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AMC 10 · 2020 · #12

Grade 7 geometry-2d
Archetype Compound Area by Decomposition / Difference
area-trianglescentroid-2-to-1isosceles-trianglespatial-visualization identify-subproblemsarea-difference ↑ Prerequisites: area-triangles
📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Triangle AMCAMC is isosceles with AM=ACAM = AC. Medians MV‾\overline{MV} and CU‾\overline{CU} are perpendicular to each other, and MV=CU=12MV=CU=12. What is the area of △AMC?\triangle AMC?

(A) 48(B) 72(C) 96(D) 144(E) 192\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192

Pick an answer.

(A)
48
(B)
72
(C)
96
(D)
144
(E)
192
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Toolkit + CCSS Solution

Understand

Restated: In isosceles triangle $\triangle AMC$ with $AM = AC$, the medians $\overline{MV}$ (from $M$ to midpoint $V$ of $\overline{AC}$) and $\overline{CU}$ (from $C$ to midpoint $U$ of $\overline{AM}$) are perpendicular and each has length $12$. Find the area of $\triangle AMC$.

Givens: $\triangle AMC$ is isosceles with $AM = AC$; $\overline{MV}$ and $\overline{CU}$ are medians; $\overline{MV} \perp \overline{CU}$; $MV = CU = 12$; Answer choices: $48,\; 72,\; 96,\; 144,\; 192$

Unknowns: Area of $\triangle AMC$

Understand

Restated: In isosceles triangle $\triangle AMC$ with $AM = AC$, the medians $\overline{MV}$ (from $M$ to midpoint $V$ of $\overline{AC}$) and $\overline{CU}$ (from $C$ to midpoint $U$ of $\overline{AM}$) are perpendicular and each has length $12$. Find the area of $\triangle AMC$.

Givens: $\triangle AMC$ is isosceles with $AM = AC$; $\overline{MV}$ and $\overline{CU}$ are medians; $\overline{MV} \perp \overline{CU}$; $MV = CU = 12$; Answer choices: $48,\; 72,\; 96,\; 144,\; 192$

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities

Tool #1 (Draw): sketch $\triangle AMC$, the two medians $\overline{MV}$ and $\overline{CU}$, label their intersection $P$ (the centroid), and mark the right angle at $P$. Tool #7 (Subproblems): instead of attacking the full triangle directly, (a) use the $2:1$ centroid property to get the two leg lengths $MP$ and $CP$, (b) compute the area of the small right triangle $\triangle MPC$, (c) scale up to $\triangle AMC$ via the fact that the three centroid sub-triangles have equal area. Tool #3 (Eliminate): confirm the final number against the five choices.

Execute — Answer: C

#1 Draw a Diagram 4.G.A.1 Step 1
  • Draw $\triangle AMC$ with the two medians $\overline{MV}$ (from $M$ to midpoint $V$ of $\overline{AC}$) and $\overline{CU}$ (from $C$ to midpoint $U$ of $\overline{AM}$).
  • They cross at the centroid $P$.
  • Mark $\angle MPC = 90^\circ$ and the equal lengths $MV = CU = 12$.
$$P = \overline{MV} \cap \overline{CU},\quad \angle MPC = 90^\circ,\quad MV = CU = 12$$

💡 Grade 4 figures: drawing segments and a right angle organizes everything you'll measure.

#7 Identify Subproblems 5.NF.B.4 Step 2
  • Apply the centroid $2:1$ rule on median $\overline{MV}$: $MP = \tfrac{2}{3} \cdot 12 = 8$ and $PV = \tfrac{1}{3} \cdot 12 = 4$.
  • The same rule on median $\overline{CU}$ gives $CP = \tfrac{2}{3} \cdot 12 = 8$ and $PU = \tfrac{1}{3} \cdot 12 = 4$.
$$MP = \tfrac{2}{3}(12) = 8,\quad CP = \tfrac{2}{3}(12) = 8$$

💡 Grade 5 fraction-times-whole: take $\tfrac{2}{3}$ of $12$ to get the longer piece of each median.

#7 Identify Subproblems 6.G.A.1 Step 3
  • Inside the original triangle, look at the small triangle $\triangle MPC$ with legs $MP = 8$ and $CP = 8$ meeting at the right angle $P$.
  • Its area is $\tfrac{1}{2} \cdot 8 \cdot 8 = 32$.
$$[MPC] = \tfrac{1}{2} \cdot MP \cdot CP = \tfrac{1}{2} \cdot 8 \cdot 8 = 32$$

💡 Grade 6 triangle area: with a right angle, the two legs are base and height directly.

