Area of a Circle - Examples, Exercises and Solutions

Remember that the formula for the area of a circle is

πR²

We replace the data we know:

π7²

π49

First we need the formula for the area of a circle:

 $\pi r^2$

In the question, we are given the diameter of the circle, but we still need the radius.

It is known that the radius is actually half of the diameter, therefore:

$r=7:2=3.5$

We substitute the value into the formula.

$\pi3.5^2=12.25\pi$

Remember that the formula for the area of a circle is

πR²

We insert the known data:

π3²

π9

First, let's remember what the formula for the area of a circle is:

$S=\pi r^2$

The problem gives us the diameter, and we know that the radius is half of the diameter therefore:

$\frac{13}{2}=6.5$

We replace in the formula and solve:

$S=\pi\times6.5^2$

$S=42.25\pi$

Since AB is just a chord and we know nothing else about the diameter or the radius, we cannot calculate the area of the circle.

Area of the circle:

$S=\pi r^2$

We insert the known data:

$25=\pi r^2$

Divide by Pi:$\frac{25}{\pi}=r^2$

Extract the root:$\sqrt{\frac{25}{\pi}}=r$

$\frac{5}{\sqrt{\pi}}=r$

Formula for the area of a circle:

$S=\pi r^2$

We complete the shape into a full circle and notice that 14 is the diameter.

A diameter is equal to 2 radii, so:$r=7$

We replace in the formula:$S=\pi\times7^2$

$S=49\pi$

The area of the entire shape equals the area of the rectangle plus the area of each of the semicircles.

Let's label each semicircle with a number:

Therefore, we can claim that:

The area of the entire shape equals the area of the rectangle plus 2A1+2A3

Let's calculate the area of semicircle A1:

$\frac{1}{2}\pi r^2$

$\frac{1}{2}\pi4^2=8\pi$

Let's calculate the area of semicircle A3:

$\frac{1}{2}\pi r^2$

$\frac{1}{2}\pi2^2=2\pi$

The area of the rectangle equals:

$4\times8=32$

Now we can calculate the total area of the shape:

$32+2\times8\pi+2\times2\pi=32+16\pi+4\pi=32+20\pi$

First, we add letters as reference points:

Let's observe points A and B.

We know that two tangent lines to a circle that start from the same point are parallel to each other.

Therefore:

$AE=AF=3$
$BG=BF=6$

From here we can calculate:

$AB=AF+FB=3+6=9$

Now we need the height of the parallelogram.

We know that F is tangent to the circle, so the diameter that comes out of point F will also be the height of the parallelogram.

It is also known that the diameter is equal to two radii.

It is known that the circumference of the circle is 25.13.

Formula of the circumference:$2\pi R$
We replace and solve:

$2\pi R=25.13$
$\pi R=12.565$
$R\approx4$

The height of the parallelogram is equal to two radii, that is, 8.

And from here it is possible to calculate the area of the parallelogram:

$\text{Lado }x\text{ Altura}$$9\times8\approx72$

Now, we calculate the area of the circle according to the formula:$\pi R^2$

$\pi4^2=50.26$

Now, subtract the area of the circle from the surface of the trapezoid to get the answer:

$72-56.24\approx21.73$