Question

AD is a diameter of a circle and AB is a chord If AD = 34 cm, AB = 30 cm the distance of AB from the centre of the circle is

• A. 17 cm
• B. 15 cm
• C. 4 cm
• D. 8 cm

Answer 1

Answer: (D)

OX is the perpendicular distance from centre O to the chord AB

$\therefore AX=XB$

Now,

$AX+XB=AB$

$\Rightarrow 2AX=AB$

$\Rightarrow 2AX=30$

$\Rightarrow AX=\frac{30}{2}$

$\Rightarrow AX=15$

Again,

$AO+OD=AD$

$\Rightarrow 2AO=AD$

$\Rightarrow 2AO=34$

$\Rightarrow AO=17$

$△AOX$ is a right angled triangle

$\therefore A{O}^2=O{X}^2+A{X}^2$

$\Rightarrow {17}^2=O{X}^2+{15}^2$

$\Rightarrow O{X}^2={17}^2-{15}^2$

$\Rightarrow O{X}^2=\sqrt{64}$

$\Rightarrow OX=8$cm