Question

AB and CD are two parallel chords of a circle such that AB = 10 cm and CD = 24cm. If the chords are on the opposite sides of the centre and the distance between them is 17 cm, then what is the radius of the circle?

Answer 1

Given AB and CD are two chords of a circles on opposite sides of the centre.

Construction: Draw perpendiculars OE and OF onto AB and CD respectively from centre O.
AE = EB = 5cm and CF = FD = 12 cm
[ Perpendicular drawn to a chord from center bisects the chord]

Given,
Distance between two chords = 17 cm
Let distance between O and F $=x$ cm
And distance between O and E $=(17-x)cm$

In ΔOEB,
$O{B}^2=O{E}^2+E{B}^2$
[Pythagoras theorem]
$=(17-x{)}^2+5^2$ ---(1)

In ΔOFD,
$OD2=O{F}^2+F{D}^2$
[Pythagoras theorem]
$=(x{)}^2+{12}^2$----------→(2)

But OB = OD ( radii of the same circle).

From 1 & 2,
$(17-x{)}^2+5^2=(x{)}^2+{12}^2$
⇒ $289+x^2-34x+25=x^2+144$
⇒ $34x=170$
∴ $x=5$
Subsitute x in equation (2);
$O{D}^2=(5{)}^2+{12}^2=169$
$OD=13$
∴ Radius of the circle is 13 cm.