in Quantitative Aptitude edited by
143 views
3 votes

A chord of length $5$ cm subtends an angle of $60^\circ$ at the centre of a circle. The length, in cm, of a chord that subtends an angle of $120^\circ$ at the centre of the same circle is

  1. $8$
  2. $6\sqrt2$
  3. $5\sqrt3$
  4. $2\pi$
in Quantitative Aptitude edited by
by
12.8k points 250 1881 2459
143 views

1 Answer

2 votes
 
Best answer

Given that, a chord length $5 \; \text{cm}$ subtends an angle of $60^{\circ}$ at the centre of a circle.

Let the radius of a circle be $`r\text{’} \; \text {cm}.$

So, $\boxed{OA = OB = OC = r}$



In $\triangle OAB,$ two sides are equal $(OA = OB = r).$

So, $\angle A = \angle B = 60^{\circ}$

Thus, $\triangle ABC$ is an equilateral triangle.

$OABC$ is a rhombus.

Area of rhombus $= \frac{d_{1} \times d_{2}}{2} \; \text{sq. units},$ where $d_{1}, d_{2}$ are the diagonal of the rhombus.

Diagonals are perpendicular, to each other. $ \boxed{d^{2}_{1} + d^{2}_{2} = 4a^{2}} $

So, $ (OB)^{2} + (AC)^{2} = 4r^{2} $

$ \Rightarrow (5)^{2} + (AC)^{2} = 4(5)^{2} $

$ \Rightarrow 25 + (AC)^{2} = 100 $

$ \Rightarrow (AC)^{2} = 75 $

$ \Rightarrow AC = \sqrt{75} = \sqrt{25 \times 3} $

$ \Rightarrow AC = 5 \sqrt{3} \; \text{cm} $

$\therefore$ The length of a chord that subtends an angle of $120^{\circ}$ at the center of the circle is $5 \sqrt{3} \; \text{cm}.$

Correct Answer $:\text{C}$

edited by
by
4k points 3 6 24
Answer:

Related questions

2 votes
1 answer
3
jothee asked in Quantitative Aptitude Mar 20, 2020
179 views
jothee asked in Quantitative Aptitude Mar 20, 2020
by jothee
12.8k points 250 1881 2459
179 views
Ask
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true