Problem 6 · 2015 AMC 8
Medium
Geometry & Measurement
pythagorean-triplearea
In ▵ABC, AB = BC = 29, and AC = 42. What is the area of ▵ABC?
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Answer: B — Area 420.
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Hint 1 of 2
Area needs a height, and there isn't one yet — so make one. The altitude from the apex B has a bonus property in an isosceles triangle: it lands dead center on the base, slicing the triangle into two identical right triangles.
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Hint 2 of 2
That gives a right triangle with hypotenuse 29 and one leg 42÷2 = 21; the height is the other leg by the Pythagorean theorem. (Spotting 21 and 29 should ring a bell — it's the 20-21-29 right triangle.)
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Approach: drop the altitude to the base; isosceles means it bisects the base
- Drop the altitude from B to base AC. Because AB = BC, this altitude hits the midpoint M, so AM = 42/2 = 21 — this symmetry is the whole point, it hands you a right triangle for free.
- In right ▵ABM: height BM = √(292 − 212) = √(841 − 441) = √400 = 20. (Recognizing the 20-21-29 triple skips the arithmetic entirely.)
- Area = ½ · base · height = ½ · 42 · 20 = 420.
- Why this transfers: an altitude to the unequal side of an isosceles triangle always bisects that side, converting any isosceles triangle into two congruent right triangles — your go-to move for its area or height.
Another way — Heron's formula (no altitude needed):
- Sides 29, 29, 42 give semiperimeter s = (29 + 29 + 42)/2 = 50.
- Area = √(s(s−a)(s−b)(s−c)) = √(50 · 21 · 21 · 8) = √176400 = 420.
- Slower here, but Heron is the fallback when no convenient altitude exists.
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