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AMC 10 · 2022 · #2

Grade 8 geometry-2d
Archetype Compound Area by Decomposition / Difference
pythagorean-theoreminteger-pythagorean-triplesarea-rectangles identify-subproblems ↑ Prerequisites: pythagorean-theorem
📏 Short solution 💡 2 insights 📊 Diagram

Problem

In rhombus ABCDABCD, point PP lies on segment AD‾\overline{AD} so that BP‾\overline{BP} ⊥\perp AD‾\overline{AD}, AP=3AP = 3, and PD=2PD = 2. What is the area of ABCDABCD? (Note: The figure is not drawn to scale.)

(A) 35(B) 10(C) 65(D) 20(E) 25\textbf{(A) }3\sqrt 5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }6\sqrt 5 \qquad \textbf{(D) }20\qquad \textbf{(E) }25

Pick an answer.

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

Understand

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

Givens: $ABCD$ is a rhombus (all four sides equal); $P$ lies on $\overline{AD}$; $AP = 3$ and $PD = 2$; $\overline{BP} \perp \overline{AD}$, so $\triangle APB$ has a right angle at $P$; Answer choices: (A) $3\sqrt{5}$, (B) $10$, (C) $6\sqrt{5}$, (D) $20$, (E) $25$

Unknowns: The area of rhombus $ABCD$

Understand

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

Givens: $ABCD$ is a rhombus (all four sides equal); $P$ lies on $\overline{AD}$; $AP = 3$ and $PD = 2$; $\overline{BP} \perp \overline{AD}$, so $\triangle APB$ has a right angle at $P$; Answer choices: (A) $3\sqrt{5}$, (B) $10$, (C) $6\sqrt{5}$, (D) $20$, (E) $25$

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems

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 $\times$ height into the parallelogram-area rule.

Execute — Answer: D

#1 Draw a Diagram 5.G.B.4 Step 1
  • Read the side length from the picture.
  • Since $P$ is on $\overline{AD}$, we have $AD = AP + PD = 3 + 2 = 5$.
  • A rhombus has all four sides equal, so $AB = 5$ as well.
$$AD = 3 + 2 = 5, \quad AB = AD = 5$$

💡 Knowing that a rhombus has four equal sides is Grade 5 "classify two-dimensional figures by properties" — the label travels around the shape.

#7 Identify Subproblems 8.G.B.7 Step 2
  • Focus on right triangle $APB$.
  • The legs are $AP = 3$ and $BP = ?$, and the hypotenuse is $AB = 5$.
  • Apply the Pythagorean theorem to find $BP$.
$$AP^2 + BP^2 = AB^2 \;\Rightarrow\; 3^2 + BP^2 = 5^2 \;\Rightarrow\; BP^2 = 25 - 9 = 16 \;\Rightarrow\; BP = 4$$

💡 The classic 3-4-5 right triangle — Grade 8 "Pythagorean theorem to find an unknown side". You can almost spot it without computing.

#7 Identify Subproblems 6.G.A.1 Step 3
  • Use the parallelogram (rhombus) area rule with $AD$ as the base and $BP$ as the height.
  • Multiply.
$$\text{Area} = AD \times BP = 5 \times 4 = 20$$

💡 A rhombus is a parallelogram, so area $=$ base $\times$ height — Grade 6 "find area by composing shapes". The answer is (D).

[1] #1 5.G.B.4 Read the side length from the picture. Since $P$ is on $\overline{AD}$, we have
[2] #7 8.G.B.7 Focus on right triangle $APB$. The legs are $AP = 3$ and $BP = ?$, and the hypot
[3] #7 6.G.A.1 Use the parallelogram (rhombus) area rule with $AD$ as the base and $BP$ as the

Review

Reasonableness: Cross-check the picture: a rhombus with side $5$ has area at most $5 \times 5 = 25$ (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 $3^2 + 4^2 = 9 + 16 = 25 = 5^2$ checks the 3-4-5 triangle exactly.

Alternative: Tool #13 (Convert to Algebra) plus the rhombus diagonal-product formula. Set up coordinates with $A = (0,0)$, $D = (5,0)$, $P = (3,0)$. Then $B = (3,4)$ (since $BP$ is vertical of length $4$) and $C = B + \vec{AD} = (8,4)$. The diagonals are $AC$ and $BD$, and $\text{Area} = \tfrac{1}{2}|AC| \cdot |BD|$. Even faster — but the base-times-height path stays at Grade 6.

CCSS standards used (min grade 8)

  • 5.G.B.4 Classify two-dimensional figures in a hierarchy based on properties (Recognizing that a rhombus has four equal sides, so $AB = AD = 5$.)
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Applying the parallelogram area rule $\text{Area} = \text{base} \times \text{height}$ to the rhombus.)
  • 8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding $BP = 4$ from the right triangle $APB$ with legs $3$ and $BP$ and hypotenuse $5$.)

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

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

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