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BELLRINGER

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W

1.

BELLRINGER

P

Y

45°

PR = ?, ∠RPS = ?

2.

95°

S

V

?

4k-2

4

132°

?

k+10

WX = ?

∠VZY = ?

Q

R

Z

T

X

- This is the newest type of segment we will learn that exists in a triangle.
- This segment has one endpoint at a vertex of the triangle, and its other endpoint perpendicular to the opposite side.

X

XY is an altitude.

Y

Every Triangle has THREE altitudes.

The altitude is called the HEIGHT of a triangle.

37°

122 + 52 = AC2

144 + 25 = AC2

169 = AC2 AC = √169 = 13

A

CA

D

53°

AF is an altitude of length 12 cm.

CF = 5cm.

EC = ?

∠ADE = ?

B

53°

Triangle Angle Sum

E

90°

But, since DE is a midsegment, EC is half the length of AC. Thus, EC = 13/2 = 6.5 cm.

F

C

What is the height of the triangle?

HYPOTENUSE

LEG

Use the pythagorean theorem: leg2 + leg2 = hypotenuse2

Therefore,

72 + height2 = 252.

49 + h2 = 625. h2 = 625 - 49

h2 = √576, h = 24 feet.

25 ft

7 ft

LEG

- See sketchpad example . . .
- 3!

The LEGS of a right triangle are ALTITUDES.

2

1

3

- AREA OF A TRIANGLE = (base)(height) / 2.
- Altitude = Height and allows us to find the area. Observe Skethpad example.

- Find the area of the triangle.

First, find the base YZ.

Use P.T. twice—

WY2 + 102 = 152

WY2 +100 = 225

WY2 = 125 WY = 11.2

A = (b ∙ h) / 2

A = (29.7)(10) / 2

A = 297 / 2

A = 148.5 cm2

X

21 cm

15 cm

Then, using PT again, WZ = 18.5 cm

10 cm

Thus, YZ = 11.2 + 18.5 = 29.7 cm

Y

Z

W

X

The altitude is 7 feet. The area of Triangle XYZ is 70 ft2

A = (b ∙ h) / 2

70 = 7b / 2

70 ∙ 2 = 7b

140 = 7b

b = 140/7

b = 20 feet.

Z

Y

What is the length of YZ?

20 FEET!

Draw 3 altitudes!

A

2.

1.

D

4(x + 2)

A

AB is an altitude. Find the area of triangle AXZ

3.

X +20

C

8 in

10 in

17 in

Z

E

B

AB = ?

X

B

3.

DO page 418, #1, 5, 6. DO page 465, #1, 4, 6, 10.