Thread: 4 questions of circle geometry (The 4 theorems)

1. 4 questions of circle geometry (The 4 theorems)

Hey, I have a few problems with these questions relating to the geometry of a circle and the 4 theorems.

1) In a circle, O is the centre of a circle. C is the midpoint of chord AB. If AB = 10 and OC = 3, find the radius of the circle to 3 sig.fig.

2) In a circle, AB and CD are two parallel chords of length 6cm and 8cm respectively. If the radius is 5, find the distance between the two chords.

3) In a circle, O is the centre of the circle. The radius is 5cm. PQ and RS are two perpendicular chords of the circle, intersecting at K. If each chord is 8cm, find the length of OK.

4) In the circle, AC is the diameter of the circle. BD is perpendicular to AC and cuts AC at E. Prove triangle CBD is an isosceles triangle.

Thanks for any help

2. 1) i can't draw it here but my drawing shows that OA=radius
and OAC is a triangle rectangle..and OA is the hypotenuse..so
OA(squared)=OC(squared)+CA(squared)
OA(squared)=5(squared)+3(squared)
OA(squared)=34
OA=(square root)34

3. Originally Posted by BG5965
Hey, I have a few problems with these questions relating to the geometry of a circle and the 4 theorems.

2) In a circle, AB and CD are two parallel chords of length 6cm and 8cm respectively. If the radius is 5, find the distance between the two chords.

Thanks for any help
Draw the perpendicular from the center to both chords.

Draw a radius from the center to an endpoint on each chord.

You have now formed two right triangles whose hypotenuses = 5.

The base of the first right triangle = 4 and the base of the second triangle = 3(the chords were bisected by the perpendicular from the center)

Use the Pathagorean theorem to find each leg.

$5^2=4^2+x^2$

$\boxed{x=3}$

$5^2=3^2+y^2$

$\boxed{y=4}$

The distance between the two chords is $y - x \Rightarrow 4-3=1$

4. Originally Posted by BG5965

4) In the circle, AC is the diameter of the circle. BD is perpendicular to AC and cuts AC at E. Prove triangle CBD is an isosceles triangle.

Thanks for any help
$\overline{BE}\cong \overline{DE}$

The perpendicular drawn from the center of a circle to a chord bisects the chord.

$\angle BEC \ \ and \ \ \angle DEC$ are right angles.

Perpendicular lines meet to form right angles.

$\overline{EC}\cong \overline{EC}$

Reflexive property of congruence.

$\triangle BEC \cong \triangle DEC$

Hypotenuse-Leg Theorem for right triangles.

$\overline{BC}\cong \overline{DC}$

CPCTC

$\triangle CBD \ \ is \ \ isosceles$

Definition of isosceles triangle.

5. i think..it's 3+4=7 and not 4-3 or maybe it relies if they are on the same side..it's 1 and if they are the opposite by the diameter...what do you think?

6. Originally Posted by franckherve1
i think..it's 3+4=7 and not 4-3 or maybe it relies if they are on the same side..it's 1 and if they are the opposite by the diameter...what do you think?
The distance from the center to the 2nd parallel is 4.

The distance from the center to the 1st parallel is 3.

The difference 4 - 3 is the distance between the parallel lines.

7. yeah but if they are one opposite sides of the diameter

8. Originally Posted by franckherve1
yeah but if they are one opposite sides of the diameter
In my opinion both solutions are valid: