Area of a Circle

Consider the first figure and suppose that one semi-circle is cut from the center near the circumference in each direction. Then the semi-circle is spread out as in the second figure. The length $$AB$$ is the length of half of the circumference of the circle.


Suppose that the other half of the circle is cut in the same way and fitted into the first, as shown by the dashes in the second figure. It is evident that if we make a large number of cuts, the figure formed will approximate a rectangle whose length is equal to one-half of the circumference and whose width is equal to the radius. We note that the rectangle is $$r \times \pi r$$ or $$\pi {r^2}$$ and this rectangle has the same area as the circle.

Rule: The area of a circle equals $$\pi $$ times the square of the radius, or the area of a circle is equal to one-fourth $$\pi $$ times the square of the diameter.

            If $$A = $$area, $$C = $$circumference, $$d = $$diameter, $$r = $$radius, then

  1. $$A = \pi {r^2}$$ (if radius is given)
  2. $$A = \frac{\pi }{4}{d^2}$$ (if diameter is given)
  3. $$A = \frac{{Cr}}{2}$$ (if circumference and radius are given)



A reinforced concrete beam is to have a total area of steel of $$1.07$$ square cm. How many bars must be used if they are to be $$\frac{7}{8}$$ cm in diameter?


Given that the total area of reinforced concrete beam is $$ = 1.07$$ square cm — (1)
$$\therefore $$ Area of the circle of $$\frac{7}{8}$$cm. diameter $$ = \pi {r^2}$$
$$ = \frac{{22}}{7} \times {\left( {\frac{7}{{16}}} \right)^2}$$
$$ = 0.601$$ square cm — (2)
$$\therefore $$ Number of bars $$ = \frac{{{\text{total area}}}}{{{\text{area of one bar}}}} = \frac{{1.07}}{{0.601}} = 1.78 \approx 2$$ (approx)



A paper is in the form of a rectangle $$ABCD$$, where $$AB = 22$$cm and $$BC = 14$$cm. A semi-circular portion with $$BC$$ as the diameter is cut off. Find the area of the remaining paper.


             Area of rectangle $$ABCD = 22 \times 14 = 308$$ square cm
Area of semi-circle $$ = \frac{1}{2}\pi {r^2}$$
$$ = \frac{1}{2} \times \frac{{22}}{7} \times {(7)^2}$$
$$ = 11 \times 7 = 77$$ square cm
$$\therefore $$ Required area of the remaining paper $$ = 308 – 77 = 231$$ square cm.