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 ofintersection of these chords (in cm) from the center of the circle

Answer 1

Given: Radius 15 cm, AB = 24 cm
And CD = 20 cm

To find FG = ?

So here we begin....

In Δ BOG,

(OG)² = (OB)² - (BG)²
(OG)² = (15)² - (12)²
(OG)² = 225 - 144 = 81.......(1)

Similarly, in Δ COF,

(OF)² = (OC)² - (CF)²
(OF)² = (15)² - (10)²
(OF)² = 225 - 100 = 125....(2)

Again in Δ GOF,

(FG)² = (OF)² + (OG)²
(FG)² = 125 + 81 = 206
(FG)² = √206

$hope \: it \: helps \: you.$

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