# High School: Geometry

### Modeling with Geometry HSG-MG.A.3

3. Apply geometric methods to solve design problems (e.g., designing an object or structure to satisfy physical constraints or minimize cost; working with typographic grid systems based on ratios).

Now we're getting to the fun part of geometry: playing with tangrams. Using those little plastic pieces to model neat designs, like a fox, a house, or even a swan. Unfortunately, those probably aren't the kind of design problems your students will be working on. We're talking about more about solving multi-step word problems.

Really, these types of problems differ depending on what students might be learning. It could involve calculating the maximum area of a cow's grazing plot that can be contained by 600 meters of fencing (22,500 square meters). Maybe they'll have to find the volume of a pizza box when given a surface area of 320 square inches, a height of 3 inches, and a square base (300 cubic inches). Or maybe they'll just stick to tangrams.

Whatever kinds of problems students face, they will need at least a basic knowledge of the shapes and the formulas that describe them. At the very least.

Students will also need the know-how to solve systems of equations, the mental fortitude to simplify algebraic expressions, and the plucky spirit of adventure to resourcefully draw on the vast pool of their geometric enlightenment as they pick and choose exactly which formulas will serve them best, and modify them as needed.

Tangrams have got nothing on these design problems.

#### Drills

1. A triangle has a perimeter of 100 centimeters and one side is 35 centimeters. The other two sides have a ratio of 5:8. What is the length of the longest side of the triangle?

40 centimeters

The first side is 35 cm, leaving the other two to be 65 cm combined, ruling out (A). Splitting 65 into 5 + 8 = 13 means each part is 65 cm ÷ 13 = 5 centimeters long. The two parts must be 5 cm × 5 = 25 cm and 5 cm × 8 = 40 cm. All told, the three sides are 25 cm, 35 cm, and 40 cm. Choice (C) was just a fluke in case the students were like, "Hey, maybe the given side is the longest."

2. An isosceles trapezoid has perimeter of 42 inches. Each of the congruent nonparallel sides is 5 inches long, and the trapezoid is 3 inches tall. How long are the two parallel sides?

12 inches and 20 inches

With the given information, we can draw this picture:

Using the Pythagorean Theorem, we can find that b1 and b3 are each 4 inches long (because 52 = 32 + 42). This leaves t and b2, which are congruent. If we add the sides we already know, they total 5 + 5 + 4 + 4 = 18, leaving 42 – 18 = 24 inches unaccounted for. These are split evenly between the top t and base b2, meaning each is 12 inches long. To find the total length of the base, we add b1 + b2 + b3 = 4 + 12 + 4 = 20 inches. So our two parallel sides are 12 inches and 20 inches long.

3. A cereal company is redesigning its cereal boxes to make them stand out on the supermarket shelves. For some unknown reason, they went with a triangular prism for the shape of their box. (It certainly stands out from the crowd, but it's so hard to pour cereal out of!) The front and back are isosceles triangles with base 10 inches and height 12 inches. The surface area of the entire box is 384 square inches. What is the depth of the box?

4 inches

The front and back are congruent, per the definition of the prism, and they both have an area of 2(½ × 10 in × 12 in) = 120 in2. This leaves the congruent left and right sides as well as the bottom of the box to have a combined area of 144 square inches. By the Pythagorean Theorem, we know that the longer edge of the triangular sides is 13 inches. If we call the depth x, then the left and right sides each have an area of 13x and the bottom has an area of 10x. That is, 144 = 13x + 13x + 10x = 36x, making x = 4 inches.

4. Shmikea, the newly imported Norwegian superstore, sells beautiful 6-foot round shag rugs. Borg's living room is a 12 by 18 foot rectangle, and his goal is to cover as much of the floor as possible with these rugs without them overlapping. How much of the floor space will be left uncovered if he goes through with his plan? (Use the approximation π = 3.14.)

