Question

In a circle, O is the centre and $\angle$COD is a right angle. $AC=BD$ and $CD$ is the tangent at $P.$ Which of the following are true, if the radius of the circle is $1m?$

• A. 105 cm
• B. 141.4 cm
• C. 138.6 cm
• D. Can't be determined

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A

105 cm

B

141.4 cm

C

138.6 cm

D

Can't be determined

$AO=BO$ ...1 [radius]

$CA=BD$ ...2 [given]

Adding the equations 1 and 2:

$AO+AC=OB+BD$

$\Rightarrow OC=OD$

$\Rightarrow \angle ODC=\angle OCD$

Since, $\angle COD={90}^{\circ }$

Therefore, $\angle ODC=\angle OCD={45}^{\circ }$

Now, join OP.

OP is perpendicular to CD. [Line joining center to tangent at point of contact]

$\tan \angle ODP=\frac{OP}{PD}$

$\Rightarrow \tan {45}^{\circ }=\frac{100}{PD}$

$\Rightarrow 1=\frac{100}{PD}$

$\Rightarrow PD=100\text{cm}$

$\sin \angle ODP=\frac{OP}{PD}$

$\Rightarrow \sin {45}^{\circ }=\frac{100}{OD}$

$\Rightarrow OD=\frac{100}{0.7071}$

$\Rightarrow OD=141.42\text{cm}$

$BD=OD-OB$

$\Rightarrow BD=141.42–100$

$\Rightarrow BD=41.42\text{cm}$

$\therefore AC+CP=100+41.42=141.42\text{cm}$