Download Right Triangles - Math Talent Quest

Right Triangles
Math Talent Quest
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Material
Though there are only a few things you need to know about right triangles (which you probably already
know anyway), they come up in so many competitions. Essentially, there is only one thing you really need
to know to handle these:
The Pythagorean Theorem: A right triangle with legs of length a and b and hypotenuse c satisfies
a2 + b2 = c2 . You probably already know this. We encourage you to look up all of the different, creative
proofs of this theorem online.
Another result of the Pythagorean Theorem that recurs very frequently, are two special types of right triangles. We call them the 30-60-90 and the 45-45-90, as these are the angles of the two types of triangles.
Until you learn trigonometry, there is no nice way to convert angles in a diagram to side lengths, unless you
have these special angles that are multiples of 15 degrees (though 15 and 75 can be a bit trickier). Thus,
when you see an angle that is a multiple of 15 degrees, you should try to drop altitudes to create 30-60-90
or 45-45-90 right triangles.
√
30-60-90 triangles: Has side lengths in the ratio of 1 to 3 to√2.
45-45-90 triangles: Has side lengths in the ratio of 1 to 1 to 2.
30◦
√
45◦
2
3
1
√
2
60◦
45◦
1
1
30-60-90
45-45-90
Of course, these triangles don’t have to have these exact side lengths. However, once you know one side
length, you can use these ratios to compute the others. Here are a few practice problems:
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Example Problems
√
1. Triangle ABC has ∠A = 75◦ , ∠B = 45◦ , and ∠C = 60◦ . If AC has length of 2 6, find the area of
triangle ABC.
Since the angles are multiples of 15, we think about using 30-60-90 triangles and 45-45-90 triangles.
The only problem is that angle A is 75 degrees, which is part of neither triangle. However, we can
break up the triangle by dropping altitude AD onto BC.
A
B
D
C
Now we get two triangles, one which is a 30-60-90 triangle and the other a 45-45-90 triangle. Now we
can use the√triangles ratios√ to √
compute
√ the other side lengths. Since ∆ACD is a 30-60-90, we have
that CD = 6 and √
AD = 6 · 3 = 3 2. Now we look at triangle ABD, a√45-45-90
We have
√ triangle.
√
√
that BD = AD = 3 2, so the area of triangle ABC is 12 · AD · BC = 12 · 3 2 · (3 2 + 6) = 9 + 3 3.
2. A circular table is pushed into the corner of a square room so that a point P on the edge of the table
(between the two walls) is 9 inches from one wall and 8 inches from the other wall. Find the radius of
the table, in inches.
Because we have a right angle in the corner, and lines tangent to a circle, this motivates us to use
Pythagorean Theorem to make use of these right angles. However, we only have two lengths, and they
don’t look like they can be used as part of a right triangle. Since we’re solving for the radius of the
table, we let the radius of the circle be r. Then we notice that we can use the lengths of 8 and 9 as
part of the sides of the right triangle shown below:
r−9
r
r−8
9
8
We can now write the following equation with Pythagorean Theorem:
2
(r − 8)2 + (r − 9)2 = r2
r2 − 16r + 64 + r2 − 18r + 81 = r2
r2 − 34r + 145 = 0
(r − 29)(r − 5) = 0
(If you aren’t familiar with how we solved for r in the equation above, we recommend that you read
about quadratics. The important part is to understand why we do certain steps, since the computation is easier.) Thus, the radius is either 29 or 5 inches. However, since there is a point on the circle
that is 9 inches away from the wall, the radius cannot be 5 inches, so the answer is 29 inches. Since
most computational geometry problems from contests use the Pythagorean Theorem in some way, you
should try to look for a way to use it. Setting variables often helps.
3. In isosceles trapezoid ABCD, BC = 30, AC = 40, and DC = 50. Find the area of the trapezoid.
The lengths of 30, 40, and 50 are the side lengths of a right triangle, so they motivate us to look for
a right triangle. Since the trapezoid is isosceles, we have that AC = BD = 40, so now we have right
triangle BCD.
A
B
30
40
D
50
E
C
To find the area of the trapezoid, it would be good to have the lengths of the two bases, and the height.
We have one of the bases. What we can do is drop an altitude from point B to side CD, with the foot
being point E. We can use similar triangles to find the height BE of the trapezoid. We can also notice
that the area of ∆BCD can be expressed in two different ways:
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1
BC · BD = CD · BE
2
2
30 · 40 = 50BE
Thus, we have that
√ BE = 24. Using the Pythagorean Theorem (or similar triangles if you wish), we
find that EC = BC 2 − BE 2 = 18. Hence, we have that the smaller base AB = CD − 2 × CE =
50 − 36 = 14. Now we can find the area:
Area =
1
(14 + 50)(24) = 768
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While this problem itself doesn’t require us to use Pythagorean Theorem to compute any lengths (since
they can be done with similar triangles), the important part is recognizing that 30, 40, and 50 form a
special triplet - namely, the sides of a right triangle! If you encounter such numbers in a math problem,
you should be on the look out to make them the sides of a triangle.
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