Question

AB and CD are two chords of a circle such that AB = 6 cm, CD = 12 cm and $AB\parallel CD$. If the distance between AB and CD is 3 cm, find the radius of the circle. [4 MARKS]

Answer 1

Concept: 2 Marks
Application: 2 Marks


Let AB and CD be two parallel chords of a circle with centre O such that AB = 6 cm and CD = 12 cm.

Construction:
Join OB and OD

Draw perpendicular bisectors OL of AB and OM of CD. Since $AB\parallel CD$, the points O, M and L are collinear.

Clearly, LM = 3 cm. Let OM = x cm and let OB = OD = r.

From right triangle OLB, we get:
Using Pythagoras Theorem:

$O{B}^2=O{L}^2+B{L}^2$ or $r^2=(x+3{)}^2+3^2$ (i)

$(\because BL=\frac{1}{2}AB=3cm)$

From right triangle ODM, we get:
Using Pythagoras Theorem:

$O{D}^2=O{M}^2+M{D}^2$ or $r^2=x^2+6^2$ (ii)

$(\because MD=\frac{1}{2}CD=6cm)$

Thus, from (i) and (ii), we get:

$(x+3{)}^2+3^2=x^2+6^2$ or $6x=18$ or $x=3cm$

$\therefore r^2=x^2+6^2=3^2+6^2=45$ or $r=\sqrt{45}=6.7cm$

Hence, the radius of the circle is 6.7cm.