Question

Question 19
Find the difference of the areas of two segments of a circle formed by a chord of length 5 cm subtending an angle of ${90}^{\circ }$ at the centre.

Answer 1

Let the radius of the circle be r.

$\therefore$ OA=OB = r cm

Given that, length of chord of a circle, AB = 5cm

And central angle of the sector $AOBA\left(\theta \right)={90}^{\circ }$

Now, in $\Delta AOB(AB{)}^2=(OA{)}^2+(OB{)}^2$ [ by Pythagoras theorem]

$(5{)}^2=r^2+r^2$

$\Rightarrow 2r^2=25$

$\therefore r=\frac{5}{\sqrt{2}}cm$

[Since, the perpendicular drawn from the centre to the chord of a circle divides the chord into two equal parts]

By Pythagoras theorem, in $\Delta ADO$

$(OA{)}^2=O{D}^2+A{D}^2$

$\Rightarrow O{D}^2=O{A}^2-A{D}^2$

$={\left(\frac{5}{\sqrt{2}}\right)}^2-{\left(\frac{5}{2}\right)}^2=\frac{25}{2}-\frac{25}{4}$

$=\frac{50-25}{4}=\frac{25}{4}$

$\Rightarrow OD=\frac{5}{2}cm$

$\therefore Areaofanisosceles\Delta AOB$

= \frac{1}{2}\times Base (=AB)\times Height(=OD)\)

$=\frac{1}{2}\times 5\times \frac{5}{2}=\frac{25}{4}c{m}^2$

Now, area of sector AOBA $=\frac{\pi r^2}{{360}^{\circ }}\times \theta =\frac{\pi \times {\left(\frac{5}{\sqrt{2}}\right)}^2}{{360}^{\circ }}\times {90}^{\circ }$

$=\frac{\pi \times 25}{2\times 4}=\frac{25\pi }{8}c{m}^2$

$\therefore$ Area of minor segment = Area of sector AOBA - Area of an isosceles $\Delta AOB$

$=(\frac{25\pi }{8}-\frac{25}{4})c{m}^2$

Now, area of the circle $=\pi r^2=\pi \left(\frac{5}{\sqrt{2}}\right)=\frac{25\pi }{2}c{m}^2$

$\therefore$ Area of major segment = Area of circle - Area of minor segment

$=\frac{25\pi }{2}-(\frac{25\pi }{8}-\frac{25}{4})$

$=\frac{25\pi }{8}(4-1)+\frac{25}{4}$

$=(\frac{75\pi }{8}+\frac{25}{4})c{m}^2$

$\therefore$ Difference of the areas of two segments of a circle = [Area of minor segment]

$=|(\frac{75\pi }{8}+\frac{25}{4})-(\frac{25\pi }{4}-\frac{25}{4})|$

$=|(\frac{75\pi }{8}+\frac{25}{4})-(\frac{25\pi }{8}-\frac{25}{4})|$

$=|\frac{75\pi -25\pi }{8}+\frac{50}{4}|=|\frac{50\pi }{8}+\frac{50}{4}|$

$=(\frac{25\pi }{4}+\frac{25}{2})c{m}^2$

Hence, the required difference of the areas of two segments is $(\frac{25\pi }{4}+\frac{25}{2})c{m}^2$