Special Right Triangles

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Lesson

Special Right Triangles 487 The Moseleys are preparing for an outdoor party at their new

home. There are a number of things that need to be done before the party. For instance, they would like to tile the ﬂ oor of their regular octagonal gazebo. The cost of the tile is $2.50 per square foot. To ensure they do not

overpay, they would like to ﬁ nd the exact area of the octagonal base. The length of a side of the octagon is 4 ft.

The Moseleys know they can fi nd the area of the gazebo fl oor by dividing the region into triangles. To fi nd the exact area of those triangles, they can use properties of some special right triangles. These you will see in this lesson.

Lengths in Isosceles Right Triangles

Certain right triangles have well-known relationships among their sides and angles. These special triangles occur in many situations that involve other polygons. For instance, by drawing a diagonal of a square, two isosceles right triangles are formed. The Isosceles Triangle Base Angles Theorem tells us that the base angles of an isosceles triangle are congruent. Because one angle is right, the Triangle-Sum Theorem tells us that the measure of each base angle of an isosceles right triangle must be 45º. For this reason, an isosceles right triangle is called a 45-45-90 triangle. In a 45-45-90 triangle, if you know the length of one side, you can ﬁ nd the lengths of the others.

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Special Right Triangles

BIG IDEA The lengths of the sides of right triangles with 30º or 45º angles are related in simple ways.

Vocabulary

45-45-90 triangle 30-60-90 triangle apothem

a. A right triangle has both legs of length 6. What is the length of its hypotenuse?

b. A triangle has sides of length 2, 5, and 6. Is the triangle a right triangle?

c. One right triangle has a leg of length 7. A second right triangle has a hypotenuse of length 7. Which triangle has the longer hypotenuse?

Mental Math

SMP_SEGEO_C08L07_487-493.indd 487

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488 Lengths and Areas

Suppose the congruent sides of one of these triangles have length x and the hypotenuse has length c, as in the triangle at the right.

By the Pythagorean Theorem, c2 = x2+x2.

So c2 = 2x2.

Taking the positive square root, c = x· √ __ 2 .

The result is a relationship among the sides of any isosceles right triangle.

Isosceles Right Triangle Theorem

In an isosceles right triangle, if a leg has length x, then the hypotenuse has length x √ __2 .

Example 1

Find the missing side lengths in each isosceles right triangle. a. b. 45° 45° A C 6.4 45° G I H 10

Solution For each triangle, label each side with x or x √ __2 . Remember, the hypotenuse is the longest side of a right triangle. Because x √ __2 > x, the x √ __2 must always be the length of the hypotenuse.

a. If x = 6.4, then AC = ? _{and BC}₌ ? _. b. If x √ __ 2 = 10, then x = __ 10 √ ¯¯ 2 = 10 · √ ¯¯ 2 _____ √ ¯¯ 2 · √ ¯¯ 2 = 10 √ ¯¯ 2 ____ ₂ = 5

√

¯¯ 2 . x = ? _{, HI}₌ ? _{, GI}₌ ? QY1

45-45-90 triangles are useful in ﬁ nding area because they allow you to ﬁ nd the height of the triangle from just one side length.

Example 2

The bottom step of the Moseleys’ pool needs to be replaced. The cost of the repair is based on the size of the step. A side view of the step is shown at the right. FABD is a trapezoid with AB −−

FD −and FD −⊥ DB .−−

The Moseleys know that AB=BD,AD= 12 in., and FD= 18 in. Round the area of the side of the step to the nearest square inch.

45° 45° 45° c x x 45° GUIDED QY1

Check your answers to Guided Example 1 using the Pythagorean Theorem.

A B D F 18" 12" SMP_SEGEO_C08L07_487-493.indd 488 SMP_SEGEO_C08L07_487-493.indd 488 4/25/08 9:48:17 AM4/25/08 9:48:17 AM

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Special Right Triangles 489

Solution You can fi nd the area of this region by fi nding the area of each triangular region and using the Additive Property of Area. ∠FDB is a right angle because it is a same-side interior angle with ∠ABD and

__

AB

__

FD . Thus, BD is the height of both triangles AFD and ABD. Because AB=DB, ADB is an isosceles right triangle. Let x = AB. Then, by the Isosceles Right Triangle Theorem,

x

√

¯¯2 = 12 x = __12 √ ¯¯ 2 x = 6

√

¯¯2 . So AB = BD =6

√

¯¯2 . Area(AFD) = _₂1(18)(6

√

¯¯2 ) = 54

√

¯¯2 in2 Area(ABD) = _₂1(6

√

¯¯2 )(6

√

¯¯2 ) = 36 in2 Area(FABD) = 54

√

¯¯2 + 36 ≈ 112 in2

Lengths in 30-60-90 Triangles

When an altitude is drawn in an equilateral triangle, two triangles with angles 30º, 60º, and 90º are formed. This special right triangle is called a 30-60-90 triangle. Again the lengths of all sides can be determined if one side length is known.

