AMC 10 · 2009 · #4
Grade 6 geometry-2dA rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remainder of the yard has a trapezoidal shape, as shown. The parallel sides of the trapezoid have lengths 15 and 25 meters. What fraction of the yard is occupied by the flower beds?
Pick an answer.
AMC 10 2009 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: A rectangular yard holds two flower beds, each a congruent isosceles right triangle sitting in the top corners. What is left is a trapezoid whose two parallel sides are $15$ meters (the shorter, top edge) and $25$ meters (the longer, bottom edge). Find what fraction of the whole yard the two triangular flower beds cover.
Givens: The yard is a rectangle; two flower beds are congruent isosceles right triangles; The leftover middle region is a trapezoid with parallel sides $15$ m and $25$ m; From the figure, the long side $25$ is the full bottom of the rectangle and the short side $15$ is the top between the two triangles; Answer choices: (A) $\frac{1}{8}$, (B) $\frac{1}{6}$, (C) $\frac{1}{5}$, (D) $\frac{1}{4}$, (E) $\frac{1}{3}$
Unknowns: The fraction (flower-bed area) / (whole yard area)
Understand
Restated: A rectangular yard holds two flower beds, each a congruent isosceles right triangle sitting in the top corners. What is left is a trapezoid whose two parallel sides are $15$ meters (the shorter, top edge) and $25$ meters (the longer, bottom edge). Find what fraction of the whole yard the two triangular flower beds cover.
Givens: The yard is a rectangle; two flower beds are congruent isosceles right triangles; The leftover middle region is a trapezoid with parallel sides $15$ m and $25$ m; From the figure, the long side $25$ is the full bottom of the rectangle and the short side $15$ is the top between the two triangles; Answer choices: (A) $\frac{1}{8}$, (B) $\frac{1}{6}$, (C) $\frac{1}{5}$, (D) $\frac{1}{4}$, (E) $\frac{1}{3}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #17 Visualize Spatial Relationships, #1 Draw a Diagram, #3 Eliminate Possibilities
The figure is a compound shape, so Tool #7 (Identify Subproblems) splits the work into three clean pieces: first find the triangle leg length from the two parallel sides, then find the total flower-bed area, then find the whole-yard area — and divide. Tool #17 (Visualize Spatial Relationships) supplies the key shortcut: the two congruent right triangles slide together into a single square, so their combined area is easy. Tool #1 (Draw a Diagram) reads the leg length straight off the picture, and Tool #3 (Eliminate Possibilities) checks the final fraction against the choices.
Execute — Answer: C
4.G.A.2 Step 1 Find each triangle's leg
- Along the top of the rectangle, the short parallel side ($15$) sits between the two triangles, while the bottom ($25$) is the full rectangle length.
- The two top legs must make up the difference: $25-15=10$ meters shared equally between the two congruent triangles, so each top leg is $10\div 2 = 5$ meters.
- Because each triangle is an isosceles right triangle, its two legs are equal, so the vertical leg is also $5$ — and that vertical leg is exactly the height (width) of the rectangle.
💡 An isosceles right triangle has two equal legs, so once you know one leg you know the other — and the height of the yard.
6.G.A.1 Step 2 Combine the two triangles into a square
- The two flower beds are congruent isosceles right triangles.
- Slide one over and turn it to meet the other along its slanted side, and together they fill a perfect $5\times 5$ square.
- So instead of computing two triangle areas, just take the area of that square: $5\times 5 = 25$ square meters of flower bed in total.
💡 Two congruent right triangles glued along their equal legs always make a square, turning a hard area into an easy one.
4.MD.A.3 Step 3 Find the whole yard's area
- The yard is a rectangle whose length is the long side $25$ and whose width is the triangle leg $5$ found in step 1.
- Multiply length by width to get the total area.
💡 A rectangle's area is just its length times its width.
4.NF.A.1 Step 4 Divide and simplify
- The fraction of the yard that is flower bed is the bed area over the yard area: $\dfrac{25}{125}$.
- Both numbers share the factor $25$, so $\dfrac{25}{125}=\dfrac{1}{5}$.
- Checking the choices, $\dfrac15$ is exactly choice (C); the other options do not match, so the answer is (C).
💡 Dividing numerator and denominator by the same number keeps the fraction's value while shrinking it to lowest terms.
4.G.A.2 Along the top of the rectangle, the short parallel side ($15$) sits between the 6.G.A.1 The two flower beds are congruent isosceles right triangles. Slide one over and 4.MD.A.3 The yard is a rectangle whose length is the long side $25$ and whose width is th 4.NF.A.1 The fraction of the yard that is flower bed is the bed area over the yard area: Review
Reasonableness: The two triangles fill just a $5\times 5=25$ patch of a $25\times 5=125$ yard, so the beds should be a small slice — and $\tfrac15 = 0.2$ is small, matching the picture where the corners are thin compared with the wide trapezoid. It also passes a sanity check: the trapezoid area is $\tfrac12(15+25)\cdot 5 = 100$, and $100+25 = 125$ equals the whole yard, so the pieces add up perfectly.
Alternative: Use the trapezoid directly. Its area is $\tfrac12(b_1+b_2)h = \tfrac12(15+25)(5) = 100$. Since the whole yard is $125$, the flower beds occupy $125-100 = 25$, and $\tfrac{25}{125} = \tfrac15$ — the same choice (C).
CCSS standards used (min grade 6)
4.G.A.2Classify two-dimensional figures based on presence of parallel or perpendicular lines (Using the isosceles-right-triangle property (two equal legs) to conclude both legs are $5$ and the rectangle height is $5$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Composing the two congruent right triangles into a $5\times 5$ square to get the total flower-bed area of $25$.)4.MD.A.3Apply area and perimeter formulas for rectangles in real-world problems (Computing the whole yard area as length times width, $25\times 5 = 125$.)4.NF.A.1Explain why a fraction is equivalent to another fraction (Simplifying $\dfrac{25}{125}$ to $\dfrac{1}{5}$ by dividing top and bottom by $25$.)
⭐ Two matching right triangles snap together into a square, so find that square's area, compare it to the whole rectangle, and reduce the fraction.
⭐ Two matching right triangles snap together into a square, so find that square's area, compare it to the whole rectangle, and reduce the fraction.
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