Competition · AMC preparation · step 4 of 4
AMC 8 · 2011 · #16
Grade 8 geometry-2dPick an answer.
AMC 8 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Only side lengths are given, so Tool #1 (Draw a Diagram) is the natural first move: sketch each isosceles triangle and drop the altitude to the unequal side. That altitude bisects the base and creates two right triangles. Tool #7 (Identify Subproblems) then splits the task — compute A from its right-triangle pieces, compute B from its right-triangle pieces, and compare. With a right triangle in each picture, the Pythagorean theorem gives the height and the area follows from 1/2 × base × height.
Split the first triangle
Sketch triangle A (25, 25, 30); the altitude bisects the base into halves of 15, making right triangles with hypotenuse 25 and leg 15.
An isosceles triangle has a line of symmetry through the apex, and the altitude to the unequal side IS that line — a Grade 4 symmetry idea.
4.G.A.3Draw A DiagramFind the first height
In that right triangle, apply the Pythagorean theorem (leg² + leg² = hyp²) to get the height h_A = 20.
Finding an unknown leg of a right triangle from the other two sides is exactly Grade 8 Pythagorean-theorem work.
In triangle A, the altitude drawn to the base of length 30 measures 20.
▸ Why?
The altitude, together with half of the base and one of the 25-length sides, forms a right triangle in which the altitude and the half-base 15 are the two legs and the 25-side is the hypotenuse.
▸ Why?
The altitude from the apex of an isosceles triangle meets the base at a right angle and cuts it into two equal halves of 15, because folding the triangle along that altitude lays one 25-side exactly onto the other and one base-half exactly onto the other.
▸ Why?
In that right triangle the legs and hypotenuse must satisfy 15² + h² = 25², so h² = 625 - 225 = 400 and the only positive length is h = 20.
▸ Why?
In a right triangle the two legs' squares always add up to the hypotenuse's square, and here the legs are 15 and h while the hypotenuse is 25, so 15² + h² = 25².
Compute the first area
Compute area A from base and height: × 30 × 20 = 300.
Area = 1/2 × base × height for a triangle is Grade 6 geometry.
6.G.A.1Identify SubproblemsFind the second height
Do the same for triangle B (25, 25, 40): the altitude splits 40 into halves of 20, so the Pythagorean theorem gives height h_B = 15.
Same Grade 8 Pythagorean move — the right triangle now has legs 20 and h_B with hypotenuse 25.
8.G.B.7Identify SubproblemsCompare the two areas
Area B = × 40 × 15 = 300 — the same as A, so the two areas are equal, choice (C).
Both triangle areas come from the same 1/2 × base × height formula; comparing the two results is a direct Grade 6 step.
6.G.A.1Identify SubproblemsThis AMC 8 problem only needs Grade 8 Pythagorean-theorem reasoning you already know!
- Split the first triangle
- Find the first height
- Compute the first area
- Find the second height
- Compare the two areas
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