AMC 8 · 2024 · #24

Grade 6 geometry-2d

Archetype Compound Area by Decomposition / Difference

area-trianglesperfect-squares area-differencecasework ↑ Prerequisites: area-trianglesmulti-digit-arithmetic

📏 Long solution 💡 5 insights 📊 Diagram

Problem

Jean has made a piece of stained glass art in the shape of two mountains, as shown in the figure below. One mountain peak is $8$ feet high while the other peak is $12$ feet high. Each peak forms a $90^\circ$ angle, and the straight sides form a $45^\circ$ angle with the ground. The artwork has an area of $183$ square feet. The sides of the mountain meet at an intersection point near the center of the artwork, $h$ feet above the ground. What is the value of $h?$

Pick an answer.

(A)

(B)

(C)

$4\sqrt{2}$

(D)

(E)

$5\sqrt{2}$

View mode:

Toolkit + CCSS Solution

Understand

Restated: A stained-glass artwork is shaped like two overlapping mountain triangles. Each peak is a $90^\circ$ angle, and each mountain's two straight sides meet the ground at $45^\circ$. The left peak is $8$ feet high, the right peak is $12$ feet high, and the total artwork covers $183$ square feet. The two mountains overlap, and their inner sides cross at a point $h$ feet above the ground. Find $h$.

Givens: Left mountain peak height: $8$ feet; Right mountain peak height: $12$ feet; Each peak angle: $90^\circ$; Each ground-side angle: $45^\circ$ (so each mountain is a 45-45-90 right triangle); Total artwork area: $183$ square feet; Answer choices: (A) $4$, (B) $5$, (C) $4\sqrt{2}$, (D) $6$, (E) $5\sqrt{2}$

Unknowns: The height $h$ (in feet) of the point where the two inner mountain sides cross

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #9 Solve an Easier Related Problem, #6 Guess and Check, #3 Eliminate Possibilities

The artwork is the union of two overlapping mountain-triangles, so Tool #1 (Draw a Diagram) is the starting point — we extend each mountain's inner side back down to the ground so we can see two complete large triangles overlapping in a smaller triangle. Tool #7 (Identify Subproblems) breaks the problem into three friendly area pieces (big triangle, other big triangle, the overlap). To find the bases of the mountains from their heights, Tool #9 (Easier Related Problem) lets us replace each mountain with a tiny $45^\circ$ unit triangle and notice that a $45^\circ$ slope means "one over, one up," so the base is exactly twice the peak height. After we get $h^2 = 25$, Tool #6 (Guess and Check) finds $h = 5$ instantly, and Tool #3 (Eliminate Possibilities) confirms that $5$ is choice (B) while the $\sqrt{2}$ choices and the others are not consistent with the area equation.

Execute — Answer: B

#1 Draw a Diagram 6.G.A.1 Step 1

Draw the artwork and extend each mountain's inner side back down to the ground.
The figure becomes two complete 45-45-90 mountain triangles that overlap in a smaller upside-down 45-45-90 triangle whose top vertex sits $h$ feet above the ground.
Labeling the picture turns the problem into: area of big-left $+$ area of big-right $-$ area of overlap $= 183$.

$$\text{Area}_{\text{left}} + \text{Area}_{\text{right}} - \text{Area}_{\text{overlap}} = 183$$

💡 Finding the area of a shape by composing or decomposing it from triangles is the core Grade 6 geometry move.

#9 Solve an Easier Related Problem 4.MD.C.6 Step 2

Find each mountain's base from its peak height.
Try the easier related problem first: a peak $1$ foot high with $45^\circ$ sides.
A $45^\circ$ slope means "one foot over for every one foot up," so the peak reaches $1$ foot left and $1$ foot right on the ground — base $= 2$ feet.
Scaling the same picture up: peak height $H$ gives base $2H$.
So the left mountain has base $2 \times 8 = 16$ feet and the right mountain has base $2 \times 12 = 24$ feet.

$$\text{base}_{\text{left}} = 2 \times 8 = 16,\quad \text{base}_{\text{right}} = 2 \times 12 = 24$$

💡 Recognizing the $45^\circ$ angle and reading "over equals up" from the picture is a Grade 4 angle skill.

#7 Identify Subproblems 6.G.A.1 Step 3

Compute the two large triangle areas using base $\times$ height $\div 2$.
Left: $\tfrac{1}{2} \times 16 \times 8 = 64$ square feet.
Right: $\tfrac{1}{2} \times 24 \times 12 = 144$ square feet.
These are simple multiplication facts — no algebra needed.

$$\text{Area}_{\text{left}} = \tfrac{1}{2}\cdot 16 \cdot 8 = 64,\quad \text{Area}_{\text{right}} = \tfrac{1}{2}\cdot 24 \cdot 12 = 144$$

💡 Computing triangle areas with the $\tfrac{1}{2}\cdot b\cdot h$ formula is exactly what Grade 6 geometry asks for.

