Problem 2 · 2016 AMC 8
Easy
Geometry & Measurement
area
In rectangle ABCD, AB = 6 and AD = 8. Point M is the midpoint of AD. What is the area of ▵AMC?
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Answer: A — Area 12.
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Hint 1 of 2
A triangle's area doesn't care WHICH side you call the base — so pick the base that makes the height free. Is there a side of this triangle that lies right along an edge of the rectangle?
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Hint 2 of 2
Use AM (part of side AD) as the base. Then the height is the perpendicular distance from C across to that side — which is just the width AB, no extra work.
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Approach: choose the base along a rectangle edge so the height is free
- M is the midpoint of AD, so the base AM = AD/2 = 8/2 = 4.
- Because AM lies along side AD, the height to it is the full width across the rectangle: the perpendicular distance from C to line AD equals AB = 6.
- Area = (1/2)(base)(height) = (1/2)(4)(6) = 12.
- Why this transfers: when a triangle shares a side with a rectangle, choose that shared side as the base — the height collapses to a known rectangle dimension and you skip the Pythagorean theorem entirely.
Another way — coordinates (the shoelace / base-height made explicit):
- Place A = (0, 0), D = (8, 0), so M = (4, 0), and C = (8, 6).
- Triangle AMC has a horizontal base AM from x = 0 to x = 4 (length 4) on the line y = 0; vertex C sits at height 6.
- Area = (1/2)(4)(6) = 12.
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