^2

Had to brush up on my trigonometry to figure this out. Of course I way overcomplicated it on my first attempt and realized a much faster and easier way to do it after wasting a lot of time.

 

[drifter]

Hmm, I seem pretty off. Did you get a 67.5/67.5/45 isosceles? Curious if my reasoning is bad, my math is bad, or both.

-edit-

Never mind, I recalculated. Prolly made a rounding error.

bhamv3, here's a hint:

Spoiler

It's all about congruent angles. Also, probably easier to visualize if the octagon is laying flat on one side instead of at an angle.

 

[Azurephoenix]

Your angles are indeed correct... not sure where the slight variance is from though.​

(I checked my answer by drawing the octagon in AutoCAD and measuring the triangle there so I'm quite confident with it).

 

[GasBandit]

Since when do they teach trigonometry in 5th grade? I was ahead of the curve and even I had to take that shit in 10th.

Hell, in 5th grade we hadn't even introduced variables (the most basic part of algebra).

 

[bhamv3]

... I still can't get it.

 

[strawman]

... I still can't get it.

You only need to know the length of one side of the blue triangle, and at least two angles for the blue triangle.

The length of the lower right side of the blue triangle isn't hard to get. It's part of another triangle with one 135 degree angle and two 30 cm sides, just to the right of it. So now you can get the length of that side.

now look at the bottom angle of the blue triangle. It's a portion of the 135 degree angle that comprises the bottom of the whole figure. Since that vertical line bisects the figure, then the blue triangle and little white triangle we just used are half of 135, or 67.5 degrees. Since we just used the little white triangle and have all the side lengths and angles for it, we know that it takes 22.5 degrees of the 67.5 degrees we have available for the bottom right half of the figure. This leave the bottom of the blue triangle with an angle of 45 degrees.

The top right line of the blue triangle is parallel to two of the faces of the figure. Let's look at the top right face. We know that it has an angle of 67.5 degrees from the vertical line. Since the top line of the blue triangle is parallel, then we know that it has an angle of 67.5 from the left side of the blue triangle.

We now know how to get the length of one side, and two angles of the blue triangle. The rest is relatively simple trigonometry.

 

[bhamv3]

Hmm, I see what you mean. But could there be an easier way to solve it, a method suitable for fifth graders?

 

[Azurephoenix]

I really don't think this question is appropriate for fifth graders. I seem to remember learning trigonometry way later than grade 5 (but I am getting doddery in my old age).

 

[GasBandit]

Yeah, I didn't even get basic geometry until 9th grade.

 

[drifter]

I'm not sure how difficult it would be for fifth graders to learn, it's just basic arithmetic comes first.

That said, I don't recall doing any trig until sophomore year in high school, and I don't think I really got exposed to trig identities until college.

 

[GasBandit]

I was on the AP track, and we did-
8th grade: Algebra
9th grade: Geometry
10th grade: Trigonometry
11th grade: Alg II
12th grade: Calculus

I don't remember what we learned math-wise in 5th grade, but googling up 5th grade math worksheets shows decimals, percentages, adding/subtracting fractions, and multiplication of numbers of 4 or more digits.

 

[Gusto]

I had algebra and geometry in grades 7 and 8, trigonometry in 9th.

 

[Azurephoenix]

To really understand principles behind trigonometry effictively (and not just regurgitate memorized formulas) you need basic algebra.

Gas, that looks pretty similar to what I learned grade wise with the exception of calculus being a separate math course option (grade 12 math here was crap like conics and statistics if I recall correctly).

bhamv3, in my opinion this question is way beyond what a 5th grader should be expected to solve.

 

[Ravenpoe]

I was on the AP track, and we did-
8th grade: Algebra
9th grade: Geometry
10th grade: Trigonometry
11th grade: Alg II
12th grade: Calculus

I don't remember what we learned math-wise in 5th grade, but googling up 5th grade math worksheets shows decimals, percentages, adding/subtracting fractions, and multiplication of numbers of 4 or more digits.

The AP track I took in my podunk school in Florida was the same order.

 

[strawman]

You don't need trigonometry, just Pythagorean's theorem. Turn all the non right angle triangles into two right angle triangles, and you can do it without sin.

I hated geometry because we had to show each step of our proofs, but I do recall learning it in sixth grade. Now kids are learning basic geometry in fifth grade, and probably earlier.

I expect that a child at the end of fifth grade could probably solve it with some hints.

 

[Azurephoenix]

Pythagorean's theorem is algebra which is likely not being taught to grade fivers.

 

[GasBandit]

Ok, I figured out how to do it with algebra and geometry alone. You have to make some more cuts in the picture...

By making the two extra slices above, you create isosceles right triangles with hypotenuse 30 and sides of X.

In an isoceles triangle, X+X=sqrt(2)*hypotenuse, because the ratio of the sides is always 1:1:sqrt(2).

Thus, X is 21.2132034356 cm.

Since the slice bisects the blue triangle as well, we can see they are really two halves of a rectangle, which means its area is really (30+x)*x.

51.2132034356*21.2132034356=1086.39610307 cm

 

[bhamv3]

With Gas's solution, basically the most advanced theory needed is Pythagoras, right? Hmm... I suppose it may be plausible that Taiwanese fifth graders would know Pythagoras...

 

[Officer_Charon]

I now feel extraordinarily stupid... it has been FAR too long since I've had to do anything other than basic arithmetic.... I think the last time I had to use Geometry was on the ASVAB... back in 2001.....

 

[GasBandit]

I now feel extraordinarily stupid... it has been FAR too long since I've had to do anything other than basic arithmetic.... I think the last time I had to use Geometry was on the ASVAB... back in 2001.....

Yeah, I had to fire up neurons I forgot I had, too. I haven't had to use a formula or equation since college. Which really puts the two times I had to take calculus into perspective.

 

[GasBandit]

I wish the question we had to solve had this graphic instead:

 

[Squidleybits]

Print out an octagon on a piece of paper.
Measure the sides and pretend that they are 30cm (remember the actual ratio)
Draw the figure.
Measure the base and height of the blue triangle.
Remember the ratio you figured out earlier.
Use formula for area of triangle.

 

[rac3r_x]

Find the area under the curve:

 

[strawman]

 

[Squidleybits]

Just so everyone understands, HCGLNS stole my iPad.

 

[GasBandit]

Find the area under the curve:

d/dx(batman) = Bruce Wayne.