AMC 10 · 2010 · #7
Grade 8 geometry-2dA triangle has side lengths 10, 10, and 12. A rectangle has width 4 and area equal to the
area of the triangle. What is the perimeter of this rectangle?
Pick an answer.
AMC 10 2010 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 triangle has sides 10, 10, and 12. A rectangle that is 4 wide covers the same area as this triangle. Find the perimeter of that rectangle.
Givens: Triangle side lengths are 10, 10, and 12; The rectangle has width 4; The rectangle's area equals the triangle's area
Unknowns: The perimeter of the rectangle
Understand
Restated: A triangle has sides 10, 10, and 12. A rectangle that is 4 wide covers the same area as this triangle. Find the perimeter of that rectangle.
Givens: Triangle side lengths are 10, 10, and 12; The rectangle has width 4; The rectangle's area equals the triangle's area
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #4 Introduce a Variable
The perimeter cannot be reached in one move. Break it into a chain: find the triangle's height, then its area, then use that area with the known width to get the rectangle's length, and finally the perimeter. Each subproblem is small and standard.
Execute — Answer: D
8.G.B.7 Step 1 Split the triangle in half
- The two sides of length 10 are equal, so the triangle is isosceles.
- Drop a straight line from the top corner down to the middle of the base of length 12.
- This cuts the base into two equal pieces of length 6 and creates two matching right triangles, each with a leg of 6 and a slanted side (hypotenuse) of 10.
💡 An isosceles triangle folds neatly in half, turning an awkward shape into two clean right triangles.
8.G.B.7 Step 2 Find the height with Pythagoras
- In one right triangle the legs are 6 and the unknown height h, and the hypotenuse is 10.
- Call the height h and apply the Pythagorean theorem: h squared plus 6 squared equals 10 squared.
- So h squared is 100 minus 36, which is 64, and h is 8.
💡 The right angle lets the two legs and the slanted side lock into one equation you can solve for the missing height.
6.G.A.1 Step 3 Compute the triangle's area
- Now use the base of 12 and the height of 8.
- The area of a triangle is one half the base times the height, so the area is one half of 12 times 8, which is 48.
💡 A triangle is exactly half of the rectangle that would box it in, so half of base times height gives the area.
4.MD.A.3 Step 4 Find the rectangle's length
- The rectangle has the same area, 48, and its width is 4.
- Since area equals width times length, the length is 48 divided by 4, which is 12.
💡 If you know the area and one side of a rectangle, dividing gives the other side.
3.MD.D.8 Step 5 Add up the perimeter
- The rectangle is 4 wide and 12 long.
- The perimeter is two widths plus two lengths, which is two times the sum of 4 and 12, giving 32.
- The answer is (D).
💡 A rectangle has two pairs of equal sides, so the perimeter is just twice the length plus twice the width.
8.G.B.7 The two sides of length 10 are equal, so the triangle is isosceles. Drop a strai 8.G.B.7 In one right triangle the legs are 6 and the unknown height h, and the hypotenus 6.G.A.1 Now use the base of 12 and the height of 8. The area of a triangle is one half t 4.MD.A.3 The rectangle has the same area, 48, and its width is 4. Since area equals width 3.MD.D.8 The rectangle is 4 wide and 12 long. The perimeter is two widths plus two length Review
Reasonableness: The height 8 fits, since 6-8-10 is a well-known right triangle. Area 48 and width 4 give length 12, which is larger than the width, as expected for a long thin rectangle. Perimeter 2(4+12)=32 matches choice (D). A quick sanity check: the rectangle's length 12 equals the triangle's base 12, which makes sense because the rectangle's height 4 is exactly half the triangle's height 8.
Alternative: Instead of Pythagoras, split the triangle into two 6-8-10 right triangles by recognizing the classic Pythagorean triple, which gives the height 8 immediately. Or use Heron's formula with semi-perimeter 16: area equals the square root of 16 times 6 times 6 times 4, which is the square root of 2304, namely 48. Either way the area is 48 and the perimeter is 32.
CCSS standards used (min grade 8)
8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the triangle's height from the 6-h-10 right triangle)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Computing the triangle's area as one half base times height)4.MD.A.3Apply area and perimeter formulas for rectangles in real-world problems (Turning the rectangle's area and width into its length)3.MD.D.8Solve real-world problems involving perimeters of polygons (Adding the rectangle's sides to get the perimeter)
⭐ Chase the area from shape to shape: find the triangle's area, hand it to the rectangle, and let width divide out the missing side.
⭐ Chase the area from shape to shape: find the triangle's area, hand it to the rectangle, and let width divide out the missing side.
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