Warm Up for Section 1.1

Simplify: (1). (2).

Use the triangle below to answer #3, #4:

(3). Find x.

(4). If a = 5, b = 3, find c.

183

5

40o

a

b

cxo

Answers for Warm Up, Section 1.1

Simplify: (1). (2).

Use the triangle below to answer #3, #4:

(3). Find x.

(4). If a = 5, b = 3, find c.

183

5

40o

a

b

cxo

233

35

50x

34c

Work for Answers to WU, Section 1.1

(1). (2).

(3). x = 90 – 40 (4). a2 + b2 = c2 = 50 52 + 32 = c2

25 + 9 = c2

34 = c2

= c

23

29

2918

3

35

3

5

3

5

34

3

3

Special Right Triangles Section 1.1

Essential Question: What is the relationship between the lengths of the legs of a 45°–45°–90° triangle and a 30°–60°–90°

triangle?

Standard: MM2G1a, b

Vocabulary

Right Triangle: A triangle containing one angle that measures exactly 90 degrees.

Hypotenuse: The longest side of a right triangle. Reference angle: The measured, or known angle in a right triangle other than the 90° angle.

Investigation 1:

With your partner, complete each step in the investigation then answer questions 1-10.

Step 1: Using the grid paper provided and a straightedge, draw a square with side length 5 cm.

Step 2: Label the vertices of the square A, B, C, and D. Label each side with its length.

Step 3: Using a straightedge, draw diagonal . AC

Investigation 1:

A B

CD

5 cm

5 cm

5 cm

5 cm

Answer the following questions:

(1). mD = ____o (2). mACD = ____o

(3). mDAC = ____o (4). DC = ____

(5). AD = ____

(6). ADC is (acute, right, obtuse).

(7). ADC is (isosceles, scalene, equilateral).

(8). Using the Pythagorean Theorem, find AC. Be sure to write your answer in simple radical form.

90 45

45 5 cm

5 cm

45°

45°

5

5

a2 + b2 = c2

52 + 52 = x2

25 + 25 = x2

50 = x2

x50

x25

x x 225

Look at two additional 45o-45o-90o triangles and determine the length of the hypotenuse, x.

Be sure to write your answer in simple radical form.

45°

45°

3

3

a2 + b2 = c2

32 + 32 = x2

9 + 9 = x2

18 = x2

x18

x23

x

Question 9: Find x

x 29

45°

45°

8

8

a2 + b2 = c2

82 + 82 = x2

64 + 64 = x2

128 = x2

x128

x28

x

Question 10: Find x

x 264

45°

45°

x

x

2x

(a). Length of hypotenuse = length of leg times . 2

(b). Length of legs = length of hypotenuse divided by .2

Summary: In a 45o-45o-90o triangle

Investigation 2:

With your partner, complete the following regarding equilateral ABC where AB =10:

Step 1: Label the length of each edge.Step 2: Label the measure of B and C.Step 3: Using a straightedge, draw and label altitude . Step 4: Label the length of and .Step 5: Label the measure of BAD and CAD.Step 6: Label the measure of ADC.Step 7: Using the Pythagorean Theorem, find AD.

AD

CD BD

10

10

10

5 5

60°

30°

60°

a2 + b2 = c2

52 + x2 = 102

25 + x2 = 10075 = x2

x75

x35

30°

A

BC D

x

x 325

Investigation 2:

Note: the two legs of a 30o-60o-90o triangle are NOT equal in measure. The longer leg will always be opposite the ___o

angle.

The shorter leg will always be opposite the ___o angle.

60

30

Consider the 30o-60o-90o right triangle created from an equilateral triangle pictured at right. (13). The long leg is segment ______

and the short leg is segment _______.

(14). Use the Pythagorean Theorem to find RT.

RT

ST12

6

60°

30°

x

R

TS

12

6

60°

30°

a2 + b2 = c2

62 + x2 = 122

36 + x2 = 144108 = x2

x108

x36x

R

TS

x 336

2x

x

60°

30°

3x

Length of hypotenuse = length of short leg times 2

Length of long leg: length of short leg times

Length of short leg: half the

length of hypotenuse or the length of the long leg divided by

3

3

Summary: In a 30o-60o-90o triangle:

Check for Understanding: Find the missing edge lengths for each triangle:

Example 13:

26

x

x2x

226 x

226 x22

x66

6

Check for Understanding: Find the missing edge lengths for each triangle:

Example 14:

60o

30o

3834 x x2

3x

382 x

22

34x

382 x

12

34

3)34(3

x

12

Check for Understanding: Find the missing edge lengths for each triangle:

Example 15:

5

5

10

x

x

2x

5x

10

252

x

Check for Understanding: Find the missing edge lengths for each triangle:

Example 14:

60o

30o

3834 x x2

3x

382 x

22

34x

382 x

12

34

3)34(3

x

12

Check for Understanding: Find the missing edge lengths for each triangle:

Example 14:

60o

30o

3834 x x2

3x

382 x

22

34x

382 x

12

34

3)34(3

x

12

Check for Understanding: Find the missing edge lengths for each triangle:

Example 14:

60o

30o

3834 x x2

3x

382 x

22

34x

382 x

12

34

3)34(3

x

12

Check for Understanding: Find the missing edge lengths for each triangle:

Example 14:

60o

30o

3834 x x2

3x

382 x

22

34x

382 x

12

34

3)34(3

x

12