A chord whose length is equal to the radius of a circle is drawn to divide the circle into two parts. If the radius of the circle is 42 cm, then what is the area of the smaller part (in $$cm^{2}$$)?

Given a chord of length r, and the lines joining the centre and the ends of the chord are also radii, these lines form an equilateral triangle.

The area of the minor part will be the area of sector OAB minus the $$\triangle\ $$le OAB will be the required answer.

The area of sector = $$\frac{\theta}{360^o}\cdot\pi\ r^2$$

Where $$\theta\ $$=$$60^o$$

Area of sector = $$\frac{60^o}{360^o}\cdot\pi\ \cdot42^2$$ = $$\frac{\pi}{6}\ \cdot42^2$$

Area of triangle OAB = $$\frac{1}{2}\cdot b\cdot h\ =\ \frac{1}{2}\cdot42\cdot\frac{42\sqrt{\ 3}}{2}\ =\ 42^2\cdot\frac{\sqrt{\ 3}}{4}$$

Area of shaded region = Area of sector OAB - area of $$\triangle\ $$le OAB = $$\frac{\pi}{6}\ \cdot42^2 - \ 42^2\cdot\frac{\sqrt{\ 3}}{4}$$ = $$42^{2} (\frac{\pi}{6} - \frac{\sqrt{3}}{4})$$