The diameter of a circle is 16 kilometers. Find the area of the circle.

Mathematics

1 answer:

Elden [556K]3 years ago

4 0

The diameter of a circle is 16 kilometers. Then the area of the circle is $201.062 \mathrm{km}^{2}$

Solution:

Given that , the diameter of a circle is 16 kilometers

We have to find the area of the circle .

Radius is half of diameter.

$\begin{array}{l}{\text { diameter }=2 \times \text { radius } \rightarrow \text { radius }=\frac{\text {diameter}}{2}} \\\\ {\rightarrow \text { radius }=\frac{16}{2} \rightarrow \text { radius }=8 \text { kilometers. }}\end{array}$

The area of circle is given as:

$\begin{array}{l}{\text { area of a circle }=\pi \times r^2 \\ {\text { Area of a circle }=\pi \times 8^{2}=\pi \times 64=64 \times 3.14=201.062}\end{array}$

Hence, the area of the circle is 201.062 $km^2$

You might be interested in

What SPC tool best indicates the shape of the data population and indicates if there are any gaps in the data

romanna [79]

The gaps in the area needs to be closed so the data you use is lenient

7 0

2 years ago

The measure of each interior angle of a regular polygon is 150 how many sides does the polygon have?

Dmitry [639]

The exterior and an interior angles always add up to 180 degrees.
180-150=One exterior angle
180-150=30
We know that all of the exterior angles add up to 360
360/30= the number of sides
360/30=12
The shape has 12 sides.

3 0

3 years ago

Identify whether the series sigma notation infinity i=1 15(4)^i-1 is a convergent or divergent geometric series and find the sum

const2013 [10]

Answer: The correct option is

(d) This is a divergent geometric series. The sum cannot be found.

Step-by-step explanation: The given infinite geometric series is

$S=\sum_{i=1}^{\infty}15(4)^{i-1}.$

We are to identify whether the given geometric series is convergent or divergent. If convergent, we are to find the sum of the series.

We have the D' Alembert's ratio test, states as follows:

Let, $\sum_{i=1}^{\infty}a_i$ is an infinite series, with complex coefficients $a_i$ and we consider the following limit:

$L=\lim_{i\rightarrow \infty}\dfrac{a_{i+1}}{a_i}.$

Then, the series will be convergent if L < 1 and divergent if L > 1.

For the given series, we have

$a_i=15(4)^{i-1},\\\\a_{i+1}=15(4)^i.$

So, the limit is given by

$L\\\\\\=\lim_{i\rightarrow \infty}\dfrac{a_{i+1}}{a_i}\\\\\\=\lim_{i\rightarrow \infty}\dfrac{15(4)^i}{15(4)^{i-1}}\\\\\\=\lim_{i\rightarrow \infty}\dfrac{15(4)^i}{15(4)^{i}4^{-1}}\\\\\\=\dfrac{1}{4^{-1}}\\\\=4>1.$