In this problem sequence, you will learn some of the geometry of the three triangles that you are likely to encounter most often: the equilateral triangle, half of it, and the half-square.

Measuring their angles and sides

1. The dotted line is a line of symmetry--the gray and white triangles precisely match each other and are each half of square DEF G. Tell how that one fact lets you figure out the measures of F DE, DEF , and EF D (even if you did not already know those angles).
   Well, you do need one more fact: the measure of the angles in a square!
   
2. Suppose that you don’t know the angles in ABC, but you do know that all three angles match. How does that one fact tell you how big those angles must be?
   Well, you do need one more fact: that the sum of the angles in any triangle is the same as the sum of the angles in any other triangle.
   
3. If you know that BAH precisely matches CAH (because they’re each half of ABC), how does that tell you the measures of BAH, AHB, and HBA?
4. Using the following measurements, how long is side HB? What fact given earlier lets you figure out that length?
   
5. Whenever you know the lengths of two sides of a right triangle, the Pythagorean theorem lets you figure out the length of the third side. How long are sides AH and DF ?

   Getting to know them well

   It is worth remembering the two numbers you just found, and knowing how they describe the geometry of the triangles. Here are some problems to help you do that.

   Equilateral triangles, isosceles right triangles, and 30^o-60^o-90^o triangles turn up quite often in real-life problems and also in school-life tests.
6. ABC is equilateral, and ADB is a right angle.
   
   1. If DB = 1, how long are AD and AB?
   2. If AB = x, how long are AD and DB?
   3. If DC = x, how long are AD and AC?
   4. If AD = 5, how long are DB and AB?
   5. For any size of AB, what is the ratio of AD to DB? And what is the ratio of AB to DB?
7. This picture shows a regular hexagon ABCDEF carved up in two different ways. On the left, you see ABCDEF made up of six equilateral triangles. On the right, you see it composed of one large equilateral triangle colored black, and three matching isosceles triangles shaded gray.
   
   The picture on the right has the annoying habit of sometimes looking like a gray cube with the front corner cut off in a special way, and the cut surface painted black. Other times it looks like the gray hexagon it was intended to be, with a black triangle in it.
   Sometimes, the picture on the left insists on becoming a cube, too!
   1. From one of the pictures, you can deduce that each interior angle of the regular hexagon--ABC, BCD, and so on--has the same measure. What is that measure? How can that be deduced from the picture?
   2. Figure out the angles in ABC (right-hand picture) and explain how you did it.
   3. If AC = inches long, how long are AB and BE?
8. Three squares of Origami paper are piled one on top of the other. The corners of the smallest square touch the midpoints of the sides of the black square. The black square fits on the largest square the same way.
   
   1. If the largest square measures 10 cm on the side, what are the lengths of the sides of the other two squares?
   2. If the smallest square measures 5 cm along its side, what is the length of its diagonal? What is the length of the largest square’s diagonal?

Hint to problem 2. The three angles are the same. Their sum is 180^o.

Hint to problem 8. The diagonals of the black square match the sides of the largest square.

Measuring their angles and sides

1. See solutions.
2. See solutions.
3. See solutions.
4. HB = inch.
5. AH = (That can also be written as AH = .)
   DF =

   Getting to know them well

6. 1. If DB = 1, then AB = 2 and AD = .
      
   2. If AB = x, DB = and AD = .
   3. If DC = x, AC = 2x and AD = x.
   4. If AD = 5, then DB = = and
      AB = = .
   5. For any size of AB, = and = 2.
      These ratios come up often and are worth remembering.
7. 1. The lefthand picture shows that each interior angle of the regular hexagon is composed of two 60^o angles, so mABC = mBCD = mCDE = ... = 120^o.
      
   2. mABC = 120^o, leaving 60^o for the sum of angles BAC and BCA. Symmetry shows that those two angles are congruent, so mBAC = mBCA = 30^o.

      Another way of computing the size of BAC is to notice that BAC + EAC + EAC = 30^o.

   3. AB is 1'' long, and BE is 2'' long.
8. 1. If the largest square’s side measures 10 cm, then the black square’s side is , or 5 and the smallest square’s side is , or 5, half that of the largest square.
      Remember!  
      = .
      
   2. If the smallest square’s side is 5 cm, its diagonal is 5 and the largest square’s diagonal is 10.

Measuring their angles and sides

1. A square’s corners are right angles, so mDEF = 90^o. Because the dotted line is a line of symmetry, F DE and F DG match, so each must be half of the right angle. That makes mF DE = 45^o. The same is true of EF D and GF D, which means that EF D also measures 45^o.
   
2. The sum of the angles in any triangle is 180^o. The three angles in ABC are all the same, so each must be 60^o.
   
3. You already know that mHBA = 60^o. Angles BAH and CAH are each half of BAC, so mBAH = 30^o. There are two ways to see that angle AHB measures 90^o. The sum of interior angles in a triangle is 180^o, and the sum of BAH and HBA is 90^o, leaving 90^o for AHB. Or you could argue that mCHB = 180^o and the line of symmetry splits it equally, making mAHB = 90^o.
4. All sides of the equilateral triangle are equal. Because the HB is half of side CB, HB = inch.
5. AH² = AB² - BH² = 1 - ()² = so
   AH = = = .

   DF ² = DE² + EF ² = 2 so
   DF = .

   Getting to know them well

6. AB² = AD² + DB² and AC² = AD² + DC².
   Because ADB is a right angle, we also know that ABD is a 30^o-60^o-90^o triangle, so DB = AB = CB = DC.
   
   1. If DB = 1, then CB = 2 so AB = 2.
      By the Pythagorean theorem, AD = .
   2. If AB = x, DB = and AD = .
   3. If DC = x, AC = 2x and AD = x.
   4. Part (c) shows that if DB = x, then AD = x, so if AD = 5, then DB = .   AB = 2DB, so AB = .
      If you care about “rationalizing the denominator,” then DB = and AB = .
   5. Part (c) shows that if DB = x, then AD = x, so for any size of AB, = = . Also, for any size of AB, = 2.
      Because these results apply for all 30^o-60^o-90^o triangles, and because these triangles are important, these ratios are worth remembering.
7.  
   
   1. The picture on the left shows that each interior angle of the regular hexagon is composed of two 60^o angles, so mABC = mBCD = mCDE = ... = 120^o.
   2. mABC = 120^o, leaving 60^o for the sum of angles BAC and BCA. Symmetry shows that those two angles are congruent, so mBAC = mBCA = 30^o.

      Another way of computing the size of BAC is to notice that BAC + EAC + EAC = 30^o.

   3. Looking at both pictures shows that AC bisects the two equilateral triangles, ABO and BCO, and is twice the altitude of those triangles. That altitude must therefore be = inches. That makes AB 1'' long, and BE 2'' long.
8. The key feature of the picture is that the diagonal of each square is the same length as the side of the next larger square on which it lies.
   Remember!   = .
   
   
   
   1. If the largest square’s side measures 10 cm, so does the black square’s diagonal. Because = , the black square’s side (and smallest square’s diagonal!) is , or 5. The same reasoning shows that the smallest square’s side is , or 5, half that of the largest square.
   2. This problem just reverses the reasoning. If the smallest square measures 5 cm along its side, its diagonal is 5. The side of the largest square is 10, so its diagonal is 10.