AMC 10 · 2022 · #2

Grade 8 geometry-2d

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

pythagorean-theoreminteger-pythagorean-triplesarea-rectangles identify-subproblems ↑ Prerequisites: pythagorean-theorem

📏 Short solution 💡 2 insights 📊 Diagram

Problem

In rhombus ABCD, a point P on side AD splits it into AP = 3 and PD = 2, and BP is perpendicular to AD. Find the area of the rhombus.

Pick an answer.

(A)

$3\sqrt 5$

(B)

(C)

$6\sqrt 5$

(D)

(E)

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The figure is already drawn for us, but the key is to mark on it what the words say: AP=3, PD=2, right angle at P. Tool #1 (Diagram) keeps the labels straight — we see that BP is exactly the height of the rhombus measured from the base AD. From there Tool #7 (Subproblems) splits the work into two clean pieces: first find BP using the right triangle APB, then plug base × height into the parallelogram-area rule.

1STEP 1

Since P lies on AD, the side AD = AP + PD = 5, and a rhombus has all sides equal, so AB = 5 too.

AD = 3 + 2 = 5, AB = AD = 5

2STEP 2

In right triangle APB the legs are AP = 3 and BP, with hypotenuse AB = 5, so Pythagoras gives BP = 4.

AP² + BP² = AB² → 3² + BP² = 5² → BP² = 25 - 9 = 16 → BP = 4

3STEP 3

A rhombus is a parallelogram, so area = base × height = AD × BP = 5 × 4 = 20.

Area = AD × BP = 5 × 4 = 20

Answer

Cross-check the picture: a rhombus with side 5 has area at most 5 × 5 = 25 (if it were a square). Our height BP = 4 is just under the side length, so the rhombus is close to but not quite a square — 20 sits nicely in the range [0, 25] and matches choice (D). Also 3² + 4² = 9 + 16 = 25 = 5² checks the 3-4-5 triangle exactly.

💡Key takeaway

This AMC 10 problem only needs Grade 8 "3-4-5 right triangle" — once BP comes out to 4, the rhombus area is just 5 × 4 = 20.

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