GEOMETRY!!FIND THE AREA OF THE YELLOW REGION!! ROUND TO THE NEAREST TENTH EXPLAIN PLEASE!!

Mathematics

2 answers:

bixtya [17]3 years ago

6 0

Hi there!

In this problem, you want to solve for the area of the yellow region. To do that, you must find the area of the square and quarter-circle first. I am assuming this is a square just because they only give you one side length.

To find the area of a circle, you use the formula: $\pi$ r^2, where r is the radius of the circle. Since this is a quarter-circle (one-fourth of a circle), you would find the area normally then divide your answer by 4.

So the formula for a quarter-circle would be: 1/4 $\pi$ r^2. Now for the square; the formula for area of a square is s^2, where s is a side length of the square.

All sides are of equal length in a square, so that's why you can take any side of the square and multiply it by itself to find the area.

Now that we know how to solve for the area of a square and quarter-circle, let's substitute in the side length that we are given, 15 cm, into the area formulas. Let's solve for the area of a square first.

{Area of a square: s^2

Substitute in 15 for the side length:

(15)^2 = 225

The area of the square is 225 cm^2.}

Now let's solve for the area of the quarter-circle.

{Area of a quarter-circle: 1/4 $\pi$ r^2

Substitute in 15 for the radius.

1/4 $\pi$ (15)^2

1/4 $\pi$ (225)

1/4(706.9) = 176.7 (rounded to nearest tenth)

The area of the quarter-circle is 176.7 cm^2.}

To finish off this problem, you want to subtract the area of the quarter-circle from the area of the square. This makes sense because you have just found the area of the square, which holds both the yellow region and the circle.

The square has two parts inside of it, the quarter-circle and the yellow region. So subtracting the quarter-circle from the square leaves you with only the yellow region left, so that is why we needed to find the area of the square and quarter-circle first. Now let's do this:

Area of square - area of q-circle = area of yellow region

225 - 176.7 = 48.3

The area of the yellow region is $\boxed {48.3~cm^2}$ .