Let's take a look at right triangles and some of their special properties today. As a quick refresher, a right triangle is a triangle with a right angle. We can call the side opposite the right angle as the
hypotenuse, and the two sides adjacent to the
90^{\circ} angle as the
legs. In the image to the right,
AC is the hypotenuse, and the two legs are
AB and
BC.
Pythagorean Theorem
One of the most famous and useful theorems in geometry is the Pythagorean Theorem. The theorem states that in right triangle
\triangle ABC with hypotenuse
c and legs
a and
b,
a^2+b^2=c^2
Lets apply this in a problem to see how it works.
Problem 1: Let right triangle \triangle ABC with right angle at B have hypotenuse of r+1, and legs of length 7 and r. Find r.
Solution: We know that the sum of the squares of the legs is equal to the square of the hypotenuse. In other words, we have
7^2 + r^2 = (r+1)^2.
Simplifying this and solving, we have
49+r^2=r^2+2r+1 \Rightarrow 48 = 2r \Rightarrow \boxed{r=24}
.
Special Right Triangles
There are some "special" right triangles whose properties are an absolute must-know in competition math.
1. 45^{\circ}-45^{\circ}-90^{\circ} Triangle, also known as an isosceles right triangle is a triangle where the two legs are congruent, and the non-right angles are both
45^{\circ}. In such triangles, the ratio of the sides is
x - x - x \sqrt{2}, where the hypotenuse is
x\sqrt{2}, and both the legs are
x.
2. 30^{\circ}-60^{\circ}-90^{\circ} Triangle is a right triangle where the hypotenuse is twice the length of the shortest side, which is opposite the
30^{\circ}. In these triangles, the ratio of the side lengths is
x - x \sqrt{3} - 2x, where the
x is opposite the
30^{\circ}, the
x \sqrt{3} is opposite the
60^{\circ}, and the hypotenuse is
2x.
Let's see these concepts in action.
Problem 2: Given that
AB = 1,
BD = BC and
EF = CF in the diagram below, find the length of
CF.
Solution: Just looking at the diagram, we see many 30 - 60 - 90 and 45 - 45 - 90 triangles in this diagram. We can work our way around, starting with triangle \triangle ABC, and eventually get to CF. From AB = 1, we can say that the hypotenuse BC = 2 because we know the hypotenuse is twice the length of the side opposite the 30^{\circ}.
Next, we are given that BD = BC, so \triangle BCD is an isosceles right triangle. Thus, we can says that BC = BD = 2, which leads us to DC = 2 \sqrt{2} because of the side length ratios of a 45-45-90 triangle.
Now, in order to find
DE, we just divide
DC by
\sqrt{3}. Doing this, we get
DE = \frac{CE}{2} = \frac{DC}{\sqrt{3}} = \frac{2 \sqrt{2}}{\sqrt{3}} = \frac{2 \sqrt{6}}{3} \Rightarrow \frac{CE}{2} = \frac{2 \sqrt{6}}{3} \Rightarrow CE = \frac{4 \sqrt{6}}{3}
.
Now, we are almost done. We know that
\triangle CEF is a
45-45-90 triangle, so in order to find
CF, we just need to divide
CE by
\sqrt{2}. Doing this, we get
CF = \frac{CE}{\sqrt{2}} = \frac{\frac{4 \sqrt{6}}{3}}{\sqrt{2}} \Rightarrow CF = \boxed{\frac{4 \sqrt{3}}{3}}
There are some important results that are derived from Pythagorean Theorem. Feel free to reach out to me if you would like to see proofs behind each of these results. If not, I would highly recommend for you to memorize the following for easy application.
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