How to find the area of a figure - HSPT Math

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Question

A square with a side length of 4 inches is inscribed in a circle, as shown below. What is the area of the unshaded region inside of the circle, in square inches?

Answer

Using the Pythagorean Theorem, the diameter of the circle (also the diagonal of the square) can be found to be 4√2. Thus, the radius of the circle is half of the diameter, or 2√2. The area of the circle is then π(2√2)2, which equals 8π. Next, the area of the square must be subtracted from the entire circle, yielding an area of 8π-16 square inches.

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Question

Assume π = 3.14

A man would like to put a circular whirlpool in his backyard. He would like the whirlpool to be six feet wide. His backyard is 8 feet long by 7 feet wide. By state regulation, in order to put a whirlpool in a backyard space, the space must be 1.5 times bigger than the pool. Can the man legally install the whirlpool?

Answer

If you answered that the whirlpool’s area is 18.84 feet and therefore fits, you are incorrect because 18.84 is the circumference of the whirlpool, not the area.

If you answered that the area of the whirlpool is 56.52 feet, you multiplied the area of the whirlpool by 1.5 and assumed that that was the correct area, not the legal limit.

If you answered that the area of the backyard was smaller than the area of the whirlpool, you did not calculate area correctly.

And if you thought the area of the backyard was 30 feet, you found the perimeter of the backyard, not the area.

The correct answer is that the area of the whirlpool is 28.26 feet and, when multiplied by 1.5 = 42.39, which is smaller than the area of the backyard, which is 56 square feet.

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Question

A manufacturer makes wooden circles out of square blocks of wood. If the wood costs $0.25 per square inch, what is the minimum waste cost possible for cutting a circle with a radius of 44 in.?

Answer

The smallest block from which a circle could be made would be a square that perfectly matches the diameter of the given circle. (This is presuming we have perfectly calibrated equipment.) Such a square would have dimensions equal to the diameter of the circle, meaning it would have sides of 88 inches for our problem. Its total area would be 88 * 88 or 7744 in2.

Now, the waste amount would be the "corners" remaining after the circle was cut. The area of the circle is πr2 or π * 442 = 1936π in2. Therefore, the area remaining would be 7744 – 1936π. The cost of the waste would be 0.25 * (7744 – 1936π). This is not an option for our answers, so let us simplify a bit. We can factor out a common 4 from our subtraction. This would give us: 0.25 * 4 * (1936 – 484π). Since 0.25 is equal to 1/4, 0.25 * 4 = 1. Therefore, our final answer is: 1936 – 484π dollars.

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Question

A manufacturer makes wooden circles out of square blocks of wood. If the wood costs $0.20 per square inch, what is the minimum waste cost possible for cutting a circle with a radius of 25 in.?

Answer

The smallest block from which a circle could be made would be a square that perfectly matches the diameter of the given circle. (This is presuming we have perfectly calibrated equipment.) Such a square would have dimensions equal to the diameter of the circle, meaning it would have sides of 50 inches for our problem. Its total area would be 50 * 50 or 2500 in2.

Now, the waste amount would be the "corners" remaining after the circle was cut. The area of the circle is πr2 or π * 252 = 625π in2. Therefore, the area remaining would be 2500 - 625π. The cost of the waste would be 0.2 * (2500 – 625π). This is not an option for our answers, so let us simplify a bit. We can factor out a common 5 from our subtraction. This would give us: 0.2 * 5 * (500 – 125π). Since 0.2 is equal to 1/5, 0.2 * 625 = 125. Therefore, our final answer is: 500 – 125π dollars.

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Question

Mary has a decorative plate with a diameter of ten inches. She places the plate on a rectangular placemat with a length of 18 inches and a width of 12 inches. How much of the placemat is visible?

Answer

First we will calculate the total area of the placemat:

$A=l\times w= 18\times 12= 216\hspace{1 mm}inches^2$

Next we will calculate the area of the circular place

$A=\pi r^2$

And

$d=2r=10$

So

$r=5\hspace{1 mm}inches$

$A=\pi r^2=\pi (5^2)=25\pi\hspace{1 mm}inches^2$

We will subtract the area of the plate from the total area

$216-25\pi\hspace{1 mm}inches^2$

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Question

What is the area of a circle with a radius of 10?

