Question

AB = 36 cm and CD = 48 cm are two parallel chords of a circle with centre O. The distance between them is 42 cm. The radius of the circle is:

• A. 30 cm
• B. 27cm
• C. 25 cm
• D. 33 cm

Answer 1

The correct option is A 30 cm

Given: MN = 42 cm, AB = 36 cm, CD = 48 cm.
Since the perpendicular from the centre to a chord bisects it,
AM = BM = 18 cm and CN = DN = 24 cm

By Pythagoras Theorem.
$x^2+{18}^2=r^2...\left(i\right)and(42-x{)}^2+{24}^2=r^2...(ii)$
Thus, we have (i) = (ii).
$\Rightarrow x^2+{18}^2=r^2...\left(i\right)and(42-x{)}^2+{24}^2=r^2...(ii)$
Thus, we have (i) = (ii).
$\Rightarrow x^2+{18}^2={42}^2+x^2-84x+{24}^2$
$\Rightarrow 84x={42}^2+{24}^2-{18}^2=2,016$
$\Rightarrow x=24$
Thus, (i) $\Rightarrow r^2={24}^2+{18}^2=900$
$\Rightarrow r=30$
Hence, the radius of the circle is 30 cm.