Two points P and Q are 3 cm apart. These two points lie on the circumference of a circle having radius 1.7 cm. What is the distance (in cm) of the line segment PQ from the centre of the circle?

Given:

PQ = 3 cm

Radius (OP) = 1.7 cm

Concept used:

The distance of a chord from the centre cuts the chord in half and perpendicular to it.

Calculation:

OM ⊥ PQ

PM = 3/2 = 1.5 cm

Now, By Pythagoras theorem

⇒ H² = P² + B²

⇒ (1.7)² = OM² + (1.5)²

⇒ OM² = 1.7² - 1.5²

⇒ OM = √{(1.7 + 1.5)(1.7 - 1.5)}

⇒ OM = √{3.2 × 0.2} = √0.64

⇒ OM = 0.8 cm

∴ The correct answer is 0.8 cm.

Given:

Two points P and Q are 3 cm apart. These two points lie on the circumference of a circle with a radius of 1.7 cm.

Concept Used:

Use the properties of circles and right triangles to find the perpendicular distance from the center of the circle to the line segment PQ.

Formula Used:

1. Pythagorean theorem:\( a^2 + b^2 = c^2 \)

2. The perpendicular distance from the center to the chord: \(d = \sqrt{r^2 - \left(\frac{chord length}{2}\right)^2} \)

Calculation:

Identify the given values:

Radius of the circle r = 1.7 cm

Length of the chord PQ = 3 cm

Calculate half of the chord length
Use the formula to find the perpendicular distance from the center to the chord:\( \frac{chord length}{2} = \frac{3}{2} = 1.5 cm \)

\(d = \sqrt{r^2 - \left(\frac{chord length}{2}\right)^2} \) \( d = \sqrt{1.7^2 - 1.5^2} \) \( d = \sqrt{2.89 - 2.25} \) \( d = \sqrt{0.64} \) \( d = 0.8 cm\)

So, the distance of the line segment PQ from the center of the circle is 0.8 cm.