AMC 8 · 2015 · #6
Easy mode Grade 8Problem
Picture a triangle ABC. Two of its sides have the same length: AB=29 and BC=29. The third side is AC=42.
What is the area of this triangle?
Diagram
Drop an altitude from B to base AC — symmetry splits the isosceles triangle into two equal right triangles.
Pick an answer.
AMC 8 2015 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: Triangle $ABC$ is isosceles with $AB = BC = 29$ and base $AC = 42$. Find its area.
Givens: $AB = BC = 29$ (the two equal legs); $AC = 42$ (the base); Answer choices: (A) $100$, (B) $420$, (C) $500$, (D) $609$, (E) $701$
Unknowns: The area of $\triangle ABC$
Understand
Restated: Triangle $ABC$ is isosceles with $AB = BC = 29$ and base $AC = 42$. Find its area.
Givens: $AB = BC = 29$ (the two equal legs); $AC = 42$ (the base); Answer choices: (A) $100$, (B) $420$, (C) $500$, (D) $609$, (E) $701$
Plan
Primary tool: #12 Draw a Picture
Secondary: #13 Use Symmetry
We have three side lengths but no height, and the area formula needs a base and a height. Tool #12 (Draw a Picture) makes the isosceles triangle visible so we can add an altitude from $B$ to the base $AC$. Tool #13 (Use Symmetry) is the key insight: because $AB = BC$, the altitude from $B$ is also the perpendicular bisector of $AC$, so it splits the base exactly in half ($21$ and $21$) and creates two congruent right triangles. From there, the Pythagorean theorem gives the height in one line, and the area formula finishes the job.
Execute — Answer: B
4.G.A.3 Step 1 - Draw $\triangle ABC$ with $AB = BC = 29$ and base $AC = 42$.
- Drop an altitude from the apex $B$ to the base, meeting $AC$ at point $D$.
- Because the triangle is isosceles, this altitude is the axis of symmetry, so $D$ is the midpoint of $AC$ and $AD = DC = 21$.
💡 Recognizing the line of symmetry in an isosceles triangle is a Grade 4 idea about symmetric figures.
8.G.B.7 Step 2 - Look at right triangle $\triangle ABD$.
- It has hypotenuse $AB = 29$, one leg $AD = 21$, and the other leg is the height $h = BD$ that we want.
- Apply the Pythagorean theorem.
💡 The altitude split the isosceles triangle into two right triangles — a Grade 8 "apply the Pythagorean theorem" setup.
8.EE.A.2 Step 3 - Solve for $h$.
- Compute the two squares and subtract.
💡 Taking the positive square root of $400$ is the Grade 8 "square roots of small perfect squares" skill — and $20\text{-}21\text{-}29$ is a well-known Pythagorean triple.
6.G.A.1 Step 4 Use the area formula with base $AC = 42$ and height $BD = 20$.
💡 Applying $\tfrac{1}{2} \times b \times h$ to a triangle is the Grade 6 area formula.
4.G.A.3 Draw $\triangle ABC$ with $AB = BC = 29$ and base $AC = 42$. Drop an altitude fr 8.G.B.7 Look at right triangle $\triangle ABD$. It has hypotenuse $AB = 29$, one leg $AD 8.EE.A.2 Solve for $h$. Compute the two squares and subtract. 6.G.A.1 Use the area formula with base $AC = 42$ and height $BD = 20$. Review
Reasonableness: Sanity-check with the bounding rectangle: a $42 \times 20$ rectangle has area $840$, and a triangle that fits inside it covers exactly half, giving $420$. That matches our answer. Also, $20\text{-}21\text{-}29$ is a standard Pythagorean triple ($20^2 + 21^2 = 400 + 441 = 841 = 29^2$), so the height is exact, not an approximation.
Alternative: Tool #17 (Use a Formula) — Heron's formula. The semi-perimeter is $s = \tfrac{29 + 29 + 42}{2} = 50$, so $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{50 \cdot 21 \cdot 21 \cdot 8} = \sqrt{176400} = 420$. Same answer, but Heron's formula is overkill once you see the symmetry trick.
CCSS standards used (min grade 8)
4.G.A.3Recognize a line of symmetry for a two-dimensional figure (Seeing that the altitude from $B$ in the isosceles triangle is the line of symmetry, so it bisects $AC$ into two segments of length $21$.)6.G.A.1Find the area of triangles and other polygons (Applying $\text{Area} = \tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} \times 42 \times 20 = 420$.)8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Using $21^2 + h^2 = 29^2$ in the right triangle $\triangle ABD$ to find the altitude.)8.EE.A.2Use square root symbols to represent solutions, evaluate square roots of small perfect squares (Computing $h = \sqrt{400} = 20$ to get the height from $h^2 = 400$.)
⭐ The symmetry of an isosceles triangle turns this into a Grade 8 Pythagorean theorem problem with the $20\text{-}21\text{-}29$ triple — then a Grade 6 area formula finishes it.
⭐ The symmetry of an isosceles triangle turns this into a Grade 8 Pythagorean theorem problem with the $20\text{-}21\text{-}29$ triple — then a Grade 6 area formula finishes it.
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