"Vertical Diameter Theorem" PPT courseware

Description

"Vertical Diameter Theorem" PPT courseware

1. The diameter perpendicular to the chord bisects the ________ and bisects the ________ subtended by the chord.

2. The diameter of the bisector chord ( ) is perpendicular to the chord and bisects the _____________ subtended by the chord.

1. (4 points) (2013·Shanghai) In ⊙O, it is known that the radius is 3 and the length of chord AB is 4, then the distance from the center O to AB is ________.

2. (4 points) As shown in the figure, in ⊙O with a radius of 13, OC is perpendicular to the chord AB and intersects at point D, and intersects ⊙O at point C. AB=24, then the length of CD is ________.

3. (4 points) (2013•Langfang) As shown in the figure, the diameter of ⊙O is AB=12, CD is the chord of ⊙O, CD⊥AB, the vertical foot is P, and BP:AP=1:5, then the length of CD is ()

A. 42B. 82C． 25D． 45

4. (4 points) As shown in the figure, among the two circles with point O as the center, the chord AB of the larger circle intersects the smaller circle at points C and D. It is known that AB=4, CD=2, and the distance from point O to chord AB is equal to 1 , then the ratio of the radii of these two circles is ()

A. 3:2 B.5:2

C.5∶2 D. 5:4

14． (9 points) As shown in the figure, both circles have point O as the center. The chord AB of the larger circle intersects the smaller circles at points C and D. Prove: AC=BD.

Draw OM⊥AB through the point O, and the vertical foot is M. According to the vertical diameter theorem, we can get MA=MB, MC=MD, so AC=BD

15． (9 points) As shown in the figure, the radius of ⊙O is 17 cm, the chords AB∥CD, AB=30 cm, CD=16 cm, the center O of the circle is located above AB and CD, find the distance between AB and CD.

Through point O, make OH⊥CD at point H, intersect AB at point G, connect OA, OC, from AB∥CD, we have OG⊥AB, ∴CH=DH, AG=BG, in Rt△OCH, according to the Pythagorean theorem We get OH=15. In Rt△OAG, we get OG=8 from the Pythagorean theorem, so the distance between AB and CD is OH-OG=7 (cm)

16. (12 points) As shown in the figure, there is an arc-shaped arch bridge somewhere. The width below the bridge is 7.2 m, and the arch is 2.4 m above the water. There is an existing ship with a width of 3 m, a square cabin top and a height of 2 m above the water. The cargo ship is going to pass here. Can the cargo ship pass this arch bridge smoothly at this time? Please explain the reason.

Can pass smoothly. Reason: According to the meaning of the question, AB=7.2, CD=2.4, let the radius of ⊙O be R, in Rt△AOD, OD=R-2.4, AD=3.6, ∴R2=(R-2.4)2+3.62, ∴ R=3.9, in Rt△OHN, if HN=1.5, then OH=ON2-HN2=3.92-1.52=3.6, ∵OD=OC-DC=3.9-2.4=1.5, ∴DH=OH-OD=3.6- 1.5=2.1(m), NF=ME=HD=2.1 m>2 m, ∴This cargo ship can pass smoothly

Keywords: vertical diameter theorem teaching courseware, Hebei Education Edition ninth grade mathematics volume PPT courseware download, ninth grade mathematics slide courseware download, vertical diameter theorem PPT courseware download, .PPT format;

For more information about the "Vertical Path Theorem" PPT courseware, please click on the "Vertical Path Theorem" ppt tab.

File Info

Update Time: 2024-09-25

This template belongs to Mathematics courseware Hebei Education Edition Ninth Grade Mathematics Volume 1 industry PPT template

Preview