circle Center o with pq parallel to RS. straight line GOF is perpendicular to PQ. RS is equal to 80 units PQ is equal to 60 units and radius of the cycle is 50 units. calculate the distance FG between the chords PQ and RS
First, let's draw the diagram with the given information:
- Center of the circle: O
- Radius of the circle: 50 units
- Chord RS: 80 units
- Chord PQ: 60 units
- Line GOF perpendicular to PQ
Since PQ is parallel to RS, we can see that triangle GOF is a right-angled triangle with GO as the hypotenuse.
Using the Pythagorean Theorem for the right-angled triangle GOF:
(GO)^2 = (GF)^2 + (FO)^2
We also know that FO = 50 units (radius of the circle) and PQ = 60 units.
Since FO is perpendicular to PQ, triangle FOA is also a right-angled triangle with hypotenuse OA as the radius.
Using Pythagorean Theorem for right-angled triangle FOA:
(OA)^2 = (OA)^2 + (FA)^2
(50)^2 = (60/2)^2 + (FA)^2
2500 = 900 + (FA)^2
(FB)^2 = 1600
FB = 40 units
Now, we can substitute FB = 40 units into the first equation:
(GO)^2 = (GF)^2 + (40)^2
(GO)^2 = (GF)^2 + 1600
Now, we know that GO = 50 units (radius of the circle), so:
(50)^2 = (GF)^2 + 1600
2500 = (GF)^2 + 1600
(GF)^2 = 900
GF = 30
Therefore, the distance FG between the chords PQ and RS is 30 units.