\triangle GDC has a base \overline{DC} of 20 and a height \overline{GD} of 10. Its area is therefore 100.
\triangle OFC has a base of 10 and has identical angles with \triangle GDC for the reasons previously stated. Based on the relative base sizes, \triangle GDC is a 2x scaled up version of \triangle OFC. Therefore the height of \triangle OFC must be 5.
Part of my thinking in reviewing early material was that it might help with that, but Iâm not so sure now. Maybe I should just try it.
Right, I think thatâs basically the same way you did it earlier in this post:
In that solution, you used the lengths of the hypotenuses (GC and OC) to figure out that GDC was 2x scaled up version of OFC. But you could have just used the bases of the triangles (DC and FC) instead.
In general, I recommend using the given information over the information that you figured out yourself, if you have a choice â that reduces possible sources of errors. And, also, I wanted it to be clear there was a solution to this problem that didnât require pythagorean.
I think youâve already passed the addition part of the book, so donât know if the rest will really have much that would be helpful in that.
Also, 2+3 is a basic enough problem that you should be able to figure out different ways to do it without going through a bunch of background material.
You can try to relate this back to the original problem now - you were able to figure out the side of the triangle using similar triangles.
The original problem also had similar triangles, without having to draw any new lines. But in that problem, the focus was on figuring out areas, not line lengths.
AH HA
If the bigger of two similar triangles is a 2x scaled up version of the smaller triangle, then its base and height will be 2 as long as the smaller similar triangle. This means the area will be 4x that of the smaller similar triangle.
The hypotenuse of the bigger similar triangle \triangle LDC is 2x that of the smaller triangle \triangle KFC.
The area of \triangle GDC = 100. \triangle GDC consists of one of the bigger similar triangles - \triangle LDC - and one of the smaller ones - \triangle DLG. Another way of saying this is that it consists of the area of 5 of the smaller similar triangles - one area of a smaller triangle and one area of a larger triangle that is equal to 4 smaller triangles.
Let s stand for the area of the smaller similar triangles.
5s = 100
s = 20
We know that the bigger of the similar triangles, \triangle LDC, is equal to 4s. We now know that the area represented by that triangle is 80. There are four interlocking such triangles, so their total area is 320. The total area of the square is 400, so the difference between the 4 interlocking big triangles and the total area gives us the shaded area. That is 400 - 320 = 80.
I. 2 + 3 = 5. This is the ânormal wayâ I would do it - just add the numbers based on number knowledge.
II. First we break things down into 1, then add.
2 + 3 = (1 + 1) + (1 + 1 + 1).
We can reorder the grouping if we want.
(1 + 1 + 1 + 1 + 1)
We donât need the parentheses anymore.
1 + 1 + 1 + 1 + 1 = 5
III. Number line method. I represent the 2 as a blue box and the 3 as a red box. Their combined values represent a distance on the number line, starting at 0.
IV. Reduce one value 1 at a time, and add 1 to the other, until you get to 0.
2 + 3 = 1 + 4
= 0 + 5
Any number plus 0 is itself so we get 5.
Yeah, that was basically the solution I was thinking of.
It requires figuring out that DLC is similar to FKC, with 2x lengths, which means the area of DLC is 4x the area of FKC. And then you can find the area of GDC, from the information given.
So then with that information, there are a few different ways you can put it together to find the area of the black part.
For figuring out that the triangles are similar in the first place, you can also use the angle rules from chapter 10 to do that. So the way I did it didnât require knowing or assuming they were 90 degree triangles. And, also, this problem would be still be solvable in a similar way if it was a rectangle, not a square. (Which would mean the angles would actually NOT be 90 degree angles). It is a little bit harder to put together, since all the triangles would not be the same, but itâs still possible in a similar way.
Assume \overline{HD} and \overline{BF} are parallel lines.Theyâre going to make the same angles when they intersect \overline{GC}, so \angle DLC and \angle FKC are the same. Since \triangle DLC and \triangle FKC share \angle C, all the angles in the triangles must be the same, and so the triangles are similar (I think you could also say that since the parallel lines also intersect \overline{DC}, that the angles formed by the intersection there must also be the same).
Right, parallel lines will make the same angle with any line they intersect, so you could use that rule for either GC or DC. And angle C is just the same angle in both cases. And once youâve shown 2 angles are the same, you know all 3 have to be the same.
You can also use the angle rules + the rotational symmetry to figure out that the angles at I, J, L, K must all be 90 degree angles. Itâs not necessary to solve the problem, but you can figure it out.
hmm not sure Iâve used that concept much before.
You donât have to think of it in that way.
You can also figure the same thing out because itâs a square, so the different sides have to be congruent with each other.
So the shaded region appears to be a square.
So the internal angles of the square must be 90 degrees. So one example is \angle ILK. That has to be 90 degrees.
So then \angle GLD is the opposite angle to \angle ILK. So it also has to be 90 degrees.
\angle ILG is a supplementary angle to \angle ILK on the same line. So it has to be 90 degrees. And the opposite angle of \angle ILG is \angle KLD, which also has to be 90 degrees.
I meant without starting with that assumption, since itâs not information you were given.
But, like, you can figure out that the triangles FKC and EJB are congruent (so all their angles & sides must be the same), since the outer shape is a square, so each side is identical. Does that make sense?
That makes sense, though Iâm not sure I see where youâre going with that quite yet.
Well my point was that you can figure out that the angles in the middle are 90 degree angles by applying the angle rules from chapter 10 of the book (10.1 and 10.2 specifically, e.g., the rules for parallel lines and transversals, which can also be found online in a lot of places), along with the knowledge that since the shape is a square, the different sides are going to be symmetrical to each other.
Hereâs one way that uses some angle stuff. I think itâs not quite what you had in mind but maybe at least overlaps.
The shaded area is a 4-sided figure. The internal angles must thus add up to 360 degrees by the formula 180(n - 2) where n represents the number of sides. So the sum of the internal angles must be 360 degrees.
Because ABCD is a square, it must have identical sides. So \triangle FKC is congruent to \triangle EJB and the other small triangles around the square.
The four small triangles have opposite angles inside the shaded area (e.g. \angle JKL is opposite to \angle FKC). These opposite angles inside the shaded area must be the same as their counterparts within the triangles. Since we know there are four triangles and they are all congruent, the value of these opposite angles inside the shaded area, of which there are four, must be the same as each other. Let a stand for one of these angles. There are four such angles inside the shaded area, so 4a = 360, or a = 90.
Yes, this works and has some overlap with the way I thought of it.
The way I was thinking of was similar, but uses the rules for parallel lines and transversals, instead of the rule for the sum of the interior angles of a polygon.
For my way, you should also be able to say why/how you know the lines are parallel.
There are a lot of things that look true in the picture: the lines look parallel, the shaded shape looks like a square, the angles at I, J, L, and K look like they are all 90 degrees.
But none of that information is actually given to you in the original problem. And if the problem was with a rectangle with a very similar length and width (e.g., 19 units wide and 20 units long), those things would all still look true. So you canât just rely on how it looks.
The fact that they tell you that ABCD is a square does let you figure all the other stuff out though, in multiple different ways.
