Back to Exercise: Apply geometry to design problems

Exercises: Apply Geometric Methods to Solve Design Problems

Show your work for each problem. When a problem asks you to round, round to the place the
problem names. Several problems ask you to explain a design decision rather than compute a
number — for those, write in complete sentences.

Grade 10·23 problems·~35 min·Common Core Math - HS Geometry·standard·hsg-mg-a-3

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Warm-Up: Review What You Know

These problems review formulas and ideas you have already learned.

A cylindrical container has radius $6$ cm and height $15$ cm. What is its volume, in cubic centimeters? Round to the nearest tenth.

A closed cylindrical can has radius $5$ cm and height $12$ cm. What is its total surface area, including the top and bottom, in square centimeters? Round to the nearest tenth.

A storage bin must fit on a shelf that is $40$ cm deep. Which inequality states this requirement, if $d$ is the depth of the bin?

$d \geq 40$ , because a bin needs to reach the back wall of the shelf in order to sit stably, so its depth has to be at least the shelf depth.

$d = 40$ , because a bin designed for a $40$ cm shelf is built to exactly that depth.

$d \leq 40$ , because the bin's depth can be anything up to the shelf depth but no more.

$40d \leq 1600$ , because the shelf's footprint sets a total area budget, and the bin's depth has to be scaled against that budget rather than against a single length.

Fluency Practice

A cylindrical juice container must hold $900$ cm $^3$ . If its radius is $6$ cm, what must its height be, in centimeters? Round to the nearest hundredth.

A designer wants a cylindrical can to hold $900$ cm $^3$ using as little metal as possible. Which quantity should the designer minimize?

The volume $\pi r^2 h$ , because a can that encloses less space is a smaller object overall, and smaller objects are built from less material.

The surface area $2\pi r^2 + 2\pi r h$ , because the metal forms the skin of the can and its area is what gets cut from the sheet.

The radius $r$ , because the circular top and bottom are the costly parts, and shrinking the radius shrinks both of them at once.

The height $h$ , because the curved side is a rectangle whose length is the height, so a shorter can needs a shorter strip of metal wrapped around it.

$A U-shaped curve of surface area against radius for a fixed-volume cylinder, with four marked points$

A cylindrical can must hold $500$ cm $^3$ . For each radius below, the height is found from $h = \dfrac{500}{\pi r^2}$ and the surface area from $SA = 2\pi r^2 + 2\pi r h$ . Complete the table by giving each surface area in square centimeters, rounded to the nearest tenth.

At $r = 3$ cm the surface area is ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ cm $^2$ . At $r = 4$ cm it is ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ cm $^2$ . At $r = 5$ cm it is ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ cm $^2$ . At $r = 6$ cm it is ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ ̲ cm $^2$ .

An open-top box has a square base of side $16$ cm and holds $2000$ cm $^3$ . How much material does it use, in square centimeters? Count the base and the four sides but not a top. Round to the nearest tenth.

$A poster divided into five equal columns separated by four narrow gutters, with the total width and one gutter width marked$

A poster is $36$ inches wide and is laid out on a $5$ -column grid. Each gap between neighbouring columns is $0.75$ inch, and the columns run to the left and right edges of the poster. How wide is each column, in inches?

A banner $26$ inches wide is split into two panels whose widths are in the golden ratio: the wider panel is $1.618$ times the narrower one, and together they fill the full width. How wide is the narrower panel, in inches? Round to the nearest hundredth.

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