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Mathematics LibreTexts

4.5: Special Right Triangles

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There are two kinds of right triangle which deserve special attention: the 306090 right triangle and the 454590 right triangle.

30°−60°−90° Right Triangles

A triangle whose angles are 30,60, and 90 is called a 306090 triangle. ABC in Figure 4.5.1 is a 306090 triangle with side AC=1.

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Figure 4.5.1: A 306090 triangIe.
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Figure 4.5.2: Draw BD and CD.

To learn more about this triangle let us draw lines BD and CD as in Figure 4.5.2. ABCDBC 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).

clipboard_e43a62a5040176eee2263715ec187b75b.png
Figure 4.5.3: ABD is equiangular with all sides equal to 2 .

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 306090 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=EF383=EF

clipboard_ebf457adce79b4af3bd3bbf861aa3ffb1.png
Figure 4.5.3:

Our conclusions about triangles ABC and DEF suggest the following theorem:

Theorem 4.5.1

In the 306090 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.

clipboard_e0fe7722615621b15eee5df801e479d4b.png
Figure 4.5.5: The hypotenuse is twice the shorter leg and the longer leg is equal to the shorter leg times the 3.

In Figure 4.5.5, s= shorter leg, L= longer leg, and hyp = hypotenuse. Theorem 4.5.1 says that

hyp=2sL=s3

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.

Example 4.5.1

Find x and y:

clipboard_ebf457adce79b4af3bd3bbf861aa3ffb1.png

Solution

B=180(60+90)=180150=30, so ABC is a 306090 triangle. By theorem 4.5.1,

hyp=2sy=2(7)=14. L=s3x=73

Answer

x=73, y=14.

Example 4.5.2

Find x and y:

屏幕快照 2020-11-17 下午9.26.53.png

Solution

B=60 so ABC is a 306090 triangle. By Theorem 4.5.1,

L=s310=x3103=x33x=103=10333=1033.hyp=2sy=2x=2(1033)=2033

Answer

x=1033,y=2033.

45°−45°−90° Right Triangles

The second special triangle we will consider is the 454590 triangle. A triangle whose angles are 45, 45, and 90 is called a 454590 triangle or an isosceles right triangle. ABC in Figure 4.5.6 is a 454590 triangle with side AC=1.

屏幕快照 2020-11-17 下午9.40.51.png
Figure 4.5.6: A 454590 triangle.

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

屏幕快照 2020-11-17 下午9.43.07.png
Figure 4.5.7: The legs of the 454590 triangle are equal.

Let x=AB (Figure 4.5.7). By the Pythagorean Theorem,

leg2+leg2=hyp212+12=x21+1=x22=x22=x

Example 4.5.3

Find x:

屏幕快照 2020-11-17 下午9.51.26.png

Solution

B=180(45+90)=180135=45. So ABC is a 454590 triangle. AC=BC=8 because these sides are opposite equal angles. By the Pythagorean Theorem,

leg2+leg2=hyp282+82=x264+64=x2128=x2x=128=642=82

Answer

x=82.

The triangles of Figure 4.5.6 and Example 4.5.3 suggest the following theorem:

Theorem 4.5.2

In the 454590 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

屏幕快照 2020-11-17 下午10.00.30.png
Figure 4.5.8: The legs are equal and the hypotenuse is equal to either leg times 2.
Example 4.5.4

Find x and y:

屏幕快照 2020-11-17 下午10.02.32.png

Solution

B=45. So ABC is an isosceles right triangle and x=y.

x2+y2=42x2+x2=162x2=16x2=8x=8=42=22

Answer: x=y=22.

Another method:

ABC is a 454590 triangle. Hence by Theorem 4.5.2,

hyp=L24=x242=x22x=42=4222=422=22

Answer

x=y=22.

Example 4.5.5

Find AB:

屏幕快照 2020-11-18 下午2.08.01.png

Solution

ADE is a 454590 triangle. Hence

hyp=L210=x2102=x22x=102=10222=1022=52AE=x=52

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).

屏幕快照 2020-11-18 下午2.11.23.png
Figure 4.5.9: Draw CF perpendicular to AB.

So BCF is a 454590 triangle congruent to ADE and therefore BF=52. CDEF is a rectangle and therefore EF=10. We have AB=AE+EF+FB=52+10+52=102+10.

Answer

AB=102+10.

Example 4.5.6

Find AC and BD:

屏幕快照 2020-11-18 下午2.15.08.png

Solution

ABCD is a rhombus. The diagonals AC and BD are perpendicular and bisect each other. AEB=90 and ABE=180(9030)=60. So AEB is a 306090 triangle.

hyp=2s4=2(BE)2=BEbd=2+2=4 L=s3AE=23AC=23+23AC=43

Answer

AC=43,BD=4.

Historical Note: Irrational Numbers

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:

1.

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2.

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3.

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4.

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5.

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6.

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7.

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8.

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9.

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10.

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11 - 14. Find x:

11.

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12.

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13.

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14.

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15 - 20. Find x and y:

15.

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16.

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17.

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18.

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19.

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20.

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21 - 22. Find x and AB:

21.

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22.

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23 - 24. Find x and y:

23.

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24.

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25. Find AC and BD:

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26. Find x,AC and BD:

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This page titled 4.5: Special Right Triangles is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Henry Africk (New York City College of Technology at CUNY Academic Works) via source content that was edited to the style and standards of the LibreTexts platform.

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