4.5: Special Right Triangles
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There are two kinds of right triangle which deserve special attention: the 30∘−60∘−90∘ right triangle and the 45∘−45∘−90∘ right triangle.
30°−60°−90° Right Triangles
A triangle whose angles are 30∘,60∘, and 90∘ is called a 30∘−60∘−90∘ triangle. △ABC in Figure 4.5.1 is a 30∘−60∘−90∘ triangle with side AC=1.


To learn more about this triangle let us draw lines BD and CD as in Figure 4.5.2. △ABC≅△DBC by ASA=ASA so AC=DC=1. △ABD is an equiangular triangle so all the sides must be equal to 2.. Therefore AB=2 (Figure 4.5.3).

Let x=BC. Let us find x. Applying the Pythagorean Theorem to △ABC,
leg2+leg2=hyp212+x2=x21+x2=4x2=3x=√3
Now suppose we are given another 30∘−60∘−90∘ triangle △DEF, with side DF=8 (Figure 4.5.4). △DEF is similar to △ABC of Figure 4.5.3 Therefore
DFAC=DEAB81=DE216=DE and DFAC=EFBC81=EF√38√3=EF

Our conclusions about triangles ABC and DEF suggest the following theorem:
In the 30∘−60∘−90∘ triangle the hypotenuse is always twice as large as the leg opposite the 30∘ angle (the shorter leg). The leg opposite the 60∘ angle (the longer leg) is always equal to the shorter leg times √3.

In Figure 4.5.5, s= shorter leg, L= longer leg, and hyp = hypotenuse. Theorem 4.5.1 says that
hyp=2sL=s√3
Note that the longer leg is always the leg opposite (furthest away from) the 60∘ angle and the shorter leg is always the leg opposite (furthest away from) the 30∘ angle.
Find x and y:
Solution
∠B=180∘−(60∘+90∘)=180∘−150∘=30∘, so △ABC is a 30∘−60∘−90∘ triangle. By theorem 4.5.1,
hyp=2sy=2(7)=14. L=s√3x=7√3
Answer
x=7√3, y=14.
Find x and y:
Solution
∠B=60∘ so △ABC is a 30∘−60∘−90∘ triangle. By Theorem 4.5.1,
L=s√310=x√310√3=x√3√3x=10√3=10√3⋅√3√3=10√33.hyp=2sy=2x=2(10√33)=20√33
Answer
x=10√33,y=20√33.
45°−45°−90° Right Triangles
The second special triangle we will consider is the 45∘−45∘−90∘ triangle. A triangle whose angles are 45∘, 45∘, and 90∘ is called a 45∘−45∘−90∘ triangle or an isosceles right triangle. △ABC in Figure 4.5.6 is a 45∘−45∘−90∘ triangle with side AC=1.

Since ∠A=∠B=45∘, the sides opposite these angles must be equal (Theorem 2.5.2, Section 2.5). Therefore AC=BC=1.

Let x=AB (Figure 4.5.7). By the Pythagorean Theorem,
leg2+leg2=hyp212+12=x21+1=x22=x2√2=x
Find x:
Solution
∠B=180∘−(45∘+90∘)=180∘−135∘=45∘. So △ABC is a 45∘−45∘−90∘ triangle. AC=BC=8 because these sides are opposite equal angles. By the Pythagorean Theorem,
leg2+leg2=hyp282+82=x264+64=x2128=x2x=√128=√64√2=8√2
Answer
x=8√2.
The triangles of Figure 4.5.6 and Example 4.5.3 suggest the following theorem:
In the 45∘−45∘−90∘ triangle the legs are equal and the hypotenuse is equal to either leg times √2.
In Figure 4.5.8, hyp is the hypotenuse and L is the length of each leg. Theorem 4.5.2 says that
hyp=√2L

Find x and y:
Solution
∠B=45∘. So △ABC is an isosceles right triangle and x=y.
x2+y2=42x2+x2=162x2=16x2=8x=√8=√4√2=2√2
Answer: x=y=2√2.
Another method:
△ABC is a 45∘−45∘−90∘ triangle. Hence by Theorem 4.5.2,
hyp=L√24=x√24√2=x√2√2x=4√2=4√2⋅√2√2=4√22=2√2
Answer
x=y=2√2.
Find AB:
Solution
△ADE is a 45∘−45∘−90∘ triangle. Hence
hyp=L√210=x√210√2=x√2√2x=10√2=10√2⋅√2√2=10√22=5√2AE=x=5√2
Now draw CF perpendicular to AB (Figure 4.5.9). ∠B=45∘ since ABCD is an isosceles trapezoid (Theorem 3.2.4, Section 3.2).

So △BCF is a 45∘−45∘−90∘ triangle congruent to △ADE and therefore BF=5√2. CDEF is a rectangle and therefore EF=10. We have AB=AE+EF+FB=5√2+10+5√2=10√2+10.
Answer
AB=10√2+10.
Find AC and BD:
Solution
ABCD is a rhombus. The diagonals AC and BD are perpendicular and bisect each other. ∠AEB=90∘ and ∠ABE=180∘−(90∘−30∘)=60∘. So △AEB is a 30∘−60∘−90∘ triangle.
hyp=2s4=2(BE)2=BEbd=2+2=4 L=s√3AE=2√3AC=2√3+2√3AC=4√3
Answer
AC=4√3,BD=4.
The Pythagoreans believed that all physical relation ships could be expressed with whole numbers. However the sides of the special triangles described in this section are related by irrational numbers, √2 and √3. An irrational number is a number which can be approximated, but not expressed exactly, by a ratio of whole numbers. For example √2 can be approximated with increasing accuracy by such ratios as 1.4=1410, 1.41=141100, 1.414=14141000, etc., but there is no fraction of whole numbers which is exactly equal to √2. (For more details and a proof, see the book by Richardson listed in the References). The Pythagoreans discovered that √2 was irrational in about the 5th century B.C. It was a tremendous shock to them that not all triangles could be measured "exactly." They may have even tried to keep this discovery a secret for fear of the damage it would do to their philosophical credibility.
The inability of the Pythagoreans to accept irrational numbers had unfortunate consequences for the development of mathematics. Later Greek mathematicians avoided giving numerical values to lengths of line segments. Problems whose algebraic solutions might be irrational numbers, such as those involving quadratic equations, were instead stated and solved geometrically. The result was that geometry flourished at the expense of algebra. It was left for the Hindus and the Arabs to resurrect the study of algebra in the Middle Ages. And it was not until the 19th century that irrational numbers were placed in the kind of logical framework that the Greeks had given to geometry 2000 years before.
Problems
1 - 10. Find x and y:
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11 - 14. Find x:
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15 - 20. Find x and y:
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21 - 22. Find x and AB:
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23 - 24. Find x and y:
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25. Find AC and BD:
26. Find x,AC and BD: