Name: Perimeter and area November 14, 2011
1. How many differently shaped rectangles with whole number sides could have an area of 360?
2. Find the area of this triangle. Assume that each of the small squares have an area of one square unit.
3. Assuming all angles are right angles, what is the area of the polygon?
4. Find the area of the shaded part of this figure.
5. If a rectangle’s length and width are both doubled, by what percent is the rectangle’s area increased?
6. The figure below is composed of a square and a regular hexagon. If the area of the square is 25, what is the perimeter of the entire figure?
7. The area of one small square is 1, and the area of the largest square is 100. What is the area of the quadrilateral region enclosed by heavier lines?
8. If the area of the figure below is 200 square units, what is the perimeter of the figure?
Cascade Ridge PTSA Math Club 3 Show your work on a separate sheet of paper.
Name: Perimeter and area November 14, 2011
BONUS PROBLEMS
9. A rectangle’s long side is five more than twice the length of the shorter side. If the perimeter is 100, what is the length of the longest side?
10. Mr. Smiley, as shown below, is made up of two circular eyes, each with radius one centimeter, and a semicircular mouth with radius 3 centimeters, inside a circular face with radius 10 centimeters. To the nearest square centimeter, how much of Mr. Smiley’s face is not taken up by his eyes or mouth?
11. The regular hexagon below is circumscribed by a circle with a circumference of 8π. Find the shaded area.
12. At a construction site, some workers are rolling a large spool of cable. If the spool is a cylinder with a diameter of 48 inches, how many yards will it roll in 12 revolutions along a smooth surface?
SOLUTIONS
1. 12
The area of a rectangle is (length × width). So we need to find all pairs of whole numbers (i.e., factors) that multiply to give 360. The 12 possibilities are:
1 × 360
2 × 180
3 × 120
4 × 90
5 × 72
6 × 60
8 × 45
9 × 40
10 × 36
12 × 30
15 × 24
18 × 20
2. 24 square units
The area of a triangle is (½ × base × height). In this case, the base is 6 (the horizontal line that forms the top of the triangle), and the height is 8. So, the area is (½ × 6 × 8) = 24
3. 448 square units
Draw a large rectangle that perfectly encompasses the polygon, and fill in the missing dimensions by adding or subtracting the other dimensions:
Now, take the area of the large encompassing rectangle, and subtract the areas of the three smaller rectangles (blue, green, and red) that have been “carved out” of the large rectangle.
Area of large rectangle = 40 × 16 = 640
Area of blue rectangle = 8 × 12 = 96
Area of green rectangle = 16 × 4 = 64
Area of red rectangle = 8 × 4 = 32
So, the area of the white polygon is 640 – 96 – 64 – 32 = 448 square units.
4. 60 sq. ft.
The area of the whole triangle would be ½ (18 x 12) = 108. The area of the unshaded rectangle is 48, so subtracting that from 108 leaves 60 sq. ft. for the shaded area.
5. 300% increase
Assume you had a 1 x 2 rectangle (area of 2) and doubled the length and width to make it a 2 x 4 rectangle (area of 8). To find the increase in area, find their differences (8 – 2 = 6). 6 is three times the original area of 2, or a 300% increase.
6. 40
If the area of the square is 25, then the length of its side must be 5. The length of the side of the hexagon is the same as the length of the side of the square. So, counting around the perimeter of the whole figure, there are 8 total sides, each with a length of 5. So 8 x 5 = 40.
7. 52 square units
The area of the complete square is 100. Subtract the areas of the 4 triangles at each corner, and you will be left with the area of the funny shaped quadrilateral. The four triangles are:
upper-left ½ x 5 x 8 20
upper-right ½ x 5 x 6 15
lower-left ½ x 2 x 7 7
lower-right ½ x 3 x 4 6
So, the area of the quadrilateral is 100 – 20 – 15 – 7 – 6 = 52.
8. 90 units
Since the figure is made up of 8 squares, and the total area is 200 square units, each square must have an area of 25 square units. Therefore, each square’s side must be 5 units long. Counting around the perimeter of the whole figure, there are 18 sides, each with length 5. So the total perimeter is 18 x 5 = 90 units.
9. 35
Using algebra, let’s call the short side x and the long side y. We know that 2x + 2y = 100. We also know that 2x + 5 = y. Transposing the 2nd equation, we get 2x = y – 5. Substituting y – 5 for 2x in the first equation, we get y – 5 + 2y = 100. Solving for y: 3y – 5 = 100; 3y = 105; y = 35.
10. 294 sq. cm.
The area of the large circle is 100π. Subtract the areas of the two eyes (1π each) and the semicircle mouth (4.5π) to get 93.5π. Using π=3.14, 93.5 x 3.1415 = 293.73, which rounds to 294 sq. cm.
11. 25.133 or 8π
Looking closely at the picture, we can tell that exactly ½ of the figure is shaded. Knowing that the circumference of the circle is 8π, using the formula C = 2πr, we can deduce that the circle’s radius is 4. Using the formula for the area of a circle (A=πr2), we find that the area of the entire circle is 16π, therefore ½ of the total area would be 8π (or 25.133).
12. 50.27 or 16π yards
If the cylinder’s diameter is 48 inches, its circumference is 48π inches. In 12 revolutions, it will travel 576π inches. 576π inches = 48π feet = 16π yards.
Cascade Ridge PTSA Math Club 3 Show your work on a separate sheet of paper.