10.2 Exploring Chord Properties

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Presentation transcript:

10.2 Exploring Chord Properties Math 9

Definition A perpendicular bisector is a line that passes through the midpoint of a line segment at 90. In a diagram a 90 angle is indicated with a small square. The symbol for perpendicular is  (an upside down T) A B

Draw two chords in the circle below Draw two chords in the circle below. For each chord construct its perpendicular bisector.

What do you notice about the point of intersection of the two perpendicular bisectors? goes through the center Do you think this will be true for any chord or any size circle? Yes

How many bisectors does a chord have How many bisectors does a chord have? 1 (How many lines pass through the midpoint of a chord?) Infinite but only 1 at a right angle

How many lines are perpendicular to a chord? Infinite but only 1 at the bisector

Properties of Chords The perpendicular bisector of any chord passes through the center of the circle What can you conclude in the diagram?

A special bisector The bisector of a chord passing through the center is perpendicular to the chord.

A special perpendicular The perpendicular to any chord passing through the center must bisect the chord. The length of this line segment is also the shortest distance between the center and the chord.

a2 + b2 = c2 a2 + 32 = 52 a2 + 9 = 25 a2 = 16 a = 16 = 4 X =2a = 8 For questions #1-4, determine the value of x: a2 + b2 = c2 a2 + 32 = 52 a2 + 9 = 25 a2 = 16 a = 16 = 4 X =2a = 8

24 ÷ 2 = 12 x2 + 122 = 132 x2 + 144 = 169 x2 = 25 x = 25 = 5

16 ÷ 2 = 8 x2 = 62 + 82 x2 = 36+ 64 x2 = 100 X = 100 = 10

6 = 4 Same as last triangle.

5. A circle has a diameter of 30 cm 5. A circle has a diameter of 30 cm. A chord in the circle is 18 cm long. What is the shortest distance from the centre of the circle to the chord? Diameter = 30 cm Radius = 15 cm a2 + b2 = c2 a2 + 92 = 152 a2 + 81 = 225 a2 = 144 a = 144 = 12 cm 18 cm

6. The circle below has a radius of 10 cm and the length of EC is 6 cm 6. The circle below has a radius of 10 cm and the length of EC is 6 cm. What is the area of ADE EB2 + 62 = 102 EB2 + 36= 100 EB2 = 64 EB = 64 = 8 = ED Area of ADE = ½ bh = ½(8)(16) = 64 cm2 10 6

7. The cross section of the pipe to the right shows some water still in the pipe. The horizontal distance across the water is 10 cm, and the inner diameter of the pipe is 14 cm. What is the maximum depth of the water in the pipe? Answer to the nearest tenth of a cm. 5 x2 + 52 = 72 x2 + 25 = 49 x2 = 24 x = 24 = 4.9 7 + 4.9 = 11.9 cm The maximum depth of the water pipe is 11.9 cm. x 7 7

Practice p. 389 #4, 5, 8, 9, 10, 12, 14