Answer

The formula for the area of a circle is $\dpi{100} \pi r^{2}$

$\dpi{100} \pi r^{2}= \pi (10)^{2}=100\pi$

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Question

Allen was running around the park when he lost his keys. He was running around aimlessly for the past 30 minutes. When he checked 10 minutes ago, he still had his keys. Allen guesses that he has been running at about 3m/s.

If Allen can check 1 square kilometer per hour, what is the longest it will take him to find his keys?

Answer

Allen has been running for 10 minutes since he lost his keys at 3m/s. This gives us a maximum distance of from his current location. If we move 1800m in all directions, this gives us a circle with radius of 1800m. The area of this circle is

$1800^2\pi \approx 10,178,760m^2$

Our answer, however, is asked for in kilometers. 1800m=1.8km, so our actual area will be $\frac{10,178,760}{1,000,000} = 10.2$ square kilometers. Since he can search 1 per hour, it will take him at most 10.2 hours to find his keys.

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Question

What is the area of a circle with a radius of 7?

Answer

To find the area of a circle you must plug the radius into in the following equation $A=\pi(r^{2})$

In this case the radius is 7 so we plug it into to get $7^{2}=49$

We then multiply it by pi to get our answer $A=49\pi$

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Question

What is the area of a square with a side length of 8?

Answer

To solve this question you must know the formula for the area of a rectangle.

The formula is

In this case the rectangle is a square so we can plug in the side length for both base and height to yield

Perform the multiplication to arrive at the answer of

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Question

Refer to the above diagram, which shows an equilateral triangle with one vertex at the center of a circle and two vertices on the circle.

What percent of the circle (nearest whole number) is covered by the triangle?

Answer

Let be the radius of the circle. The area of the circle is

$A_{1} = \pi r^{2}$

is also the sidelength of the equilateral triangle. The area of the triangle is

$A_{2} = \frac{\sqrt{3}}{4} ; r^{2}$

The percent of the circle covered by the triangle is:

$\frac{A_{2} }{A_{1}} \cdot 100= \frac{\frac{\sqrt{3}}{4}r^{2}}{\pi r^{2}}\cdot 100 = \frac{100 \sqrt{3}}{4 \pi } \approx 14$

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Question

Julie wants to seed her rectangular lawn, which measures 265 feet by 215 feet. The grass seed she wants to use gets 400 square feet of coverage to the pound; a fifty-pound bag sells for $66.00, and a ten-pound bag sells for $20.00. What is the least amount of money Julie should expect to spend on grass seed?

Answer

The area of Julie's lawn is $265 \cdot215 = 56,975$ square feet. The amount of grass seed she needs is $56,975 \div 400 \approx 142.4$ pounds. This requires three fifty-pound bags, the most economical option since it is cheaper to buy a fifty-pound bag for $66 than five ten-pound bags for $100.00. Julie will spend $$66 \cdot3 = $198$

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Question

A square is 9 feet long on each side. How many smaller squares, each 3 feet on a side can be cut out of the larger square?

Answer

Each side can be divided into three 3-foot sections. This gives a total of $3\times3=9$ squares. Another way of looking at the problem is that the total area of the large square is 81 and each smaller square has an area of 9. Dividing 81 by 9 gives the correct answer.

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Question

Above is a figure that comprises a red square and a white rectangle. The ratio of the length of the white rectangle to the sidelength of the square is . What percent of the entire figure is red?

Answer

To make this easier, we will assume that the rectangle has length 5 and the square has sidelength 3. Then the area of the entire figure is

$(3 + 5) \times 3 = 24$ ,

and the area of the square is

$3 \times 3 = 9$

The square, therefore, takes up

$\frac{9}{24} \times 100 = \frac{900}{24} = 37.5 %$

of the entire figure.

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Question

Note: Figure NOT drawn to scale.

What percent of the above figure is white?

Answer

The large rectangle has length 80 and width 40, and, consequently, area

$80 \times 40 = 3,200$ .

The white region is a rectangle with length 30 and width 20, and, consequently, area

$30 \times 20 = 600$ .

The white region is

$\frac{600}{3,200} \times 100 = 18 \frac{3}{4} %$

of the large rectangle.

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Question

Find the area of a square if the length is .

Answer

The area of a square is:

Substitute the length and simplify.

