Question

what happens to the length of a chord as its perpendicular distance from the center decreases?
a. it decreases.
b. it becomes a tangent.
c. it increases.
d. it stays the same.

what is the longest chord in a circle?
a. any chord
b. a tangent line
c. a chord perpendicular to a radius
d. the diameter

if a radius is drawn to the point of tangency, what is the angle formed between the radius and the tangent?
a. 90°
b. 60°
c. 45°
d. 180°

if a chord is 8 cm long and its perpendicular distance from the center is 3 cm, what is the radius of the circle?
a. 6 cm
b. 8 cm
c. 5 cm
d. 3 cm

Explanation:

Step1: Recall the formula for the length of a chord

The formula for the length of a chord \( l = 2\sqrt{r^{2}-d^{2}}\), where \(r\) is the radius of the circle and \(d\) is the perpendicular distance from the center of the circle to the chord.

Step2: Analyze the relationship between chord length and \(d\)

If \(d\) decreases (while \(r\) is constant), then \(r^{2}-d^{2}\) increases. Since \(l = 2\sqrt{r^{2}-d^{2}}\), the value of \(l\) (the length of the chord) increases.

Step1: Define a chord and diameter

A chord is a line segment with both endpoints on the circle. The diameter is a chord that passes through the center of the circle.

Step2: Compare lengths

By the property of a circle, the length of a chord \(l = 2\sqrt{r^{2}-d^{2}}\). When \(d = 0\) (the chord passes through the center, i.e., it is a diameter), \(l=2r\). For any other chord (\(d>0\)), \(l = 2\sqrt{r^{2}-d^{2}}<2r\).

Step1: Recall the tangent - radius theorem

The tangent - radius theorem states that a radius drawn to the point of tangency is perpendicular to the tangent line.

Step2: Determine the angle

If two lines are perpendicular, the angle between them is \(90^{\circ}\).

Step1: Use the chord - radius relationship

Given chord length \(l = 8\space cm\), so half - chord length \(x=\frac{l}{2}=\frac{8}{2} = 4\space cm\), and perpendicular distance \(d = 3\space cm\).

Step2: Apply the Pythagorean theorem

By the Pythagorean theorem \(r=\sqrt{x^{2}+d^{2}}\) (where \(r\) is the radius). Substitute \(x = 4\) and \(d = 3\): \(r=\sqrt{4^{2}+3^{2}}=\sqrt{16 + 9}=\sqrt{25}=5\space cm\)

Answer:

c. It increases.