Question

A boat in a circular lake lies at its centre . The perpendicular distance of the boat is $10$ m from a bridge lying in $40$ m distance across the circular lake. Find the distance that the boat will have to travel to reach to the extreme point of left side of bridge is $m\sqrt{5}$ m. Then, $m$ is

• A. $20$ m
• B. $10$ m
• C. $-10$ m
• D. both B & C

Answer 1

Answer: (B)

$10$ m

Given- $O$ is the centre of a circular lake.

The distance of the boat is $=ON=10$ m from a bridge $AB=40$ m.

$ON\perp AB$ at $N$

The distance of $A$ from $O$ is $m\sqrt{5}$ m

To find out- $m=?$

Solution-

$ON\perp AB$ at $N$ and $AB$ is a chord of the circle with centre $O$.

$\therefore AN=\frac{1}{2}\times AB=\frac{1}{2}\times 40$ m $=20$ m, since the perpendicular from the centre of a circle bisects the latter.

Now $ON\perp AB$ at $N$.

$\therefore \Delta AON$ is a right one with $OA$ as hypotenuse.

So, by pythagoras theorem, we have

${OA}^2={ON}^2+{AN}^2$

$\Rightarrow {\left(m\sqrt{5}\right)}^2={10}^2+{20}^2$

$\Rightarrow m=\pm 10$

Ans- Option B.