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
A point on one side of a rhombus splits it into pieces of length 3 and 2. The segment from a neighbouring vertex to that point is perpendicular to the side. Find the area of the rhombus.

Pick an answer.

(A)
$3\sqrt 5$
(B)
10
(C)
$6\sqrt 5$
(D)
20
(E)
25
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

Find the side length

All four sides are equal.

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

Find the height

Pythagoras gives the height.

AP² + BP² = AB² → 3² + BP² = 5² → BP² = 25 - 9 = 16 → BP = 4
3STEP 3

Take the area

Base times height gives 20.

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

More like this

Same archetype — closest grade level first.