AMC 8 · 2011 · #16

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference

area-trianglespythagorean-theoremisosceles-triangle identify-subproblems ↑ Prerequisites: pythagorean-theoremarea-triangles

📏 Medium solution 💡 3 insights

Problem

Let $A$ be the area of the triangle with sides of length $25, 25$ , and $30$ . Let $B$ be the area of the triangle with sides of length $25, 25,$ and $40$ . What is the relationship between $A$ and $B$ ?

$\textbf{(A) } A = \dfrac9{16}B \qquad\textbf{(B) } A = \dfrac34B \qquad\textbf{(C) } A=B \qquad \textbf{(D) } A = \dfrac43B \qquad \textbf{(E) }A = \dfrac{16}9B$

Pick an answer.

(A)

$A = \dfrac{9}{16}B$

(B)

$A = \dfrac{3}{4}B$

(C)

A = B

(D)

$A = \dfrac{4}{3}B$

(E)

$A = \dfrac{16}{9}B$

View mode:

Toolkit + CCSS Solution

Understand

Restated: Triangle $A$ has sides $25, 25, 30$. Triangle $B$ has sides $25, 25, 40$. Both are isosceles. Compare the areas $A$ and $B$.

Givens: Triangle $A$: sides $25, 25, 30$; Triangle $B$: sides $25, 25, 40$; Both are isosceles (two sides of length $25$); Answer choices: (A) $A=\tfrac{9}{16}B$, (B) $A=\tfrac{3}{4}B$, (C) $A=B$, (D) $A=\tfrac{4}{3}B$, (E) $A=\tfrac{16}{9}B$

Unknowns: The relationship between the two areas $A$ and $B$

Understand

Restated: Triangle $A$ has sides $25, 25, 30$. Triangle $B$ has sides $25, 25, 40$. Both are isosceles. Compare the areas $A$ and $B$.

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems

Only side lengths are given, so Tool #1 (Draw a Diagram) is the natural first move: sketch each isosceles triangle and drop the altitude to the unequal side. That altitude bisects the base and creates two right triangles. Tool #7 (Identify Subproblems) then splits the task — compute $A$ from its right-triangle pieces, compute $B$ from its right-triangle pieces, and compare. With a right triangle in each picture, the Pythagorean theorem gives the height and the area follows from $\tfrac{1}{2} \times \text{base} \times \text{height}$.

Execute — Answer: C

#1 Draw a Diagram 4.G.A.3 Step 1

Draw triangle $A$ with sides $25, 25, 30$.
Drop the altitude from the apex (between the two $25$ sides) to the base of length $30$.
Because the triangle is isosceles, this altitude bisects the base into two segments of $15$, creating two congruent right triangles with hypotenuse $25$ and one leg $15$.

$$\text{half-base} = \tfrac{30}{2} = 15$$

💡 An isosceles triangle has a line of symmetry through the apex, and the altitude to the unequal side IS that line — a Grade 4 symmetry idea.

#7 Identify Subproblems 8.G.B.7 Step 2

In that right triangle, use the Pythagorean theorem ($\text{leg}^2 + \text{leg}^2 = \text{hyp}^2$) to find the height $h_A$.

$$h_A^2 + 15^2 = 25^2 \;\Rightarrow\; h_A^2 = 625 - 225 = 400 \;\Rightarrow\; h_A = 20$$

💡 Finding an unknown leg of a right triangle from the other two sides is exactly Grade 8 Pythagorean-theorem work.

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

Compute area $A$ from base and height.

$$A = \tfrac{1}{2} \times 30 \times 20 = 300$$

💡 Area $= \tfrac{1}{2} \times \text{base} \times \text{height}$ for a triangle is Grade 6 geometry.

#7 Identify Subproblems 8.G.B.7 Step 4

Repeat the picture for triangle $B$ with sides $25, 25, 40$.
The altitude bisects the base of $40$ into two segments of $20$, giving right triangles with hypotenuse $25$ and one leg $20$.
Apply the Pythagorean theorem to find $h_B$.

$$h_B^2 + 20^2 = 25^2 \;\Rightarrow\; h_B^2 = 625 - 400 = 225 \;\Rightarrow\; h_B = 15$$

💡 Same Grade 8 Pythagorean move — the right triangle now has legs $20$ and $h_B$ with hypotenuse $25$.

#7 Identify Subproblems 6.G.A.1 Step 5

Compute area $B$ and compare to $A$.

$$B = \tfrac{1}{2} \times 40 \times 15 = 300 \;\Rightarrow\; A = B \;\Rightarrow\; \textbf{(C)}$$

💡 Both triangle areas come from the same $\tfrac{1}{2} \times \text{base} \times \text{height}$ formula; comparing the two results is a direct Grade 6 step.

[1] #1 4.G.A.3 Draw triangle $A$ with sides $25, 25, 30$. Drop the altitude from the apex (betw

[2] #7 8.G.B.7 In that right triangle, use the Pythagorean theorem ($\text{leg}^2 + \text{leg}^

[3] #7 6.G.A.1 Compute area $A$ from base and height.

[4] #7 8.G.B.7 Repeat the picture for triangle $B$ with sides $25, 25, 40$. The altitude bisect

[5] #7 6.G.A.1 Compute area $B$ and compare to $A$.

Review

Reasonableness: The two triangles share the same pair of $25$ sides, so the right-triangle pieces both have hypotenuse $25$. In triangle $A$ the legs are $15$ and $20$; in triangle $B$ the legs are $20$ and $15$ — the same $15$-$20$-$25$ right triangle (a $3$-$4$-$5$ scaled by $5$) in both pictures. Each full isosceles triangle is built from two of these right triangles, so it makes sense that both areas come out equal to $300$. Answer (C) $A=B$ is consistent.

Alternative: Tool #3 (Eliminate Possibilities): the two triangles share the same two $25$ sides and are built from two copies of the $15$-$20$-$25$ right triangle, just glued along different legs. That symmetry forces the areas to match, ruling out every choice except (C). No arithmetic needed beyond recognizing the shared $15$-$20$-$25$ pieces.

CCSS standards used (min grade 8)

4.G.A.3 Recognize a line of symmetry for a two-dimensional figure (Justifying that the altitude from the apex of an isosceles triangle bisects the unequal side, since that altitude is the triangle's line of symmetry.)
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Computing each area as $\tfrac{1}{2} \times \text{base} \times \text{height}$ once the height is known, and comparing the two results.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the altitude of each isosceles triangle from the hypotenuse $25$ and the half-base ($15$ or $20$) of the right-triangle piece.)

⭐ This AMC 8 problem only needs Grade 8 Pythagorean-theorem reasoning you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 8)

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