Question

two chords of length 20 cm and 24 cm are drawn perpendicular to each other in a circle of radius is 15 cm. what is the distance between the points of intersection of these chords (in cm) from the center of the circle?

Answer 1

Let the chord AB be 20 cm and the chord CD , 24 cm, and of the center of the circle of radius, 15 cm.

The mid point of AB from O = [15^2- 10^2]^0.5 = [225–100]^0.5 = 125^0.5 = 11.18 cm

The mid point of CD from O = [15^2- 12^2]^0.5 = [225–144]^0.5 = 81^0.5 = 9 cm

The point of intersection of the chords AB and CD = [125 + 81]^0.5 = 206^0.5 = 14.3527 cm.

Answer 2

$\huge\sf{\underline{\underline{Solution:}}}$

Given:

• Radius = 15 cm,

• AB = 24 cm,

• CD = 20 cm.

To find:

• FG

______________________________

In ΔBOG

By Pythagoras theorem,

(OG)² = (OB)² - (BG)²

(OG)² = (15)² - (12)²

(OG)² = 225 - 144 = 81 ___(1)

Similarly, in ΔCOF

(OF)² = (OC)² - (CF)²

(OF)² = (15)² - (10)²

(OF)² = 225 - 100 = 125 ___(2)

And in ΔGOF

(FG)² = (OF)² + (OG)²

(FG)² = 125 + 81 = 206

FG = √206

Thus, distance between the points of intersection of these chords from the center of the circle is √206 cm.