← Sensim Lab Sensim Math

AMC 8 · 2017 · #22

Grade 8 geometry-2dalgebra
Archetype Compound Area by Decomposition / Difference
similar-trianglespythagorean-theoremratio-proportionarea-triangles identify-subproblemsconvert-to-algebra ↑ Prerequisites: pythagorean-theoremsimilar-trianglesratio-proportion
📏 Medium solution 💡 4 insights 📊 Diagram

Problem

In the right triangle ABCABC, AC=12AC=12, BC=5BC=5, and angle CC is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?

(A) 76(B) 135(C) 5918(D) 103(E) 6013\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{13}{5}\qquad\textbf{(C) }\frac{59}{18}\qquad\textbf{(D) }\frac{10}{3}\qquad\textbf{(E) }\frac{60}{13}

Pick an answer.

(A)
$frac{7}{6}$
(B)
$frac{13}{5}$
(C)
$frac{59}{18}$
(D)
$frac{10}{3}$
(E)
$frac{60}{13}$
View mode:

Toolkit + CCSS Solution

Understand

Restated: Right triangle $ABC$ has a right angle at $C$ with legs $AC = 12$ and $BC = 5$. A semicircle is drawn inside the triangle so that its flat diameter sits on leg $AC$ (with one endpoint at $C$) and its curved edge just touches the hypotenuse $AB$. Find the radius $r$ of that semicircle.

Givens: $\triangle ABC$ is right-angled at $C$; Legs $AC = 12$ and $BC = 5$; A semicircle is inscribed with its diameter on $AC$, touching $AB$; Answer choices: (A) $\tfrac{7}{6}$, (B) $\tfrac{13}{5}$, (C) $\tfrac{59}{18}$, (D) $\tfrac{10}{3}$, (E) $\tfrac{60}{13}$

Unknowns: The radius $r$ of the inscribed semicircle

Understand

Restated: Right triangle $ABC$ has a right angle at $C$ with legs $AC = 12$ and $BC = 5$. A semicircle is drawn inside the triangle so that its flat diameter sits on leg $AC$ (with one endpoint at $C$) and its curved edge just touches the hypotenuse $AB$. Find the radius $r$ of that semicircle.

Givens: $\triangle ABC$ is right-angled at $C$; Legs $AC = 12$ and $BC = 5$; A semicircle is inscribed with its diameter on $AC$, touching $AB$; Answer choices: (A) $\tfrac{7}{6}$, (B) $\tfrac{13}{5}$, (C) $\tfrac{59}{18}$, (D) $\tfrac{10}{3}$, (E) $\tfrac{60}{13}$

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #13 Convert to Algebra, #3 Eliminate Possibilities

The asy figure is the seed, but the key move is Tool #1 (Draw a Diagram): add the unseen pieces — the center $O$ on $AC$, the radius $OC = r$ to leg $BC$, and the radius $OT$ drawn perpendicular to the hypotenuse $AB$. Once those are on the picture, Tool #7 (Identify Subproblems) splits the work into two clean pieces — first find $AB$ with the Pythagorean theorem, then notice the small right triangle $\triangle AOT$ tucked inside the big right triangle $\triangle ABC$. The two triangles share angle $A$ and both have a right angle, so they are similar. The similar-triangle proportion gives a single linear equation in $r$, which Tool #13 (Convert to Algebra) finishes. (We could lean on Tool #6 Guess & Check against the answer choices as a fast verification, and we do exactly that in the Review.)

Execute — Answer: D

#7 Identify Subproblems 8.G.B.7 Step 1
  • Find the hypotenuse $AB$ with the Pythagorean theorem on the right triangle $\triangle ABC$.
  • The legs are $12$ and $5$, a classic Pythagorean triple.
$$AB = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$$

💡 The Pythagorean theorem turns the two legs of a right triangle into the hypotenuse — a Grade 8 right-triangle fact.

#1 Draw a Diagram 4.G.A.1 Step 2
  • Place the center $O$ of the semicircle on leg $AC$ with $OC = r$.
  • Mark $T$ as the point where the semicircle just touches the hypotenuse $AB$.
  • By the tangent-radius rule, $OT \perp AB$ and $OT = r$.
  • Also, $AO = AC - OC = 12 - r$.
$$OC = r, \quad OT = r, \quad OT \perp AB, \quad AO = 12 - r$$

💡 Adding the center, the radius to the tangent point, and the right-angle mark to the figure is just labeling lines and angles — a Grade 4 geometry skill.

