AMC 10 · 2014 · #9
Grade 8 geometry-2d
Pick an answer.
There is no area formula for a general quadrilateral from four sides, so Tool #7 (Identify Subproblems) is the move: draw the diagonal AC and treat ABCD as two triangles. That single cut is chosen because AC is the hypotenuse of the right triangle ABC, so the right angle at B both gives the first area and hands over the length the second triangle is missing. Tool #5 (Look for a Pattern) then recognises 5, 12, 13 in the second triangle and — using the converse of the Pythagorean theorem, not the theorem itself — turns those three lengths into a second right angle. Tool #4 (Introduce a Variable) closes the argument by placing coordinates and solving for D, which proves the convex figure actually exists and shows exactly where the word "convex" is load-bearing.
Cut along the diagonal AC
The diagonal splits it into two triangles.
One diagonal converts an unmanageable quadrilateral into two triangles, and convexity is what lets the two areas be added instead of subtracted.
6.G.A.1Identify SubproblemsPythagoras gives AC = 5
The right angle gives that diagonal directly.
The right angle is the only bridge from the given sides to the diagonal, so cross it first.
8.G.B.7Identify SubproblemsSides 5, 12, 13 prove the angle
The other triangle turns out to be right too.
The Pythagorean theorem reads an angle forward into the sides; its converse reads the sides back into the angle, and this problem needs the backward direction.
Sides that satisfy the Pythagorean relation prove the angle between the two shorter ones is right.
▸ Why?
The relation between the squares holds exactly when the angle is right, so it reads both forwards and backwards.
▸ Why?
Once the angle is right, the two legs are already a base and its matching height, so the area needs no extra work.
Second triangle has legs 5 and 12
Its area follows immediately.
In a right triangle the two legs are already a base and its matching height, so no extra construction is needed.
7.G.B.6Identify SubproblemsAdd the two pieces
Adding gives 36, choice (D).
Adding the parts back is only valid because the diagonal split them cleanly in the first step.
6.G.A.1Identify SubproblemsCheck the convex figure exists
Coordinates confirm the shape exists.
Constructing one concrete copy of the figure proves the answer is not resting on a picture that might have been impossible to draw.
8.G.B.8Introduce A VariableCut the quadrilateral along AC: the right angle gives AC=5, then the sides 5, 12, 13 prove a second right angle, and 6+30=36.
- Cut along the diagonal AC
- Pythagoras gives AC = 5
- Sides 5, 12, 13 prove the angle
- Second triangle has legs 5 and 12
- Add the two pieces
- Check the convex figure exists
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