#9 Solve an Easier Related Problem 6.G.A.1 Step 4

The overlap is itself a 45-45-90 triangle with peak (pointing down) at the crossing point, height $h$, and base $2h$ on the ground (same $45^\circ$ slope trick as the mountains, just upside-down).
Its area is $\tfrac{1}{2} \times 2h \times h = h^2$.

$$\text{Area}_{\text{overlap}} = \tfrac{1}{2}\cdot (2h)\cdot h = h^2$$

💡 Reusing the same triangle-area reasoning on the smaller overlap is composing/decomposing shapes, a Grade 6 standard.

#7 Identify Subproblems 4.NBT.B.4 Step 5

Plug all three areas into the inclusion-exclusion equation: $64 + 144 - h^2 = 183$.
Add the two big areas: $64 + 144 = 208$.
Subtract from both sides: $h^2 = 208 - 183 = 25$.

$$64 + 144 - h^2 = 183 \;\Rightarrow\; h^2 = 208 - 183 = 25$$

💡 Adding and subtracting whole numbers like $64+144$ and $208-183$ is Grade 4 multi-digit arithmetic.

#6 Guess and Check 3.OA.C.7 Step 6

Solve $h^2 = 25$ by Guess and Check.
Try $h = 4$: $4 \times 4 = 16$, too small.
Try $h = 5$: $5 \times 5 = 25$.
That's a match.
Try $h = 6$: $6 \times 6 = 36$, too big.
So $h = 5$ feet.
Compare with the choices: $5$ is (B).
The $\sqrt{2}$ options (C) and (E) would give $h^2 = 32$ and $50$, neither equals $25$, so they are eliminated.
(A) $4$ gives $16$ and (D) $6$ gives $36$ — both wrong.
Only (B) survives.

$$5^2 = 25 \;\Rightarrow\; h = 5 \;\Rightarrow\; \textbf{(B)}$$

💡 Knowing $5 \times 5 = 25$ instantly is a basic Grade 3 multiplication fact.

[1] #1 6.G.A.1 Draw the artwork and extend each mountain's inner side back down to the ground.

[2] #9 4.MD.C.6 Find each mountain's base from its peak height. Try the easier related problem f

[3] #7 6.G.A.1 Compute the two large triangle areas using base $\times$ height $\div 2$. Left:

[4] #9 6.G.A.1 The overlap is itself a 45-45-90 triangle with peak (pointing down) at the cross

[5] #7 4.NBT.B.4 Plug all three areas into the inclusion-exclusion equation: $64 + 144 - h^2 = 18

[6] #6 3.OA.C.7 Solve $h^2 = 25$ by Guess and Check. Try $h = 4$: $4 \times 4 = 16$, too small.

Review

Reasonableness: Sanity check the size. The two big triangles together have area $64 + 144 = 208$ square feet, and the artwork is $183$, so the overlap takes away $25$ square feet — a small bite, exactly what we'd expect from a slim wedge near the middle. The overlap peak sits at $h = 5$ feet, which is below both mountain peaks ($8$ and $12$) and above the ground — geometrically possible. Plugging back: $64 + 144 - 5^2 = 208 - 25 = 183$. The units are all square feet, consistent throughout. The $\sqrt{2}$ choices were red herrings tied to slanted-side lengths, not the vertical height the problem actually asked for.

Alternative: An alternative is Tool #3 (Eliminate Possibilities): plug each choice into $64 + 144 - h^2 = 183$ and keep only the one that works. (A) $4 \Rightarrow 208 - 16 = 192 \ne 183$. (B) $5 \Rightarrow 208 - 25 = 183$ ✓. (C) $4\sqrt{2} \Rightarrow 208 - 32 = 176 \ne 183$. (D) $6 \Rightarrow 208 - 36 = 172 \ne 183$. (E) $5\sqrt{2} \Rightarrow 208 - 50 = 158 \ne 183$. Only (B) survives — no algebra and no square roots required.

CCSS standards used (min grade 6)

6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Computing each mountain's area as $\tfrac{1}{2}\cdot b\cdot h$ and finding the overlap area $h^2$ by decomposing the artwork into two big triangles minus their overlapping triangle (inclusion-exclusion of areas).)
4.MD.C.6 Measure angles in whole-number degrees using a protractor (Recognizing the $45^\circ$ ground angles to conclude that each mountain's base equals twice its peak height ("one over, one up").)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Adding $64 + 144 = 208$ and subtracting $208 - 183 = 25$ to isolate $h^2$.)
3.OA.C.7 Fluently multiply and divide within 100 (Using the multiplication fact $5 \times 5 = 25$ to solve $h^2 = 25$ by guess-and-check, and computing $16 \times 8 = 128$ and $24 \times 12 = 288$ before halving.)

⭐ This AMC 8 problem only needs Grade 6 triangle-area thinking you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

More like this