AMC 8 · 2016 · #25

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference

pythagorean-theoremarea-trianglessimilar-triangles identify-subproblemsconvert-to-algebra ↑ Prerequisites: pythagorean-theoremarea-triangles

📏 Medium solution 💡 4 insights 📊 Diagram

Problem

A semicircle is inscribed in an isosceles triangle with base $16$ and height $15$ so that the diameter of the semicircle is contained in the base of the triangle as shown. What is the radius of the semicircle?

$\textbf{(A) }4 \sqrt{3}\qquad\textbf{(B) } \dfrac{120}{17}\qquad\textbf{(C) }10\qquad\textbf{(D) }\dfrac{17\sqrt{2}}{2}\qquad \textbf{(E)} \dfrac{17\sqrt{3}}{2}$

Pick an answer.

(A)

$4\sqrt{3}$

(B)

$\dfrac{120}{17}$

(C)

(D)

$\dfrac{17\sqrt{2}}{2}$

(E)

$\dfrac{17\sqrt{3}}{2}$

View mode:

Toolkit + CCSS Solution

Understand

Restated: An isosceles triangle has base $16$ and height $15$. A semicircle sits inside the triangle with its diameter lying along the base, and its curved side touches both slanted legs. Find the radius of that semicircle.

Givens: Isosceles triangle with base $= 16$; Height (from apex to base) $= 15$; Semicircle's diameter lies on the base of the triangle; Semicircle is tangent to both slanted legs; Answer choices: (A) $4\sqrt{3}$, (B) $\dfrac{120}{17}$, (C) $10$, (D) $\dfrac{17\sqrt{2}}{2}$, (E) $\dfrac{17\sqrt{3}}{2}$

Unknowns: The radius $r$ of the semicircle

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems

Tool #1 (Draw a Diagram) is the geometry workhorse: sketch the triangle, drop the height $CD$ from the apex to the midpoint of the base, and mark where the semicircle's edge touches a leg. Two facts pop out of that picture — (i) the height $CD$ splits the isosceles triangle into two right triangles with legs $8$ and $15$, so the leg of the isosceles triangle is the famous $8\text{-}15\text{-}17$ hypotenuse, and (ii) the radius drawn to the tangent point is perpendicular to the leg, which means $r$ is exactly the altitude from $D$ to the hypotenuse of the small right triangle. Tool #7 (Identify Subproblems) then turns the hard "find the radius" question into the easy subproblem "find the area of a right triangle two different ways" — once with legs $8 \times 15$ and once with hypotenuse $17$ and height $r$. Setting them equal solves for $r$ in one line, no algebra heavier than a linear equation.

Execute — Answer: B

#1 Draw a Diagram 4.G.A.2 Step 1

Draw the isosceles triangle $ABC$ with base $AB = 16$ and apex $C$.
Drop the height $CD$ from $C$ to the midpoint $D$ of $AB$, so $CD = 15$ and $AD = DB = 8$.
The semicircle sits centered at $D$ with radius $r$, tangent to leg $AC$ at some point $E$.
Mark the right angle $\angle ADC = 90^\circ$ at the foot of the height, and the right angle $\angle DEC = 90^\circ$ where the radius meets the tangent leg.

$$AD = 8,\quad CD = 15,\quad \angle ADC = 90^\circ$$

💡 Drawing the height and labeling the right angles is a Grade 4 "classify shapes by their properties" move — it makes the hidden right triangles visible.

#1 Draw a Diagram 8.G.B.7 Step 2

Use the Pythagorean Theorem on the right triangle $\triangle ADC$ to find the leg $AC$ of the isosceles triangle.
With legs $8$ and $15$, this is the famous $8\text{-}15\text{-}17$ triple.

$$AC^2 = AD^2 + CD^2 = 8^2 + 15^2 = 64 + 225 = 289 \;\Rightarrow\; AC = 17$$

💡 Applying $a^2 + b^2 = c^2$ to a right triangle is the Grade 8 Pythagorean Theorem standard, and recognizing $8\text{-}15\text{-}17$ saves the square-root step.

