This document discusses using shortcuts and properties of right triangles to solve for unknown side lengths. It explains that right isosceles triangles have two congruent legs and a 45-45-90 angle relationship. The shortcut for these triangles is that the hypotenuse is equal to one leg times the square root of 2. It also discusses 30-60-90 triangles having a short leg opposite the 30 degree angle, a hypotenuse twice as long as the short leg, and a long leg equal to the short leg times the square root of 3. Several examples are provided to demonstrate using these properties and shortcuts to find missing side lengths of right triangles.
2. These triangles have two congruent
legs which identifies the triangles as
isoceles.
Also, if you are given that one acute
angle is 45o then the other acute angle
is also 45o.
You can use Pythagorean Theorem to
solve any right isosceles triangle.
a2 + a2 = c2
2a2 = c2
√2 a2 =√ c2 so a√ 2 = c. This gives
us the shortcuts to use.
3. Example 1: Find the length of the diagonal
of a square with a side length of 21.
The diagonal is the hypotenuse of the right
isosceles triangle. You could use
Pythagorean theorem as shown on the
21 previous slide.
Or you could use the shortcut which says
the hypotenuse is the leg length x √2.
21
That means the hypotenuse here is
21√2.
4. Example 2: Given the hypotenuse of a right
isosceles triangle is 14, find the length of one leg.
You can solve this with
Pythagorean theorem:
a2 + a2 = 142
2a2 = 196
a2 = 98
14
a = √98 or 7√2
a
a
5. The 30-60-90 triangles also have a short-cut set of
relationships.
The shortest side is opposite the 300 angle. That side
length is denoted “a”. Then the hypotenuse is
double of the short leg, so it’s length is 2a.
The longer leg is a√3 in length.
We know this works because we can verify it by Pythagorean theorem:
a2 + (a√3)2 = (2a)2
a2 + 3a2 = 4a2
4a2 = 4a2
6. Example 3: Find the lengths of the other
two sides of the triangle if the
x short leg is 7 units.
7
30o
y
The x represents the hypotenuse so it is double of the short side, or 14 units.
The y represents the long leg, so its length is 7√3 units.
7. Example 4: Find the lengths of the other
two sides of the triangle if the
20 hypotenuse is 20 units.
x
30o
y
The x represents the short leg which is half of the hypotenuse, so it is 10 units.
The y represents the long leg, so its length is 10√3 units.
8. Example 5: Find the lengths of the other
two sides of the triangle if the
y long leg is 15 units.
x
30o
15 The easiest way to solve this problem is
Pythagorean theorem. What we know is that y = 2x because the hypotenuse
is twice as long as the short leg. X2 + 152 = (2x)2
x2 + 225 = 4x2
225 = 3x2
75 = x2
5√3 = x so y = 10√3
