If I read the question correctly, the information about the 80x80 fencing and the grass region are all irrelevant information.

In order to find the area of the side walk we are going to use the formula A=pi(r^2) where r = the radius. The diameter of the pool is 60 feet and the the sidewalk is 5 feet bigger all around meaning we had 5 feet to both sides of this 60 feet giving you an overall diameter of 70 feet (or 60 + 5*2). To find the radius take half of the diameter. 70/2 = 35.

Now we plug in 35 to the equation. A = Pi(35^2) = 1225pi. Leave your answer like this as we are not finished and will need to round at the end for an accurate answer. This is NOT the end, that number is the area of the sidewalk AND the pool.

The sidewalk is not a full circle, just a border so we now have to find the area of the actual pool and subtract it from this 1225pi for the sidewalk's area. Same formula, Area = pi(r^2). The diameter of the pool is 60 meaning the radius is 30 (or 60/2). Plug into the equation, pi(30^2) = 900pi.

Now do 1225pi - 900pi to get 325pi. Break out your calculator and hit the approximation button. 325pi is approximately 1021.018 if rounded to 3 decimal places.

The area of the 5-foot wide sidewalk encircling the pool is approximately 1021.018 square feet.

5 0

3 years ago

PLEASE PLEASE ANSWER!

muminat

The area of the shaded region is $$8(\pi \ -\sqrt{3})\ \text{cm}^2$ .

Solution:

Given radius = 4 cm

Diameter = 2 × 4 = 8 cm

Let us first find the area of the semi-circle.

Area of the semi-circle = $\frac{1}{2}\times \pi r^2$

$$=\frac{1}{2}\times \pi\times 4^2$

$$=\frac{1}{2}\times \pi\times 16$

Area of the semi-circle = $$8\pi$ cm²

Angle in a semi-circle is always 90º.

∠C = 90°

So, ABC is a right angled triangle.

Using Pythagoras theorem, we can find base of the triangle.

$AC^2+BC^2=AB^2$

$AC^2+4^2=8^2$

$AC^2=64-16$

$AC^2=48$

$AC=4\sqrt{3}$ cm

Base of the triangle ABC = $4\sqrt{3}$ cm

Height of the triangle = 4 cm

Area of the triangle ABC = $\frac{1}{2}\times b \times h$

$$=\frac{1}{2}\times 4\sqrt{3} \times 4$

Area of the triangle ABC = $8\sqrt{3}$ cm²

Area of the shaded region

= Area of the semi-circle – Area of the triangle ABC

= $$8\pi \ \text{cm}^2-8\sqrt{3}\ \text{cm}^2$

= $$8(\pi \ -\sqrt{3})\ \text{cm}^2$

Hence the area of the shaded region is $$8(\pi \ -\sqrt{3})\ \text{cm}^2$ .

3 0

2 years ago

Describe the change in the graph of the parabola f(x) when it transforms into g(x) =

saw5 [17]

Answer:

(d) The parabola g(x) will open in the same direction of f(x), and the parabola will be wider than f(x).

Step-by-step explanation:

We assume you intend ...

f(x) = equation of a parabola

g(x) = 2/3·f(x)

Multiplying a function by a factor of 2/3 will cause it to be compressed vertically to 2/3 of its original height. When the function is a parabola, this has the effect of making it appear wider than before the compression.

The compression factor is positive, so points on the graph remain on the same side of the x-axis. The direction in which the graph opens is not changed.

The attachment shows parabolas that open upward and downward, along with the transformed version.

8 0

2 years ago

Statement: If two distinct planes intersect, then their intersection is a line. Which geometry term does the statement represent

klemol [59]

Answer:

Postulate.

Step-by-step explanation:

Postulate is a statement, also known as an axiom, which is taken to be true without proof.

8 0

3 years ago

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