AMC 8 · 2015 · #6

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

area-trianglespythagorean-theoremline-symmetry identify-subproblems ↑ Prerequisites: area-trianglesperfect-squares

📏 Short solution 💡 3 insights

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Problem

Triangle ABC is isosceles with AB = BC = 29 and base AC = 42. Find its area.

Pick an answer.

(A)

100

(B)

420

(C)

500

(D)

609

(E)

701

AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Picture

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.

1STEP 1

Drop an altitude from apex B to base AC; by symmetry it meets the midpoint, so AD = DC = 21.

AD = DC =

\frac{AC}{2}

\frac{42}{2}

= 21

2STEP 2

In right triangle ABD the hypotenuse is 29 and a leg is 21, so apply the Pythagorean theorem to find height h.

AD² + BD² = AB² → 21² + h² = 29²

3STEP 3

Compute 29² - 21² = 400, so h = 20.

h² = 29² - 21² = 841 - 441 = 400 → h = √(400) = 20

4STEP 4

Plug base 42 and height 20 into half × base × height to get area = 420.

Area =

\frac{1}{2}

× 42 × 20 = 21 × 20 = 420 → (B)

Answer

420

Sanity-check with the bounding rectangle: a 42 × 20 rectangle has area 840, and a triangle that fits inside it covers exactly half, giving 420. That matches our answer. Also, 20-21-29 is a standard Pythagorean triple (20² + 21² = 400 + 441 = 841 = 29²), so the height is exact, not an approximation.

💡Key takeaway

The symmetry of an isosceles triangle turns this into a Grade 8 Pythagorean theorem problem with the 20-21-29 triple — then a Grade 6 area formula finishes it.

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