46.44 ft2

Borg's maximum floor coverage will come from creating a 2 × 3 array of rugs in the space, since two will fit in the 12-foot dimension and three will fit in the 18-foot dimension. Each rug has an area of A = πr2 = 3.14 × (3 ft)2 = 28.26 ft2, so six rugs will have an area of 6 × 28.26 ft2 = 169.56 ft2. The whole room is 12 ft × 18 ft = 216 ft2, so the uncovered portion will be 216 ft2 – 169.56 ft2 = 46.44 ft2.

5. A cube has surface area of 486 square feet. What is its volume?

729 cubic feet

By definition, the cube has six congruent faces and edges of length l. Also by the definition of a cube, SA = 6l2, or 486 ft2 = 6l2. Simplify and solve for l = 9 ft. Volume of a cube is V = l3, or V = (9 ft)3 = 729 cubic feet.

6. A major convention center, whose floors are covered by square carpet tiles with side lengths of 9 inches laid out in a 48 by 96 array, held a wild conference over the weekend. A local cleaning company was hired to clean the carpet tiles, given an offer of \$0.05 per carpet square. The cleaners accepted the job, but not until after the convention center agreed to their counter of \$0.10 per square foot. The cleaners were happy, because they knew they would be getting paid more than the convention center originally offered. How much more, exactly?

\$28.80

There are 96 × 48 = 4,608 carpet tiles, which would net the cleaners 4,608 × \$0.05 = \$230.40 at the conference center's offer of \$0.05 per tile. Those tiles cover a total of 4,608 × (0.75 ft)2 = 2,592 square feet, which means the cleaners will get 2,592 × \$0.10 = \$259.20 at their offer of \$0.10 per square foot. Their price earns them \$259.20 – \$230.40 = \$28.80 more than that of the convention center.

7. A rectangle's length equals four-thirds its width. Its perimeter equals four times its length minus ten. What are its dimensions?

15 wide by 20 long

We can set up the following equations based on the information in the question: l = 43w and P = 4l – 10. These can be rewritten 3l = 4w and 2l + 2w = 4l – 10. In both equations, put all variables on one side: 3l – 4w = 0 and 2l – 2w = 10. Solve this system of equations by subtracting multiplying second equation by 2 and subtracting it from the first, resulting in –l = -20, or l = 20. If we replace this for l in the first equation, we get 3(20) – 4w = 0 and can then solve for w = 15.

8. The city council has agreed to build a half-pipe in a neighborhood park. It will be 24 feet long, consisting of two 8-foot quarter-pipe sections separated by an 8-foot flat section. They'll be supporting the ramps with plywood cutouts. The half-pipe will be 12 feet long, with plywood supports every 16 inches. What will be the total area of plywood supporting the ramp sections of the structure? (Use the approximation π = 3.14.)

275.2 ft2

Each plywood cutout will be the result of cutting a quarter circle out of a square. The circle has radius of 8 ft, and the square has side length of 8 ft. So the area of each cutout will be A = r2 – ¼πr2 = (8 ft)2 – ¼π(8 ft)2 = 13.76 ft2. Two of these pieces occur together, meaning that every 16 inches there will be 27.52 ft2 of plywood. Spacing them 16 inches apart for a length of 12 ft × 12 in/ft = 144 inches means that there will be 10 supports (one at the front plus 14416 = 9), for a total of 275.2 square feet.

9. As the delivery person for a local bakery, it is your job to fold up the boxes that you deliver the cakes in. The baker has told you that you will need an open-top box that is 5 inches deep with a square base. It should have a volume of 720 cubic inches. But all you have is one big piece of cardboard. You will need to start by cutting it down to a square, and then you will cut smaller squares out of each corner and fold up the sides. How long should the sides of the square you start with be?

22 inches

The volume of the box will be V = lwh = 720 in3. Since we know the height of the box (5 inches) and that the bottom is a square, we can substitute the volume formula to be V = 5x2 = 720 ft3. Solving for x gives us x = 12 inches. However, in order to create this box from a square of cardboard, we'll have to account for twice the depth as well. That means we'll need a cardboard square with a side length of 12 in + 2(5 in) = 22 inches.

10. Bruiser's 10-foot long leash is tethered to the corner of his expansive 7 by 5 foot dog-mansion. How much outdoor territory can he mark? (Use the approximation π = 3.14.)