Question

Find the difference of the areas if two segments of a circle formed by a chord of length 5cm Subtending an angle of 90degree at the Centre ​

Answer 1

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

The figure is in the picture I've attached.

$\bf{Given}\begin{cases}\sf{AB\:(chord)=5\:cm}\\\sf{\angle AOB = {90}^{\circ}}\\ \sf{The\:angle\:is\:formed\:at\:the\:centre} \end{cases}$

$\bf{To\:Find:-}$

• The difference of the areas in both the segments

$\bf{Solution:-}$

Here, we are able to see that ∆ AOB is a right angle triangle.

So, Applying Pythagoras theorem,

$\tt{{OA}^{2}+{OB}^{2}={AB}^{2}}$

$\leadsto\tt{{(r)}^{2}+{(r)}^{2}={(5)}^{2}}$

$\leadsto\tt{{2(r)}^{2}={(5)}^{2}}$

$\leadsto\tt{{(r)}^{2}=\dfrac{25}{2}}$

$\leadsto\tt{r=\sqrt{\dfrac{25}{2}}}$

$\leadsto{\boxed{\tt{r=\dfrac{5}{\sqrt{2}}}}}$

Let us now find the area of the circle,

Formula = πr²

So, $\tt{\dfrac{22}{7}\times\dfrac{5}{\sqrt{2}}\times\dfrac{5}{\sqrt{2}}}$

$\leadsto\tt{\dfrac{11\:\:\:\cancel{22} \times 5 \times 5}{7 \times \cancel{2}}}$

$\leadsto\tt{\dfrac{275}{7}}$

$\leadsto{\boxed{\tt{39.285\:{cm}^{2}}}}$

We know that, Area of minor segment = Area of of sector AOB - Area of ∆ AOB.

So, $\tt{\dfrac{\pi {r}^{2} \theta}{360}-\dfrac{1\times 5\times5}{2\times\sqrt{2}\times\sqrt{2}}}$

$\leadsto\tt{\dfrac{39.185\times 90}{360}-6.248}$

$\leadsto\tt{9.821-6.248}$

$\leadsto{\boxed{\tt{3.573\:{cm}^{2}}}}$

Now, finding the area of major segment.

The area of major segment=Area of circle-Area of minor segment.

So, $\tt{39.285-3.573}$

$\leadsto{\boxed{\tt{35.712\:{cm}^{2}}}}$

Now, finding the difference between the areas of both the segments.

The final answer that is the difference between the areas of both the segments=Area of major segment-Area of minor segment.

$\leadsto\tt{35.712-3.573}$

$\huge\leadsto{\boxed{\red{\tt{32.139\:{cm}^{2}}}}}$

$\rule{200}4$

Answer 2

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