AMC 10 · 2007 · #21
Grade 8 geometry-2dRight △ABC has AB=3,BC=4, and AC=5. Square XYZW is inscribed in △ABC with X and Y on AC,W on AB, and Z on BC. What is the side length of the square?
Pick an answer.
AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: In right triangle $ABC$ with legs $AB=3$ and $BC=4$ and hypotenuse $AC=5$, a square $XYZW$ is placed so that one whole side $XY$ lies flat on the hypotenuse $AC$, while the opposite two corners reach the legs: $W$ touches $AB$ and $Z$ touches $BC$. Find the side length of that square.
Givens: Right triangle with $AB=3$, $BC=4$, $AC=5$ (the right angle is at $B$, opposite the hypotenuse $AC$); Square $XYZW$ has side $XY$ on the hypotenuse $AC$; The other two vertices land on the legs: $W$ on $AB$ and $Z$ on $BC$; Answer choices: (A) $\frac{3}{2}$, (B) $\frac{60}{37}$, (C) $\frac{12}{7}$, (D) $\frac{23}{13}$, (E) $2$
Unknowns: The side length of the inscribed square
Understand
Restated: In right triangle $ABC$ with legs $AB=3$ and $BC=4$ and hypotenuse $AC=5$, a square $XYZW$ is placed so that one whole side $XY$ lies flat on the hypotenuse $AC$, while the opposite two corners reach the legs: $W$ touches $AB$ and $Z$ touches $BC$. Find the side length of that square.
Givens: Right triangle with $AB=3$, $BC=4$, $AC=5$ (the right angle is at $B$, opposite the hypotenuse $AC$); Square $XYZW$ has side $XY$ on the hypotenuse $AC$; The other two vertices land on the legs: $W$ on $AB$ and $Z$ on $BC$; Answer choices: (A) $\frac{3}{2}$, (B) $\frac{60}{37}$, (C) $\frac{12}{7}$, (D) $\frac{23}{13}$, (E) $2$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #4 Introduce a Variable, #1 Draw a Diagram
The square rests on the hypotenuse, so the useful measurement is not a leg but the height of the triangle measured perpendicular to that hypotenuse. That splits the work into two subproblems (Tool #7): first find the altitude from $B$ down to $AC$, then see how the square eats into that altitude. The altitude comes free from area (the same triangle area written two ways). For the second part, name the side $s$ (Tool #4): the top edge of the square cuts off a smaller triangle at the top that is a scaled copy of the whole triangle, and matching the scale factor gives one equation in $s$.
Execute — Answer: B
6.G.A.1 Step 1 Find the height onto the hypotenuse
- Because the square sits on $AC$, the direction that matters is straight out from $AC$.
- That distance is the altitude from the right-angle vertex $B$ to the hypotenuse.
- Get it from area.
- The area of the right triangle is $\frac12\cdot 3\cdot 4 = 6$ using the two legs as base and height.
- The same area is also $\frac12\cdot AC\cdot h = \frac12\cdot 5\cdot h$ using the hypotenuse as base and its altitude $h$.
- Setting the two equal, $\frac12\cdot 5\cdot h = 6$, so $h = \frac{12}{5}$.
💡 One triangle has one area, so measuring it from the hypotenuse must give the same $6$ — that pins down the height onto the hypotenuse.
8.G.A.4 Step 2 Match the small top triangle to the whole
- The side $WZ$ of the square is parallel to the hypotenuse and sits one square-side above it.
- Above the square is a smaller triangle with the same apex $B$ and the same angles as triangle $ABC$, so it is a scaled-down copy.
- Let $s$ be the square's side.
- The whole triangle has base $AC=5$ and height $h=\frac{12}{5}$.
- The small top triangle has height $\frac{12}{5}-s$ (the full height minus the square) and its base is exactly the top of the square, length $s$.
- Similar triangles keep the same base-to-height ratio, so $\frac{s}{\frac{12}{5}-s} = \frac{5}{\frac{12}{5}}$.
💡 The top edge of the square caps off a mini version of the same triangle, and a copy keeps the same width-to-height proportions.
8.EE.C.7 Step 3 Solve for the side length
- Cross-multiply the proportion: $s\cdot\frac{12}{5} = 5\left(\frac{12}{5}-s\right)$.
- The right side is $12 - 5s$.
- So $\frac{12}{5}s = 12 - 5s$.
- Bring the $s$ terms together by adding $5s$ to both sides: $\frac{12}{5}s + 5s = 12$.
- Writing $5s = \frac{25}{5}s$, the left side is $\frac{12+25}{5}s = \frac{37}{5}s$.
- Thus $\frac{37}{5}s = 12$, and multiplying both sides by $\frac{5}{37}$ gives $s = \frac{60}{37}$.
- That is choice (B).
💡 One proportion becomes one linear equation, and gathering the $s$ terms leaves the side length by itself.
6.G.A.1 Because the square sits on $AC$, the direction that matters is straight out from 8.G.A.4 The side $WZ$ of the square is parallel to the hypotenuse and sits one square-si 8.EE.C.7 Cross-multiply the proportion: $s\cdot\frac{12}{5} = 5\left(\frac{12}{5}-s\right Review
Reasonableness: The side must be smaller than the height onto the hypotenuse, $\frac{12}{5}=2.4$, and $\frac{60}{37}\approx 1.62$ is comfortably below that. A quick sanity check on the formula: for a square on a base $b$ with height $h$, the side is $\frac{bh}{b+h}$. Here $\frac{5\cdot\frac{12}{5}}{5+\frac{12}{5}} = \frac{12}{\frac{37}{5}} = \frac{60}{37}$, matching. The value also sits sensibly among the choices, all clustered between $1.5$ and $2$.
Alternative: Skip the altitude and use coordinates. Place $B=(0,0)$, $A=(3,0)$, $C=(0,4)$; the hypotenuse $AC$ has length $5$. The distance from $B$ to line $AC$ is $\frac{2\cdot\text{Area}}{AC}=\frac{12}{5}$, the same $h$. A square with a side on $AC$ and side length $s$ leaves a similar apex triangle of height $h-s$; equal ratios give $s=\frac{bh}{b+h}=\frac{60}{37}$ directly, confirming the answer without redrawing.
CCSS standards used (min grade 8)
6.G.A.1Find the area of triangles and other polygons by composing or decomposing shapes (Writing the triangle's area two ways to find the altitude onto the hypotenuse, $h=\frac{12}{5}$.)8.G.A.4Understand and use similarity of two-dimensional figures (Recognizing the small triangle above the square as a scaled copy of the whole triangle and setting equal base-to-height ratios.)8.EE.C.7Solve linear equations in one variable (Turning the similar-triangle proportion into $\frac{37}{5}s=12$ and solving for $s=\frac{60}{37}$.)
⭐ When a square sits on the slanted side, measure the triangle's height straight out from that side, then the leftover triangle on top is a smaller copy that hands you one equation.
⭐ When a square sits on the slanted side, measure the triangle's height straight out from that side, then the leftover triangle on top is a smaller copy that hands you one equation.
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