#7 Identify Subproblems 7.G.B.6 Step 4
  • The three segments $PA$, $PM$, $PC$ from the centroid split $\triangle AMC$ into three triangles $\triangle APM$, $\triangle MPC$, $\triangle APC$, and a classic centroid property says these three have equal area.
  • So $[AMC] = 3 \cdot [MPC] = 3 \cdot 32 = 96$.
$$[AMC] = 3 \cdot [MPC] = 3 \cdot 32 = 96$$

💡 Grade 7 area combining: three equal pieces add up to the whole triangle.

#3 Eliminate Possibilities 4.NBT.A.2 Step 5
  • Match the area $96$ to the choices: option (C).
  • The others would correspond to forgetting the $\tfrac{2}{3}$ centroid factor ($48, 72, 144$) or doubling instead of tripling ($192$).
$$[AMC] = 96 \;\Rightarrow\; \textbf{(C)}$$

💡 Grade 4 comparison: only one option equals $96$.

[1] #1 4.G.A.1 Draw $\triangle AMC$ with the two medians $\overline{MV}$ (from $M$ to midpoint
[2] #7 5.NF.B.4 Apply the centroid $2:1$ rule on median $\overline{MV}$: $MP = \tfrac{2}{3} \cdo
[3] #7 6.G.A.1 Inside the original triangle, look at the small triangle $\triangle MPC$ with le
[4] #7 7.G.B.6 The three segments $PA$, $PM$, $PC$ from the centroid split $\triangle AMC$ into
[5] #3 4.NBT.A.2 Match the area $96$ to the choices: option (C). The others would correspond to f

Review

Reasonableness: Sanity check via a back-of-envelope: a triangle whose medians are each $12$ and meet at right angles has the same area no matter how it's oriented. Using the well-known shortcut "area of a triangle equals $\tfrac{4}{3}$ times the area of the triangle formed by its medians", the median triangle here has legs $12$ and $12$ meeting at $90^\circ$, area $\tfrac{1}{2} \cdot 12 \cdot 12 = 72$, so $[AMC] = \tfrac{4}{3} \cdot 72 = 96$. ✓ Matches.

Alternative: Tool #1 (Draw) plus coordinates. Place $P = (0,0)$ at the centroid with $\overline{CU}$ along the $x$-axis and $\overline{MV}$ along the $y$-axis. Then $C = (8, 0)$, $U = (-4, 0)$, $M = (0, -8)$, $V = (0, 4)$. Since $U$ is the midpoint of $\overline{AM}$, $A = 2U - M = (-8, 0) - (0, -8) = (-8, 8)$. Area by the shoelace rule: $[AMC] = \tfrac{1}{2}|x_A(y_M - y_C) + x_M(y_C - y_A) + x_C(y_A - y_M)| = \tfrac{1}{2}|(-8)(-8 - 0) + 0 + 8(8 - (-8))| = \tfrac{1}{2}|64 + 128| = 96$. Same answer (C).

CCSS standards used (min grade 7)

  • 4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures (Sketching $\triangle AMC$, both medians, and the right angle at the centroid $P$.)
  • 4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols (Matching the computed area $96$ to the answer choices.)
  • 5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a whole number (Computing $\tfrac{2}{3} \cdot 12 = 8$ for the longer piece of each median.)
  • 6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Finding $[MPC] = \tfrac{1}{2} \cdot 8 \cdot 8$ from the two perpendicular legs.)
  • 7.G.B.6 Solve real-world problems involving area, surface area, and volume (Combining the three equal centroid sub-triangles into $[AMC] = 3 \cdot 32 = 96$.)

⭐ This AMC 10 problem only needs Grade 7 area you already know! The two medians meet at the centroid $P$ and the centroid grabs $\tfrac{2}{3}$ of each median — so $MP = CP = \tfrac{2}{3} \cdot 12 = 8$. Because $\angle MPC = 90^\circ$, the small triangle $MPC$ has area $\tfrac{1}{2} \cdot 8 \cdot 8 = 32$. The centroid splits the whole triangle into three equal-area pieces, so $[AMC] = 3 \cdot 32 = \mathbf{96}$, answer (C).

⭐ This AMC 10 problem only needs Grade 7 area you already know! The two medians meet at the centroid $P$ and the centroid grabs $\tfrac{2}{3}$ of each median — so $MP = CP = \tfrac{2}{3} \cdot 12 = 8$. Because $\angle MPC = 90^\circ$, the small triangle $MPC$ has area $\tfrac{1}{2} \cdot 8 \cdot 8 = 32$. The centroid splits the whole triangle into three equal-area pieces, so $[AMC] = 3 \cdot 32 = \mathbf{96}$, answer (C).

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