SSC CHSL Model Paper 16

Given chords AB = 6 cm, CD = 12 cm and AB ∥ CD
Draw OP ⊥ AB. Let it intersect CD at Q and AB at P
∴ AP = PB = 3 cm and CQ = DQ = 6 cm
[Since perpendicular drawn from the centre of the chord bisects the chord]
Let OD = OB = r
In right ΔOQD,${\mathit{r}}^2={\mathit{x}}^2+62$ [By Pythagoras theorem ]
${\mathit{r}}^2={\mathit{x}}^2+36$ → (1)
In right ΔOPB, ${\mathit{r}}^2={\left(\mathit{x}+3\right)}^2+32$ [By Pythagoras theorem ]
${\mathit{r}}^2={\mathit{x}}^2+6\mathit{x}+9+9$ $={\mathit{x}}^2+6\mathit{x}+18$ → (2)
From (1) and (2) we get
${\mathit{x}}^2+36={\mathit{x}}^2+6\mathit{x}+18$
⇒ $6\mathit{x}=18$
x = 3
Put x = 3 in (1), we get
${\mathit{r}}^2=32+36=9+36=45$
∴ $\mathit{r}=\sqrt{4}5=3\sqrt{5}$ cm