30-60-90 Triangle Theorem

In a 30-60-90 triangle, if the length of the shorter leg is x, then the length of the longer leg is x √ __3 and the length of the hypotenuse is 2x.

Given ABC with m∠A = 30, m∠B = 60, m∠C = 90. The shorter leg is opposite the smaller acute angle, so let BC = x. Prove 1. AB = 2x 2. AC = x √ __3 Drawing A B D x√⎯2 =₁₂ x x 45° B C D A 60° 60° 30° 30° x√⎯3 2x x B C A 60° 30° SMP_SEGEO_C08L07_487-493.indd 489 SMP_SEGEO_C08L07_487-493.indd 489 4/25/08 9:48:22 AM4/25/08 9:48:22 AM

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Proof The idea is to think of ABC as half an equilateral triangle and use the Pythagorean Theorem.

Proof of 1 Refl ect ABC over AC . Let D = r_AC(B). Refl ections preserve distance, so CD = x. Because refl ections preserve angle measure, the image ADC is a 30-60-90 right triangle with m∠ACD = 90, m∠ADC = 60, and m∠CAD = 30. Thus, B, C, and D are collinear, and the big triangle ABD has three 60º angles, making it equilateral with AB = BD = 2x.

Proof of 2 Now, using AB, you can apply the Pythagorean Theorem to ABC and fi nd AC.

(AC)2_{+ (}_BC₎2_{= (}_AB₎2

(AC)2₊_x2_{= (2}_x₎2

(AC)2₊_x2_{= 4}_x2

(AC)2_{= 3}_x2

Taking the positive square roots of each side, AC = x √ __3 .

QY2

Finding the Areas of Regular Polygons

All regular polygons can be triangulated from their center. Suppose you begin with a regular n-gon with side length s. If you know the length a of a segment connecting the center of the polygon to the midpoint of a side, then you have enough information to create a formula for the area of the regular polygon.

Activity 1

Refer to the three regular polygons drawn at the right. (One drawing is incomplete.)

Step 1 If you draw segments joining the center of the regular polygon to its vertices, you create triangles. Explain how you know that these triangles are isosceles.

Step 2 Explain how you know that a is the height of each of these isosceles triangles.

Step 3 In terms of s and a, what is the area of each isosceles triangle?

Step 4 Write a formula for the area of any regular n-gon in terms of s, a, and n.

Step 5 Write a formula for the perimeter p of any regular n-gon with side length s.

Step 6 Use your answers to Steps 3 and 4 to create a formula for the area of any regular polygon in terms of a and p.

QY2

If the shortest side of a 30-60-90 triangle has length 70 cm, what are the lengths of the other two sides? a s a s s a B C D A 60° 60° 30° 30° 2x x x SMP_SEGEO_C08L07_487-493.indd 490 SMP_SEGEO_C08L07_487-493.indd 490 4/25/08 9:48:25 AM4/25/08 9:48:25 AM

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Special Right Triangles 491 In the Activity, the length a is known as the apothem. The apothem

of a regular polygon is the segment joining the center of the polygon to the midpoint of any side. In Step 6, you were deducing the

following formula for the area of a regular polygon.

Regular Polygon Area Formula

The area of a regular polygon is half the product of the length of its apothem a and its perimeter p.

A= __1

2ap

Sometimes special right triangles are formed by the apothem.

Example 3

Find the area of a regular hexagon with side length 10.

Solution Draw regular hexagon HIJKLM with apothem OP . −− The perimeter of the hexagon is 6(10) = 60. To fi nd the apothem, notice that KP = 5 because it is half of the side length. Also notice that m∠KOJ = ___ 360₆ = 60, so m∠POK = 30. Thus, KPO is a 30-60-90 triangle in which 5 is the length of the shorter leg. So, OP = 5

√

¯¯ 3 .

