AMC 8 · 2016 · #25
Grade 8 geometry-2dProblem
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?
Pick an answer.
AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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
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) $\dfrac{17\sqrt{3}}{2}$
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
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.
💡 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.
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.
💡 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.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.
💡 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.
6.G.A.1 Step 4 Area of $\triangle ADC$ using the two legs $AD$ and $CD$ as base and height.
💡 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.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 $r$.
💡 "Area is the same no matter which side you call the base" turns geometry into a Grade 7 one-step linear equation $17r = 120$.
4.G.A.2 Draw the isosceles triangle $ABC$ with base $AB = 16$ and apex $C$. Drop the hei 8.G.B.7 Use the Pythagorean Theorem on the right triangle $\triangle ADC$ to find the le 7.G.B.6 Because the semicircle is tangent to leg $AC$ at $E$, the radius $DE$ is perpend 6.G.A.1 Area of $\triangle ADC$ using the two legs $AD$ and $CD$ as base and height. 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.2Classify 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.1Find 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.6Solve 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.4Use 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.7Apply 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!
⭐ 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!
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