Question 19 of 50intermediate🔧 ApplyNumerical4 marks

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.

Correct Answer

7 cm

Exercise: Exercise Set 5.3 | Q: 3 | (Page 10)
For More Understanding

Explanation

According to Theorem 5 from the provided context, the perpendicular from the centre of a circle to a chord bisects the chord. This creates right-angled triangles where the radius is the hypotenuse, half the chord length is one leg, and the distance from the centre to the midpoint of the chord is the other leg. Using the relationship derived from the Pythagorean theorem (chord length = 2r2−d2\sqrt{r² - d²}), we can find the distance of each chord from the centre. For the 6 cm chord, the distance is 4 cm, and for the 8 cm chord, the distance is 3 cm. Since the chords are on opposite sides of the centre, the total distance between their midpoints is the sum of their individual distances from the centre.

Solution Steps

  1. Step 1: Let O be the centre of the circle. Let AB be the chord of length 6 cm and CD be the chord of length 8 cm. Let M and N be the midpoints of AB and CD, respectively. By Theorem 5, the perpendicular from the centre bisects the chord, so OM ⊥ AB and ON ⊥ CD.

  2. Step 2: In right ΔOAM, AM = 6/2 = 3 cm and radius OA = 5 cm. Using the Pythagorean theorem, OM = OA2−AM2\sqrt{OA² - AM²} = 52−32\sqrt{5² - 3²} = 25−9\sqrt{25 - 9} = 4 cm.

  3. Step 3: In right ΔOCN, CN = 8/2 = 4 cm and radius OC = 5 cm. Using the Pythagorean theorem, ON = OC2−CN2\sqrt{OC² - CN²} = 52−42\sqrt{5² - 4²} = 25−16\sqrt{25 - 16} = 3 cm.

  4. Step 4: Since the chords are on opposite sides of the centre, the distance between their midpoints is the sum of their distances from the centre: MN = OM + ON = 4 cm + 3 cm = 7 cm.

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