AMC 10 · 2017 · #15

Grade 8 geometry-2d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
similar-trianglescoordinate-geometryarea-trianglespythagorean-theorem identify-subproblemsconvert-to-algebra ↑ Prerequisites: similar-trianglespythagorean-theorem
📏 Medium solution 💡 3 insights
Problem
A rectangle ABCD has side AB = 3 and side BC = 4. Drop a perpendicular from corner B onto the diagonal AC, and call the point where it lands E. Find the area of triangle AED.

Pick an answer.

(A)
1
(B)
$\frac{42}{25}$
(C)
$\frac{28}{15}$
(D)
2
(E)
$\frac{54}{25}$

AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

Pin the rectangle onto a coordinate grid so every point has an address. Then the messy phrase 'area of triangle AED' breaks into easy subproblems: find how far E sits from the diagonal's start, then read off a base and a height. With A and D both on one axis, the area collapses to a single half-base-times-height computation.

1STEP 1

Place the rectangle on a grid

Put A at the origin with AB on the x-axis and AD on the y-axis, giving A=(0,0), B=(3,0), C=(3,4), D=(0,4).

A=(0,0), B=(3,0), C=(3,4), D=(0,4)
2STEP 2

Length of the diagonal

Triangle ABC is right-angled at B with legs 3 and 4, so the Pythagorean theorem gives diagonal AC = 5.

AC=√(3²+4²)=√(25)=5
3STEP 3

How far E is along AC

BE is the altitude to hypotenuse AC, splitting ABC into two triangles similar to it; matching sides gives AE = 95\frac{9}{5}.

AE=AB2AC\frac{AB²}{AC}=325\frac{3²}{5}=95\frac{9}{5}
4STEP 4

Find the x-coordinate of E

Along AC, each unit moves 35\frac{3}{5} in x, so with AE = 95\frac{9}{5} the x-coordinate of E is x_E = 2725\frac{27}{25}.

x_E=AE·ABAC\frac{AB}{AC}=95\frac{9}{5}·35\frac{3}{5}=2725\frac{27}{25}
5STEP 5

Area of triangle AED

AD (length 4) lies on the y-axis as base; the height is x_E = 2725\frac{27}{25}, so the area is (12\frac{1}{2})(4)(2725\frac{27}{25}) = 5425\frac{54}{25}.

[△ AED]=12\frac{1}{2}· AD· x_E=12\frac{1}{2}· 4·2725\frac{27}{25}=5425\frac{54}{25}
Answer
5425\frac{54}{25}
The whole rectangle has area 3 times 4 = 12, and triangle AED is a thin sliver near corner A, so a small area is expected. 5425\frac{54}{25} is about 2.16, well under 12 and close to the answers clustered around 2, which fits. A coordinate-area check on A=(0,0), E=(2725\frac{27}{25},3625\frac{36}{25}), D=(0,4) gives (12\frac{1}{2})|0(3625\frac{36}{25}-4)+2725\frac{27}{25}(4-0)+0(0-3625\frac{36}{25})| = (12\frac{1}{2})(10825\frac{108}{25}) = 5425\frac{54}{25}, confirming the result.
💡Key takeaway

Put the shape on a grid, then the area of a triangle is just half its base times how far the last point reaches sideways.

  • Place the rectangle on a grid
  • Length of the diagonal
  • How far E is along AC
  • Find the x-coordinate of E
  • Area of triangle AED

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