Big Ideas Math Geometry, 2014
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Big Ideas Math Geometry, 2014 View details
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Exercise 17 Page 486

Practice makes perfect
a

Special right triangles are 45^(∘)-45^(∘)-90^(∘) triangles and 30^(∘)-60^(∘)-90^(∘) triangles. Let's identify them one by one.

45^(∘)-45^(∘)-90^(∘) Triangles

We will start by identifying all 45^(∘)-45^(∘)-90^(∘) triangles. Let's consider the given triangles.

Recall that a 45^(∘)-45^(∘)-90^(∘) triangle is an isosceles right triangle. We can see that Triangle A is an isosceles right triangle, so it is a 45^(∘)-45^(∘)-90^(∘) triangle. At the same time, we can see that Triangles B and D cannot be isosceles right triangles, since their legs are different lengths.

To decide whether or not Triangles C and E are 45^(∘)-45^(∘)-90^(∘) triangles, we have to determine the remaining lengths of the legs of these triangles. To do so we will use the Pythagorean Theorem.

Triangle Pythagorean Theorem The Length of the Leg
C 6^2+b_C^2=(6sqrt(2))^2 b_C=6
E 5^2+b_E^2=10^2 b_E=5sqrt(3)

We can see that Triangle C is an isosceles right triangle, so it is a 45^(∘)-45^(∘)-90^(∘) triangle.

30^(∘)-60^(∘)-90^(∘) Triangles

Now we will identify all 30^(∘)-60^(∘)-90^(∘) triangles. If the shorter leg of a 30^(∘)-60^(∘)-90^(∘) triangle has the length x, the longer leg will have the length xsqrt(3) and the length of the hypotenuse will be 2x. Let's check if any of the Triangles B, D, and E meets these conditions. We will start with Triangle B.

The shorter leg of Triangle B is 3 and the longer leg is 3sqrt(3). We will check if the hypotenuse measures 2*3=6. To do it, let's use the Pythagorean Theorem.

3^2+(3sqrt(3))^2=c_B^2
â–¼
Solve for c_B
c_B^2=3^2+(3sqrt(3))^2
c_B^2=9+27
c_B^2=36
c_B=sqrt(36)
c_B=6

The hypotenuse has the desired length of 6. Therefore, Triangle B is a 30^(∘)-60^(∘)-90^(∘) triangle. Next we will take a look at Triangle D.

The shorter leg of Triangle D is 3 and the longer leg is 4. Since the length of the longer leg is not equal to the length of the shorter leg times sqrt(3), Triangle D is not a 30^(∘)-60^(∘)-90^(∘) triangle. Finally let's consider Triangle E.

We can see that the shorter leg of Triangle E is 5, its hypotenuse is 10=2* 5, and earlier we determined that the longer leg is 5sqrt(3). Thus, Triangle E is a 30^(∘)-60^(∘)-90^(∘) triangle.

Conclusion

We determined that Triangles A and C are 45^(∘)-45^(∘)-90^(∘) triangles, and that Triangles B and E are 30^(∘)-60^(∘)-90^(∘) triangles. In conclusion, Triangles A, B, C, and E are special right triangles.

b

Note that all 45^(∘)-45^(∘)-90^(∘) triangles are similar by the Angle-Angle (AA) Similarity Theorem. This is also true for 30^(∘)-60^(∘)-90^(∘) triangles — any two 30^(∘)-60^(∘)-90^(∘) triangles are similar. In Part A we determined which of the given triangles are 45^(∘)-45^(∘)-90^(∘) triangles and which are 30^(∘)-60 -90^(∘) triangles.

45^(∘)-45^(∘)-90^(∘)triangles: AandC 30^(∘)-60^(∘)-90^(∘)triangles: BandE Therefore, Triangle A is similar to Triangle C, and Triangle B is similar to Triangle E. Since Triangle D is not a 45^(∘)-45^(∘)-90^(∘) triangle nor a 30^(∘)-60^(∘)-90^(∘) triangle, it is not similar to any of the other given triangles.
c

We are asked to find the lengths of the altitudes of Triangles B and C. Let's do it one at a time.

Triangle B

Recall that in Part A we identified Triangle B as a 30^(∘)-60^(∘)-90^(∘) triangle. We know that the shorter leg of Triangle B is 3, its longer leg is 3sqrt(3), and its hypotenuse is 6. Now, let's find the altitude h_B of triangle B.

To find the altitude, we will use the Right Triangle Similarity Theorem.

Right Triangle Similarity Theorem

If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and to each other.

This theorem tells us that the altitude h_B of Triangle B divides it into two 30^(∘)-60^(∘)-90^(∘) triangles.

Note that h_B is also the shorter leg of a 30^(∘)-60^(∘)-90^(∘) triangle whose hypotenuse is 3sqrt(3). By the 30^(∘)-60^(∘)-90^(∘) Triangle Theorem, the length of the hypotenuse of a 30^(∘)-60^(∘)-90^(∘) triangle is two times the length of the shorter leg of the triangle. We will use this information to find h_B.

hypotenuse=2*shorter leg
3sqrt(3)=2* h_B
â–¼
Solve for h_B
2* h_B=3sqrt(3)
h_B=3sqrt(3)/2

The altitude of Triangle B is 3sqrt(3)2.

Triangle C

In Part A we identified Triangle C as a 45^(∘)-45^(∘)-90^(∘) triangle. Each of the legs of Triangle C is 6, and its hypotenuse is 6sqrt(2). Now we will find the altitude h_C of Triangle C.

By the Right Triangle Similarity Theorem, the altitude h_C divides Triangle C into two 45^(∘)-45^(∘)-90^(∘) triangles.

We can see that h_C is one of the legs of a 45^(∘)-45^(∘)-90^(∘) triangle whose hypotenuse is 6. By the 45^(∘)-45^(∘)-90^(∘) Triangle Theorem, the length of the hypotenuse of a 45^(∘)-45^(∘)-90^(∘) triangle is sqrt(2) times the length of each leg. Let's use this information to find h_C.

hypotenuse=leg*sqrt(2)
6= h_C*sqrt(2)
â–¼
Solve for h_C
h_C*sqrt(2)=6
h_C=6/sqrt(2)
h_C=6sqrt(2)/sqrt(2)*sqrt(2)
h_C=6sqrt(2)/(sqrt(2))^2
h_C=6sqrt(2)/2
h_C=3sqrt(2)

The altitude of Triangle C is 3sqrt(2).