Area(HIJKLM) = _ ₂1 ap

= _ ₂1 (5

√

¯¯ 3 )(60)

= 150

√

¯¯ 3

Activity 2

Find the area of the fl oor of the Moseleys’ gazebo. Recall that the gazebo fl oor is a regular octagon with side length 4 ft. Before you begin, it is important to realize that the method used in Example 3 does not work in this case. This is because the measure of each of the vertex angles of the isosceles triangles formed by joining the center to the vertices is ___360₈ = 45. Thus, when the apothem is drawn and it bisects that angle, two angles with measure 22.5 are created. This will not result in special right triangles. We provide a hint in the diagram at the right to get you started.

READING MATH

The word apothem

is pronounced with emphasis on the fi rst syllable and with a short “a” as in “apple”: A-poth-em M H O J P I L K 30° 60° 10 A C O F T J P I 45° 2 4 2 SMP_SEGEO_C08L07_487-493.indd 491 SMP_SEGEO_C08L07_487-493.indd 491 4/25/08 9:48:29 AM4/25/08 9:48:29 AM

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Questions

COVERING THE IDEAS

1. What are the measures of the three angles in an isosceles right triangle?

2. Fill in the Blank In an isosceles right triangle, the hypotenuse is ? _{times the length of a leg.}

3. Describe the relationship between the lengths of the sides of a 30-60-90 triangle.

4. If an isosceles triangle has a 45º angle, must it be a right triangle? In 5−8, fi nd the exact length of the missing sides.

5. 6. A E C 9.52 B I 9 30° G

7. a 45-45-90 triangle with hypotenuse 16

8. a 30-60-90 triangle with hypotenuse 26

9. State a formula for the area of a regular polygon.

10. Find the exact area of a regular hexagon with side length 12 cm.

11. Find the area of a regular heptagon with side length 15 m and apothem with length approximately 10.32 m, to the nearest tenth of a square meter.

12. Recalculate the area of the ﬂ oor of the Moseleys’ gazebo, but this time do it by dividing the octagon into two trapezoids and a rectangle as shown at the right.

APPLYING THE MATHEMATICS

13. In the parallelogram at the right,

____

WE

____

EO . If WF= 10, ﬁ nd the exact area of the parallelogram.

14. ABDC is a trapezoid with

___

AB

____

CD . Using the information in the ﬁ gure below, ﬁ nd the exact perimeter of ABCD. B A D C 45° 30° 10 in. 14 in.

15. Find the area of XYZ at the right in terms of h.

16. Find the area of an equilateral triangle with side s.

L F W 45° E O Y X W Z h 45° 30° SMP_SEGEO_C08L07_487-493.indd 492 SMP_SEGEO_C08L07_487-493.indd 492 4/25/08 9:48:33 AM4/25/08 9:48:33 AM

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Special Right Triangles 493 REVIEW

17. A square is

inscribed in a circle with diameter 14. What is the area of the square?

(Lesson 8-6)

18. HLAF is a convex kite. HA= 30,

HL= 25, FA= 17. Find the area of the kite. (Lessons 8-6, 8-5, 6-5)

19. In the ﬁ gure at the right

___

AC

____

DE ,DE= 30,

AC= 50, the area of BDE is 450, and the area of ABC is 1250. Find the height of ADEC. (Lessons 8-5, 8-4)

20. Is the quadrilateral below a parallelogram? How do you know? (Lesson 7-8)

4

3.4

5.3 5.6

21. A periscope is a device that uses mirrors to allow a person in a submarine to see objects on the surface. Use the ﬁ gure at the right to illustrate how this is possible. (Lesson 4-4)

EXPLORATION

22. There are many triangles that could be considered special. One such triangle is called the heptagonal triangle. It is formed by the side length a, a shorter diagonal length b, and a longer diagonal length c of a regular heptagon.

a. Use a DGS to create a regular heptagon.

b. Measure c, a, and b. Calculate

(

__ _ac

)

2+

(

__ a b

)

2

+

(

__ b_c

)

2. Your answer should be very close to a whole number. Which whole number?

c. Measure the three angles of this triangle. How do they seem to be related? B x A D C 14 _L F H A 25 30 17 E D A B C a c b QY ANSWERS 1. a. 6.42₊_6.42₌ 40.96 + 40.96 = 81.92; (6.4 √ __ 2 )2₌ (40.96)(2) = 81.92 b. (5 √ __2 )2₊₍₅_√___{2 )}2₌ 50 + 50 = 100; 102₌₁₀₀ 2. 70 √ __ 3 cm and 140 cm SMP_SEGEO_C08L07_487-493.indd 493 SMP_SEGEO_C08L07_487-493.indd 493 4/25/08 9:48:36 AM4/25/08 9:48:36 AM