Madhya Pradesh Police Exam Model Paper 1

Given:
Length of each chord = 8 cm
Distance between chords = 6 cm
Formula:
Pythagoras Theorem
Hypotenuse = H, Base = B and Perpendicular = P
${\mathit{H}}^2={\mathit{B}}^2+{\mathit{P}}^2$
Calculation:AB and CD are two chords length of each is 8 cm.O is the center of the circle and M is mid point of AB
and Q is the midpoint of CD
Let OB = OD = r (radius of circle)
MQ = 6 cm (Given)
Let OM = x, then
OQ = (6 - x) cm
In right angled ΔMOB
$\mathit{O}{\mathit{B}}^2=\mathit{O}{\mathit{M}}^2+\mathit{M}{\mathit{B}}^2$
$\Rightarrow {\mathit{r}}^2={\mathit{x}}^2+4^2$
$\Rightarrow {\mathit{x}}^2={\mathit{r}}^2-16$ ----(1)
In right angled ΔQOD
$\mathit{O}{\mathit{D}}^2=\mathit{O}{\mathit{Q}}^2+\mathit{Q}{\mathit{D}}^2$
$\Rightarrow {\mathit{r}}^2={\left(6-\mathit{x}\right)}^2+4^2$
$\Rightarrow {\mathit{r}}^2=36+{\mathit{x}}^2-12\mathit{x}+16$
$\Rightarrow {\mathit{x}}^2={\mathit{r}}^2+12\mathit{x}-52$ ----(2)
From equation (1) and equation (2)
${\mathit{r}}^2-16={\mathit{x}}^2={\mathit{r}}^2+12\mathit{x}-52$
$\Rightarrow 12\mathit{x}=36$
$\mathit{x}=3$
Now,
${\mathit{r}}^2={\mathit{x}}^2+4^2$
$\Rightarrow {\mathit{r}}^2=3^2+4^2$
$\Rightarrow {\mathit{r}}^2=9+16$
$\Rightarrow \mathit{r}=\sqrt{25}$
$\Rightarrow \mathit{r}=5$
∴ Length of radius of the circle is 5 cm.