#7 Identify Subproblems 7.G.B.6 Step 3

Because the semicircle is tangent to leg $AC$ at $E$, the radius $DE$ is perpendicular to $AC$.
So inside the small right triangle $\triangle ADC$, the segment $DE$ of length $r$ is the altitude from $D$ to the hypotenuse $AC$.
Now treat that one right triangle as a subproblem and compute its area two different ways.

$$DE = r,\quad DE \perp AC$$

💡 Splitting off the right triangle $\triangle ADC$ and recognizing $r$ as its altitude to the hypotenuse is the Tool #7 subproblems move — Grade 7 "area of triangles" reasoning applied to a piece of the figure.

#7 Identify Subproblems 6.G.A.1 Step 4

Area of $\triangle ADC$ using the two legs $AD$ and $CD$ as base and height.

$$\text{Area} = \tfrac{1}{2} \times AD \times CD = \tfrac{1}{2} \times 8 \times 15 = 60$$

💡 Area $= \tfrac{1}{2} \times \text{base} \times \text{height}$ on a right triangle is the Grade 6 area formula, no formula manipulation needed.

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

Area of the same triangle $\triangle ADC$ using hypotenuse $AC$ as the base and $DE = r$ as the height.
Set the two area expressions equal and solve the resulting linear equation for $r$.

$$\tfrac{1}{2} \times 17 \times r = 60 \;\Rightarrow\; 17r = 120 \;\Rightarrow\; r = \dfrac{120}{17} \;\Rightarrow\; \textbf{(B)}$$

💡 "Area is the same no matter which side you call the base" turns geometry into a Grade 7 one-step linear equation $17r = 120$.

[1] #1 4.G.A.2 Draw the isosceles triangle $ABC$ with base $AB = 16$ and apex $C$. Drop the hei

[2] #1 8.G.B.7 Use the Pythagorean Theorem on the right triangle $\triangle ADC$ to find the le

[3] #7 7.G.B.6 Because the semicircle is tangent to leg $AC$ at $E$, the radius $DE$ is perpend

[4] #7 6.G.A.1 Area of $\triangle ADC$ using the two legs $AD$ and $CD$ as base and height.

[5] #7 7.EE.B.4 Area of the same triangle $\triangle ADC$ using hypotenuse $AC$ as the base and

Review

Reasonableness: Sanity check the size: $\tfrac{120}{17} \approx 7.06$. The semicircle's diameter would be $\approx 14.1$, which fits inside the base of length $16$ (with a little gap on each side) — exactly what the picture in the problem shows. The radius is also smaller than the height $15$, so the curved top doesn't poke through the apex. Both checks pass, so $r = \tfrac{120}{17}$ is geometrically reasonable.

Alternative: Tool #3 (Eliminate Possibilities) on the answer choices. The radius must be less than the height $15$ and less than half the base $8$ (since the diameter $2r$ must fit in the base $16$, so $r < 8$). Compute each choice numerically: (A) $4\sqrt{3} \approx 6.93$, (B) $\tfrac{120}{17} \approx 7.06$, (C) $10$, (D) $\tfrac{17\sqrt{2}}{2} \approx 12.02$, (E) $\tfrac{17\sqrt{3}}{2} \approx 14.72$. Choices (C), (D), (E) violate $r < 8$ and are eliminated. Between (A) and (B), only (B) makes $\tfrac{1}{2} \cdot 17 \cdot r = 60$ exact.

CCSS standards used (min grade 8)

4.G.A.2 Classify two-dimensional figures based on the presence or absence of parallel/perpendicular lines and angles (Drawing the isosceles triangle, the height $CD$, and marking the right angles $\angle ADC$ and $\angle DEC$ that the diagram depends on.)
6.G.A.1 Find the area of triangles and other polygons by composing into rectangles or decomposing into triangles (Computing the area of right triangle $\triangle ADC$ as $\tfrac{1}{2} \times 8 \times 15 = 60$.)
7.G.B.6 Solve real-world and mathematical problems involving area, volume, and surface area of two- and three-dimensional objects (Setting up the area of $\triangle ADC$ a second way using hypotenuse $17$ and altitude $r$, then equating it to $60$.)
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities to solve problems (Solving the linear equation $17r = 120$ to get $r = \tfrac{120}{17}$.)
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Finding the leg $AC = 17$ of the isosceles triangle from $AC^2 = 8^2 + 15^2$.)

⭐ This AMC 8 problem #25 only needs Grade 8 Pythagorean Theorem and the trick that a triangle's area is the same no matter which side you pick as the base!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 8)

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