Question

There is a circle with radius 6 cm. A chord is drawn in it. Find the angle subtended by the minor segment of the chord at the centre of the circle if the area of segment is 22.1 ${cm}^2$.

• A. 60
• B. 90
• C. 120
• D. 180

Answer 1

The correct option is C

120

(since OS is perpendicular to PQ)

OQ = OP (radius of the circle)

$\angle$OSP = $\angle$OSQ = ${90}^{\circ }$

OS is the common side to both the triangles

Thus, $\Delta$PSO is congruent to $\Delta$QSO by RHS congruence.

This means that corresponding $\angle$POS and $\angle$QOS are equal.

So, $\angle$POS = $\angle$QOS = $\theta$

$\angle$POQ =$\angle$POS +$\angle$QOS = $2\theta$

Also PS = SQ (corresponding sides of 2 congruent triangles)

Area of segment PSQR = Area of sector POQR - Area of $\Delta$POQ

Area of sector = $\frac{2\theta }{{360}^{\circ }}\times \pi r^2$ (Where r is radius)

= $\frac{2\theta }{{360}^{\circ }}\times \pi {\left(6\right)}^2$

= $0.63\theta$

In $\Delta$OSQ, $\angle$OSQ = ${90}^{\circ }$

$sin\theta$ = $\frac{opposite}{hypotenuse}$ (SQ is opposite side for angle θ since the opposite side has no common vertex with angle)

= $\frac{SQ}{OQ}$

$\Rightarrow$ $sin\theta$ = $\frac{SQ}{6}$

Hence, SQ = $6sin\theta$

Similarly, OS = $6cos\theta$

PQ = PS + SQ = 2SQ (Since PS = SQ)

⇒ PQ = $12sin\theta$

Area of triangle OPQ = $\frac{1}{2}\times base\times height$

= $\frac{1}{2}\times PQ\times OS$

= $\frac{1}{2}\times 12sin\theta \times 6cos\theta$

= $36\left(sin\theta \right)\left(cos\theta \right)$

Area of segment = Area of sector – Area of triangle

= $0.63\theta$ - $36\left(sin\theta \right)\left(cos\theta \right)$

$22.1$ = $0.63\theta$ - $36\left(sin\theta \right)\left(cos\theta \right)$ ...........(1)

Here we can get the value of $\theta$ by substituting values of answer options. But take care! The answer is not $\theta$. What is asked is $\angle$POQ = $2\theta$

By substituting $2\theta$ = ${120}^{\circ }$, $\theta$ = ${60}^{\circ }$

When $\theta$ = ${60}^{\circ }$ the above equation (1) is satisfied. Hence, the angle subtended by segment at centre is ${120}^{\circ }$.

Note that sin (${120}^{\circ }$) = sin (${90}^{\circ }$+${30}^{\circ }$), which is sin (${30}^{\circ }$) = $\frac{1}{2}$. This is because sin (${90}^{\circ }$ + $\theta$ ) = sin($\theta$)

sin (${180}^{\circ }$) = 0, when we substitute $2\theta$ as ${180}^{\circ }$