Older television screens have an aspect ratio of 4:3. That is, the ratio of the width to the height is 4:3. The aspect ratio of many movies is not 4:3, so they are sometimes shown on a television screen by "letterboxing" - darkening strips of equal height at the top and bottom of the screen, as shown. Suppose a movie has an aspect ratio of 2:1 and is shown on an older television screen with a 27-inch diagonal. What is the height, in inches, of each darkened strip?
Try it yourself first — the explanation is most useful after you’ve attempted it.
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Toolkit + CCSS Solution
Understand
Restated: A television screen has width-to-height ratio $4:3$ and a $27$-inch diagonal. A movie with width-to-height ratio $2:1$ is shown on it so the movie fills the full width, leaving two dark strips of equal height (top and bottom). Find the height of one strip.
Givens: The screen's width-to-height ratio is $4:3$; The screen's diagonal is $27$ inches; The movie's width-to-height ratio is $2:1$; The movie fills the full screen width, and the leftover dark area is split into two equal strips (top and bottom); Answer choices: (A) $2$, (B) $2.25$, (C) $2.5$, (D) $2.7$, (E) $3$
Unknowns: The height, in inches, of one darkened strip
Understand
Restated: A television screen has width-to-height ratio $4:3$ and a $27$-inch diagonal. A movie with width-to-height ratio $2:1$ is shown on it so the movie fills the full width, leaving two dark strips of equal height (top and bottom). Find the height of one strip.
Givens: The screen's width-to-height ratio is $4:3$; The screen's diagonal is $27$ inches; The movie's width-to-height ratio is $2:1$; The movie fills the full screen width, and the leftover dark area is split into two equal strips (top and bottom); Answer choices: (A) $2$, (B) $2.25$, (C) $2.5$, (D) $2.7$, (E) $3$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
The ratios $4:3$ and $2:1$ fix the shapes but not the actual sizes, so Tool #4 (Introduce a Variable) names one scale factor $k$ and writes every length in terms of it. The picture (Tool #1) shows a right triangle made by the width, height, and diagonal, which turns the single number $27$ into an equation for $k$. Then Tool #7 (Identify Subproblems) breaks the goal into a chain: find the screen size, find the movie height, then find the leftover and halve it.
Execute — Answer: D
#4 Introduce a Variable 6.RP.A.3Step 1
Name the screen size with one variable
The width and height are in the ratio $4:3$, so write the width as $4k$ and the height as $3k$ for some positive number $k$.
Using the same $k$ keeps the ratio exactly $4:3$ no matter what $k$ turns out to be.
$$\text{width} = 4k, \quad \text{height} = 3k$$
💡 One scale factor times the ratio numbers rebuilds every actual length while locking the shape in place.
#7 Identify Subproblems 8.G.B.7Step 2
Use the diagonal to solve for k
The width, height, and diagonal of the rectangle form a right triangle, so by the Pythagorean Theorem $(4k)^2 + (3k)^2 = 27^2$.
That gives $16k^2 + 9k^2 = 25k^2$, so the diagonal is $\sqrt{25k^2} = 5k$.
💡 Whatever height the movie does not use is dark, and the two equal strips each get half of it.
[1]
#4 6.RP.A.3The width and height are in the ratio $4:3$, so write the width as $4k$ and the
[2]
#7 8.G.B.7The width, height, and diagonal of the rectangle form a right triangle, so by th
[3]
#4 5.NBT.B.7Put $k = 5.4$ back in. The width is $4k = 4 \times 5.4 = 21.6$ inches and the he
[4]
#7 6.RP.A.3The movie fills the full width, so its width is also $21.6$ inches. Its ratio is
[5]
#7 5.NBT.B.7The movie is $10.8$ inches tall inside a screen that is $16.2$ inches tall, so t
Review
Reasonableness: The strip height $2.7$ is small next to the $16.2$-inch screen, which fits two thin bands framing a tall movie. Checking the total: two strips of $2.7$ plus the movie's $10.8$ gives $2.7 + 10.8 + 2.7 = 16.2$, exactly the screen height. The figure's strips run from $0$ to $2.7$ and from $13.5$ to $16.2$, each $2.7$ tall, confirming (D).
Alternative: Skip solving for $k$ by using ratios directly. Height and width are in ratio $3:4$, so height $= \tfrac{3}{4}\times$ width; the movie height is $\tfrac{1}{2}\times$ width. The gap between them is $\left(\tfrac{3}{4} - \tfrac{1}{2}\right)\times$ width $= \tfrac{1}{4}\times$ width, and one strip is half of that, $\tfrac{1}{8}\times$ width. Since width $= \tfrac{4}{5}\times 27 = 21.6$, one strip $= \tfrac{21.6}{8} = 2.7$, again (D).
CCSS standards used (min grade 8)
6.RP.A.3 Use ratio and rate reasoning to solve real-world problems (Turning the $4:3$ and $2:1$ ratios into actual lengths, and getting the movie height as half its width.)
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Relating the $4k$ width, $3k$ height, and $27$-inch diagonal to solve $5k = 27$ for the scale factor.)
5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths (Computing $4\times5.4$, $3\times5.4$, $21.6\div2$, and $(16.2-10.8)\div2$ to reach $2.7$.)
⭐ A $4:3$ screen is a scaled $3\text{-}4\text{-}5$ triangle, so a $27$-inch diagonal makes it $16.2$ tall; the $10.8$-tall movie leaves $5.4$ of darkness split into two $2.7$-inch strips.
⭐ A $4:3$ screen is a scaled $3\text{-}4\text{-}5$ triangle, so a $27$-inch diagonal makes it $16.2$ tall; the $10.8$-tall movie leaves $5.4$ of darkness split into two $2.7$-inch strips.
