Two parallel chords of lengths 6 cm and 8 cm are on opposite sides of the centre of a circle. If the radius of the circle is 5 cm, find the distance between the midpoints of the chords.
Step-by-step solution
To find: The distance MN between the midpoints M of AB and N of CD
Idea: The line from the centre to the midpoint of a chord is perpendicular to the chord. So OM and ON are the distances of the chords from O, and each forms a right triangle with a radius and half the chord. As the chords are parallel and on opposite sides, M, O, N lie on one line.
- Let M, N be the midpoints of AB and CD. Then OM ⟂ AB and ON ⟂ CD (the line from the centre to the midpoint of a chord is perpendicular to it). AM = 3 cm and CN = 4 cm.½ mark
- In right ΔOMA: OM = √(OA2 − AM2) = √(52 − 32) = √16 = 4 cm.1 mark
- In right ΔONC: ON = √(OC2 − CN2) = √(52 − 42) = √9 = 3 cm.½ mark
- AB ∥ CD, so OM and ON are both perpendicular to the same direction and M, O, N lie on one straight line. The chords are on opposite sides of O, so MN = OM + ON = 4 + 3 = 7 cm.1 mark
Check: If the chords were on the same side of the centre, the distance would be 4 − 3 = 1 cm. Also, the longer chord (8 cm) is nearer the centre (3 cm) than the shorter one (4 cm), as it should be.
Answer to write in the exam
Let M, N be the midpoints of AB = 6 cm and CD = 8 cm; OA = OC = 5 cm.
OM ⟂ AB, ON ⟂ CD (line from centre to midpoint of a chord); AM = 3 cm, CN = 4 cm
OM = √(52 − 32) = √16 = 4 cm
ON = √(52 − 42) = √9 = 3 cm
AB ∥ CD, on opposite sides of O ⇒ M, O, N collinear and MN = OM + ON
∴ MN = 4 + 3 = 7 cm
Common mistakes that cost marks
- Using the full chord instead of half: √(52 − 62) is impossible. Use half the chord, 3 cm and 4 cm.
- Subtracting the distances (4 − 3 = 1 cm). That is for chords on the same side; here they are on opposite sides, so add.
- Pairing the wrong numbers: the 6 cm chord is at 4 cm, and the 8 cm chord is at 3 cm.
How this can come in the exam
In a circle of radius 5 cm, parallel chords of 6 cm and 8 cm lie on the same side of the centre. The distance between them is
- 1 cm
- 7 cm
- 2 cm
- 5 cm
Show answer
(A) 1 cm
Distances from the centre: 4 cm and 3 cm. On the same side, the gap is 4 − 3 = 1 cm.
In a circle of radius 25 cm, two parallel chords of lengths 14 cm and 48 cm lie on the same side of the centre. Find the distance between them.
Show answer
Distance of the 14 cm chord: √(252 − 72) = √576 = 24 cm (1 mark). Distance of the 48 cm chord: √(252 − 242) = √49 = 7 cm (1 mark). Same side: distance = 24 − 7 = 17 cm (1 mark).Try one yourself
In a circle of radius 17 cm, parallel chords of 16 cm and 30 cm lie on opposite sides of the centre. Find the distance between them.
Show answer
Distances: √(172 − 82) = 15 cm and √(172 − 152) = 8 cm. Opposite sides: 15 + 8 = 23 cm.
More questions like this
- Take a paper circle. Fold the circle from the boundary, inwards. Open the fold. The crease is now a chord (see the figure B). Now fold the paper again, so that the end points of the chord meet. Open the fold (see the figure C).
Measure the lengths of the parts into which the chord is divided. The chord gets bisected where the folds intersect. Measure the angle between the creases. The crease of the second fold is along the perpendicular from the centre to the chord. Measure the distance from the centre to the midpoint of the chord. It is the distance from the centre to the chord.
Now draw another chord of the same length. How will you do this? We will let you figure this out yourself. Join the centre to the midpoint of the new chord and measure its length. Is it the same as distance from the centre to the first chord? - Chords of a circle having the same length are all at the same distance from the centre of the circle.
- Use the Baudhāyana–Pythagoras theorem to show why chords of a circle having the same length are all at the same distance from the centre of the circle.
- Consider the figure. If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF.
- If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF. Solve this using the Baudhāyana–Pythagoras theorem.