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
Answers
Answer: √206 cm
Step by step explanation:
Refer the attachment for figure.
Given:
Radius = 15 cm,
AB = 24 cm,
CD = 20 cm.
To find: FG
Solution:
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.
Answer:
2√34 cm.
Step-by-step explanation:
Figure (i) is considered for the instructions given and figure (ii) is considered for the constructions which are required to solve the problem.
Its no more confusing, to find it we should have to apply the formula of Pythagorean triplates and to find the length the of the sides produced by the distance between the diameter and the chords, we have to subtract the length of the chords from the length of the diameters.
Don't forget that, diameter is the largest chord of the circle.
Refer to the attachments for details and solution.