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Question

Your backyard is wide and long, what is its area?

Answer

The area of a rectangle is

.

So for this you just multiple those two values together to get,

$A=L\cdot w$

.

Remeber that the units of area are squared.

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Question

If a trianlge has a base of and a height of , the area is . True or false?

Answer

The area of a triangle is

$A=\frac{1}{2}baseheight$ .

is equal to the base multiplied by the height.

So you still need to multiple by $\frac{1}{2}$ or divide by .

This gives you an actual area of .

Therefore the statement is false.

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Question

A square has area 64. Which of the following gives 25% of the length of one of its diagonals?

Answer

A square with area 64 has as the length of one of its sides the square root of this area, which is $\sqrt{64} = 8$ . The length of a diagonal of a square is $\sqrt{2}$ times this sidelength, so the square has diagonals of length $8 \sqrt{2}$ .

25% of this is

$8 \sqrt{2} \cdot \frac{25}{100} = \frac{8 \sqrt{2}}{1} \cdot \frac{1}{4} = \frac{8\sqrt{2}}{4} = 2 \sqrt{2}$ .

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Question

Above is a target. The radius of the smaller quarter-circles is half that of the larger quarter-circles.

A blindfolded man throws a dart at the above target. Assuming the dart hits the target, and that no skill is involved, what is the probability that the dart will land in a blue region?

Answer

Call the radius of one of the smaller quarter-circles 1 (the reasoning is independent of the actual radius). Then the area of each quarter-circle is

$\frac{1}{4} \cdot \pi r ^{2} = \frac{1}{4} \cdot \pi \cdot 1 ^{2} = \frac{1}{4} \pi$ .

Each of the four wedges of one such quarter-circle has area

$\frac{1}{4} \cdot \frac{1}{4} \pi = \frac{1}{16} \pi$ .

The radius of each of the larger quarter-circles is 2, so the area of each is $\frac{1}{4} \cdot \pi r ^{2} = \frac{1}{4} \cdot \pi \cdot 2 ^{2} = \pi$

Each of the three wedges of one such quarter-circle has area

$\frac{1}{3} \pi$ .

The blue regions comprise one larger wedge and one smaller wedge; their total area is

$\frac{1}{3} \pi + \frac{1}{16} \pi = \frac{16}{48} \pi + \frac{3}{48} \pi = \frac{19 \pi}{48}$

The total area of the target is

$\frac{1}{4}\pi + \frac{1}{4} \pi + \pi + \pi = \frac{5 \pi}{2}$

The blue regions together comprise

$\frac{19 \pi}{48} \div \frac{5 \pi}{2}$

$=\frac{19 \pi}{48} \cdot \frac{2}{5 \pi}$

$=\frac{19 }{24} \cdot \frac{1}{5 }$

$=\frac{19 }{120}$

of the area of the circle, making this the probability the dart will land in a blue region.

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Question

Above is a target. The radius of the smaller quarter-circles is half that of the larger quarter-circles.

A blindfolded man throws a dart at the above target. Assuming the dart hits the target, and that no skill is involved, give the odds against the dart landing in the yellow region.

Answer

Call the radius of one of the smaller quarter-circles 1 (the reasoning is independent of the actual radius). Then the area of each quarter-circle is

$\frac{1}{4} \cdot \pi r ^{2} = \frac{1}{4} \cdot \pi \cdot 1 ^{2} = \frac{1}{4} \pi$ .

Each of the four wedges of one such quarter-circle has area

$\frac{1}{4} \cdot \frac{1}{4} \pi = \frac{\pi}{16}$ .

The yellow region is one such wedge.

The radius of each of the larger quarter-circles is 2, so the area of each is $\frac{1}{4} \cdot \pi r ^{2} = \frac{1}{4} \cdot \pi \cdot 2 ^{2} = \pi$

The total area of the target is

$\frac{1}{4}\pi + \frac{1}{4} \pi + \pi + \pi = \frac{5 \pi}{2}$

Therefore, the yellow wedge is

$\frac{\pi}{16} \div \frac{5 \pi}{2} = \frac{\pi}{16} \cdot \frac{2}{5 \pi}= \frac{1}{8} \cdot \frac{1}{5 }= \frac{1}{40}$

of the target, and the odds against the dart landing in that region are 39 to 1.

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