Question

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

• A. BD = 41.42 cm
• B. OD = 141.42 cm
• C. PD = 100 cm
• D. AC + CP = 141.42 cm

Answer 1

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

AC + CP = 141.42 cm

Given that AC = BD.

OA = OB (radius)

Adding the above equations,

AC + OA = BD + OB

$⟹$OC = OD

$⟹\angle ODC=\angle OCD$ (Property of isosceles triangles)

Given $\angle COD={90}^{\circ }$

Therefore, $\angle ODC=\angle OCD={45}^{\circ }$ (Base angles are equal)

Join OP.

Since a tangent at any point of a circle is perpendicular to the radius at the point of contact, we have OP $\perp$ CD.

Consider right angled triangle ODP,
$tan\angle ODP=\frac{OP}{PD}$

$Tan{45}^{\circ }=\frac{100}{PD}(\because OP=1m=100cm)$

$⟹1=\frac{100}{PD}$

$\Rightarrow PD=100\text{cm}$
Similarly we can show that PC = 100 cm

Now, $sin\angle ODP=\frac{OP}{OD}$

$⟹sin{45}^{\circ }=0.7071=\frac{100}{OD}$

i.e., $OD=\frac{100}{0.7071}$

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

But, $BD=OD-OB$

$\Rightarrow BD=141.42–100=41.42\text{cm}$

$AC+CP=BD+100$ (since AC = BD and PC= 100 cm)
$=41.42+100=141.42cm$