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HSPT Math : How to find the area of a figure

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Example Question #21 : Geometry

On a map, one half of an inch represents miles of real distance.

Woodley County is a perfectly rectangular region which measures 45 miles along its northern border and 40 miles along its eastern border. In terms of , what is the area of the portion of the map that would be taken up by the county?

Possible Answers:

$\frac{N^{2}}{450}$

$\frac{450}{N^{2}}$

$\frac{1}{1,800 N^{2}}$

$1,800 N^{2}$

Correct answer:

$\frac{450}{N^{2}}$

Explanation:

One half of an inch represents miles of real distance, so one inch represents twice this, or miles.

If we let represent the width of the county as represented on the map, then, since the county is 45 miles wide, the proportion statement for map inches and real miles is

$\frac{X}{45}= \frac{1}{2N}$

$\frac{X}{45}\cdot 45 = \frac{1}{2N} \cdot 45$

$X= \frac{45}{2N}$ inches.

By similar reasoning, the height of the county as represented on the map is

$\frac{40}{2N} = \frac{20}{N}$ .

The area is the product of these dimensions, which is

$\frac{45}{2N} \cdot \frac{20}{N} = \frac{900}{2N^{2}}= \frac{450}{N^{2}}$

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Example Question #22 : Geometry

Spinner target

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 is thrown 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 inside one of the red regions?

Possible Answers:

$\frac{3}{4}$

$\frac{1}{2}$

$\frac{2}{3}$

$\frac{5}{9}$

Correct answer:

$\frac{1}{2}$

Explanation:

The answer to this question can be seen if we just rearrange the wedges of this target. If we exchange the upper right red wedge with the lower left green wedge, and the upper left green and blue wedges with the lower right red wedges, the target looks like this:

Spinner target

It can be seen that the probability of the dart landing in the red region of this target is $\frac{1}{2}$ , since exactly half this target is red. This is the same probability of the dart landing in the red region in the original target.

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Example Question #21 : Geometry

Spinner target

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?

Possible Answers:

$\frac{19 }{120}$

$\frac{7}{48}$

$\frac{7}{12}$

$\frac{19 }{60}$

Correct answer:

$\frac{19 }{120}$

Explanation:

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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Example Question #121 : Plane Geometry

What is the area of the figure below?

Possible Answers:

83in^2

73in^2

35in^2

45in&2

48in^2

Correct answer:

83in^2

Explanation:

To find the area of the figure above, we need to slip the figure into two rectangles.

12.5

Using our area formula, $A=l\times w$ , we can solve for the area of both of our rectangles

$A=6\times 8$ $A=7\times 5$

A=48in^2 A=35in^2

To find our final answer, we need to add the areas together.

48in^2+35in^2=83in^2

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Example Question #21 : Rectangles

What is the area of the figure below?

Possible Answers:

39in^2

18in^2

21in^2

42in^2

45in^2

Correct answer:

39in^2

Explanation:

To find the area of the figure above, we need to slip the figure into two rectangles.

11.5

Using our area formula, $A=l\times w$ , we can solve for the area of both of our rectangles

$A=7\times 3$ $A=6\times 3$

A=21in^2 A=18in^2

To find our final answer, we need to add the areas together.

21in^2+18in^2=39in^2

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Example Question #22 : Rectangles

What is the area of the figure below?

Possible Answers:

52in^2

28in^2

40in^2

26in^2

58in&2

Correct answer:

52in^2

Explanation:

To find the area of the figure above, we need to slip the figure into two rectangles.

10.5

Using our area formula, $A=l\times w$ , we can solve for the area of both of our rectangles

$A=8\times 3$ $A=7\times 4$

A=24in^2 A=28in^2

To find our final answer, we need to add the areas together.

24in^2+28in^2=52in^2

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Example Question #11 : Find Areas Of Rectilinear Figures: Ccss.Math.Content.3.Md.C.7d

What is the area of the figure below?

Possible Answers:

82in^2

36in^2

90in^2

57in^2

126in^2

Correct answer:

126in^2

Explanation:

To find the area of the figure above, we need to slip the figure into two rectangles.

9.5

Using our area formula, $A=l\times w$ , we can solve for the area of both of our rectangles

$A=10\times 9$ $A=9\times 4$

A=90in^2 A=36in^2

To find our final answer, we need to add the areas together.

90in^2+36in^2=126in^2

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Example Question #24 : Rectangles

What is the area of the figure below?

Possible Answers:

50in^2

15in^2

25in^2

35in^2

60in^2

Correct answer:

50in^2

Explanation:

To find the area of the figure above, we need to slip the figure into two rectangles.

8.5

Using our area formula, $A=l\times w$ , we can solve for the area of both of our rectangles

$A=7\times 5$ $A=5\times 3$

A=35in^2 A=15in^2

To find our final answer, we need to add the areas together.

35in^2+15in^2=50in^2

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Example Question #131 : Geometry

What is the area of the figure below?

Possible Answers:

68in^2

52in^2

16in^2

36in^2

42in^2

Correct answer:

52in^2

Explanation:

To find the area of the figure above, we need to slip the figure into two rectangles.

7.5

Using our area formula, $A=l\times w$ , we can solve for the area of both of our rectangles

$A=6\times 6$ $A=4\times 4$

A=36in^2 A=16in^2

To find our final answer, we need to add the areas together.

36in^2+16in^2=52in^2

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Example Question #26 : Rectangles

What is the area of the figure below?

6.5

Possible Answers:

15in^2

50in^2

35in^2

65in^2

25in^2

Correct answer:

65in^2

Explanation:

To find the area of the figure above, we need to slip the figure into two rectangles.

6.5

Using our area formula, $A=l\times w$ , we can solve for the area of both of our rectangles

$A=5\times 10$ $A=5\times 3$

A=50in^2 A=15in^2

To find our final answer, we need to add the areas together.

50in^2+15in^2=65in^2

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HSPT Math : How to find the area of a figure

Study concepts, example questions & explanations for HSPT Math

All HSPT Math Resources

Example Questions

Example Question #21 : Geometry

Example Question #22 : Geometry

Example Question #21 : Geometry

Example Question #121 : Plane Geometry

Example Question #21 : Rectangles

Example Question #22 : Rectangles

Example Question #11 : Find Areas Of Rectilinear Figures: Ccss.Math.Content.3.Md.C.7d

Example Question #24 : Rectangles

Example Question #131 : Geometry

Example Question #26 : Rectangles

All HSPT Math Resources

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