#7 Identify Subproblems 8.G.A.5 Step 3
  • Spot the two similar right triangles.
  • The small triangle $\triangle AOT$ and the big triangle $\triangle ABC$ both contain angle $A$ and both have a right angle (at $T$ and at $C$).
  • By Angle-Angle, the two triangles are similar, with corresponding sides $OT \leftrightarrow BC$ and $AO \leftrightarrow AB$.
$$\triangle AOT \sim \triangle ABC \;\Rightarrow\; \dfrac{OT}{BC} = \dfrac{AO}{AB}$$

💡 Recognizing AA similarity from a shared angle and a right angle is the Grade 8 informal-similarity-argument standard.

#13 Convert to Algebra 7.RP.A.2 Step 4

Substitute the known and unknown lengths into the proportion and solve for $r$.

$$\dfrac{r}{5} = \dfrac{12 - r}{13} \;\Longrightarrow\; 13r = 5(12 - r) \;\Longrightarrow\; 13r = 60 - 5r \;\Longrightarrow\; 18r = 60 \;\Longrightarrow\; r = \dfrac{60}{18} = \dfrac{10}{3}$$

💡 Cross-multiplying a proportion to solve for an unknown is the Grade 7 proportional-relationship move.

#3 Eliminate Possibilities 4.NF.A.2 Step 5

Match the value $r = \tfrac{10}{3}$ to the choice list: this is exactly option (D).

$$r = \dfrac{10}{3} \;\Rightarrow\; \textbf{(D)}$$

💡 Identifying which listed fraction equals our answer is a Grade 4 fraction-comparison step.

[1] #7 8.G.B.7 Find the hypotenuse $AB$ with the Pythagorean theorem on the right triangle $\tr
[2] #1 4.G.A.1 Place the center $O$ of the semicircle on leg $AC$ with $OC = r$. Mark $T$ as th
[3] #7 8.G.A.5 Spot the two similar right triangles. The small triangle $\triangle AOT$ and the
[4] #13 7.RP.A.2 Substitute the known and unknown lengths into the proportion and solve for $r$.
[5] #3 4.NF.A.2 Match the value $r = \tfrac{10}{3}$ to the choice list: this is exactly option (

Review

Reasonableness: The semicircle must fit inside a triangle whose shorter leg is $5$, so the radius must be less than $5$. Our answer $r = \tfrac{10}{3} \approx 3.33$ is comfortably below $5$, and it's also large enough that the semicircle visibly reaches across most of the figure — matching the asy picture (where the arc spans roughly from $x = 5.33$ to $x = 12$, giving radius $\tfrac{12 - 5.33}{2} \approx 3.33$). The value is consistent with the area cross-check too: splitting $\triangle ABC$ (area $30$) into $\triangle BCO$ (area $\tfrac{5r}{2}$) and $\triangle ABO$ (area $\tfrac{13r}{2}$) gives $\tfrac{18r}{2} = 30 \Rightarrow r = \tfrac{10}{3}$. Same answer, different path.

Alternative: Tool #6 (Guess and Check) against the multiple-choice list. The candidate radii are $\tfrac{7}{6} \approx 1.17$, $\tfrac{13}{5} = 2.6$, $\tfrac{59}{18} \approx 3.28$, $\tfrac{10}{3} \approx 3.33$, $\tfrac{60}{13} \approx 4.62$. The required proportion $\tfrac{r}{5} = \tfrac{12 - r}{13}$ becomes $13r + 5r = 60$, i.e. $r = \tfrac{60}{18}$, which is exactly (D). Plugging the other choices into $13r + 5r$ does not give $60$, so they are eliminated.

CCSS standards used (min grade 8)

  • 4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures (Adding the center $O$, the radius $OT$, and the right-angle mark at $T$ to the given figure so the geometry becomes visible.)
  • 4.NF.A.2 Compare two fractions with different numerators and different denominators (Matching the computed $r = \tfrac{10}{3}$ against the five answer-choice fractions to identify option (D).)
  • 7.RP.A.2 Recognize and represent proportional relationships between quantities (Cross-multiplying the similar-triangle proportion $\tfrac{r}{5} = \tfrac{12 - r}{13}$ and solving the linear equation for $r$.)
  • 8.G.A.5 Use informal arguments to establish facts about angle sum and exterior angles (Justifying that $\triangle AOT \sim \triangle ABC$ by the Angle-Angle criterion (shared angle $A$ plus a right angle in each).)
  • 8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Computing the hypotenuse $AB = \sqrt{12^2 + 5^2} = 13$ of the big right triangle.)

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

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

More like this

Same archetype — closest grade level first.