Question

Two chords AB, CD of length 5 cm, 11 cm respectively of a circle are parallel. If the distance between AB and CD is 3 cm. find the radius of the circle.Assume that the chords are on the same side of the center.

• A. 5.02cm
• B. 6.02cm
• C. 7.02cm
• D. 8.02cm

Answer 1

The correct option is B

6.02cm

Given: Chord AB=5 cm, chord CD = 11 cm and AB $\parallel$ CD.

Perpendicular distance ML between AB and CD = 3 cm.

To find: Radius of the circle.

Construction:Join OB, OD and draw perpendicular bisectors OL of AB and OM of CD.

ln right angled triangle OMD, ${OD}^2$ = ${MD}^2$ + ${OM}^2$ [By PythagorasTheorem]

Let OM = x cm

$r^2$ = ${\left(\frac{11}{2}\right)}^2$ + $x^2$ ..........(i)
And In right triangle OLB, ${OB}^2$ = ${BL}^2$ + ${OL}^2$ [By Pythagorastheorem]

$r^2$ = ${\left(\frac{5}{2}\right)}^2$ + ${3+x}^2$ .........(ii)

From (i) and (ii), we get

$r^2$ = ${\left(\frac{11}{2}\right)}^2$ +$x^2$ - ${\left(\frac{5}{2}\right)}^2$ + ${3+x}^2$

$\frac{121}{4}$ + $x^2$ - $\frac{25}{4}$ + 9 + $x^2$ + 6x

$x^2$ - $x^2$ + 6x = $\frac{121}{4}$ - $\frac{25}{4}$ - 9

6x = $\frac{121-25-36}{4}$ - $\frac{60}{4}$ - 15

x = $\frac{15}{6}$ cm - $\frac{5}{2}$ cm

From (1), $r^2$ = ${\left(\frac{11}{2}\right)}^2$ + ${\left(\frac{5}{2}\right)}^2$ - $\frac{121+25}{4}$ - $\frac{146}{4}$

r = $\frac{\left(\sqrt{146}\right)}{2}$ cm