AMC 8 · 2006 · #5
Grade 5 geometry-2dProblem
Points A,B,C and D are midpoints of the sides of the larger square. If the larger square has area 60, what is the area of the smaller square?
Pick an answer.
AMC 8 2006 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: Inside a larger square, points $A, B, C, D$ are the midpoints of its four sides. Joining them in order makes a smaller (tilted) square. The larger square has area $60$. What is the area of the smaller square?
Givens: $A, B, C, D$ are midpoints of the four sides of the larger square; $ABCD$ is the smaller (tilted) square formed by connecting them; Area of the larger square $= 60$; Answer choices: (A) $15$, (B) $20$, (C) $24$, (D) $30$, (E) $40$
Unknowns: The area of the smaller square $ABCD$
Understand
Restated: Inside a larger square, points $A, B, C, D$ are the midpoints of its four sides. Joining them in order makes a smaller (tilted) square. The larger square has area $60$. What is the area of the smaller square?
Givens: $A, B, C, D$ are midpoints of the four sides of the larger square; $ABCD$ is the smaller (tilted) square formed by connecting them; Area of the larger square $= 60$; Answer choices: (A) $15$, (B) $20$, (C) $24$, (D) $30$, (E) $40$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #8 Use Symmetry
The picture is already given, but it pays off to add two more lines: the diagonals $AC$ and $BD$ of the smaller square. Those diagonals are horizontal and vertical, and they split the larger square into $8$ small right triangles that are all congruent by the symmetry of midpoints. Once we see those $8$ equal triangles, the answer comes from counting: $4$ of them tile the smaller square and $4$ tile the leftover corners, so the smaller square is exactly half of the larger. No algebra, no Pythagorean theorem.
Execute — Answer: D
4.G.A.1 Step 1 - Draw the two diagonals of the smaller square: segment $AC$ from top midpoint to bottom midpoint (vertical), and segment $BD$ from right midpoint to left midpoint (horizontal).
- They cross at the center of the larger square.
💡 Drawing the diagonals of a square is a Grade 4 "draw lines and segments" move, and it turns this picture into something we can count.
4.G.A.3 Step 2 - Notice what these two cuts do.
- Together with the four sides of the inner square, they slice the larger square into $8$ small right triangles.
- By the symmetry of taking midpoints, every one of those $8$ triangles is congruent.
💡 Rotating the picture $90^\circ$ around the center sends each triangle to another one, so they must all have the same area.
3.G.A.2 Step 3 - Count which triangles belong to the smaller square.
- The two diagonals $AC$ and $BD$ split the smaller square into $4$ of the $8$ triangles.
- The other $4$ triangles sit outside the smaller square, filling the four corners of the larger square.
💡 When equal pieces fill a whole, counting how many you have gives you the fraction — Grade 3 area-as-equal-parts reasoning.
5.NF.B.4 Step 4 Take half of the larger square's area to get the smaller square's area.
💡 Multiplying a whole-number area by the fraction $\tfrac{1}{2}$ is the Grade 5 "fraction of a quantity" step.
4.G.A.1 Draw the two diagonals of the smaller square: segment $AC$ from top midpoint to 4.G.A.3 Notice what these two cuts do. Together with the four sides of the inner square, 3.G.A.2 Count which triangles belong to the smaller square. The two diagonals $AC$ and $ 5.NF.B.4 Take half of the larger square's area to get the smaller square's area. Review
Reasonableness: Cross-check with a side length. If the larger square has side $s$, then $s^2 = 60$. Each side of the smaller square is the hypotenuse of a right triangle with legs $s/2$ and $s/2$, so its squared length is $(s/2)^2 + (s/2)^2 = s^2/2 = 30$. The area of the smaller square equals the square of its side, namely $30$. That matches answer (D) and confirms the "half of $60$" reasoning from the diagram. The answer also has to be less than $60$ and clearly more than a quarter of it, which rules out (A), (B), and (E) on sight.
Alternative: Tool #9 (Try an Easier Problem): pick a friendly larger square with side $2$, so its area is $4$. The midpoints are at $(1,0), (2,1), (1,2), (0,1)$. The smaller square has side $\sqrt{1^2 + 1^2} = \sqrt{2}$, so its area is $(\sqrt{2})^2 = 2$ — exactly half of $4$. The ratio (smaller : larger) $= 1:2$ holds for any size, so the answer for area $60$ is $60/2 = 30$.
CCSS standards used (min grade 5)
4.G.A.1Draw points, lines, line segments, rays, angles, and perpendicular and parallel lines (Adding the diagonals $AC$ and $BD$ to the given figure so the larger square gets cut into countable congruent pieces.)4.G.A.3Recognize a line of symmetry for a two-dimensional figure (Using the square's symmetry under $90^\circ$ rotation to conclude that all $8$ resulting right triangles are congruent.)3.G.A.2Partition shapes into parts with equal areas; express the area of each part as a unit fraction of the whole (Counting that $4$ of the $8$ equal triangles tile the smaller square, so the smaller square is $\tfrac{4}{8} = \tfrac{1}{2}$ of the larger square.)5.NF.B.4Apply and extend previous understandings of multiplication to multiply a fraction by a whole number (Computing $\tfrac{1}{2} \times 60 = 30$ to turn the fractional relationship into the final area.)
⭐ When midpoints of a square's sides are joined, the inner square is always half the area of the outer one. Adding two diagonals to the picture makes that fact countable instead of computable.
⭐ When midpoints of a square's sides are joined, the inner square is always half the area of the outer one. Adding two diagonals to the picture makes that fact countable instead of computable.
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