AMC 8 · 2005 · #23
Grade 8 geometry-2dProblem
Isosceles right triangle ABC encloses a semicircle of area 2π. The circle has its center O on hypotenuse AB and is tangent to sides AC and BC. What is the area of triangle ABC?
Solution
First, we notice half a square so first let's create a square. Once we have a square, we will have a full circle. This circle has a diameter of 4 which will be the side of the square. The area would be 4⋅4=16. Divide 16 by 2 to get the original shape and you get 8
Pick an answer.
AMC 8 2005 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: Isosceles right triangle $ABC$ has its right angle at $C$ and encloses a semicircle whose area is $2\pi$. The center $O$ of the semicircle lies on the hypotenuse $\overline{AB}$, and the semicircle is tangent to the two legs $\overline{AC}$ and $\overline{BC}$. Find the area of triangle $ABC$.
Givens: $\triangle ABC$ is an isosceles right triangle with the right angle at $C$, so $\angle A = \angle B = 45^\circ$ and $AC = BC$; A semicircle of area $2\pi$ sits inside the triangle; The semicircle's center $O$ lies on the hypotenuse $\overline{AB}$; The semicircle is tangent to both legs $\overline{AC}$ and $\overline{BC}$; Answer choices: (A) $6$, (B) $8$, (C) $3\pi$, (D) $10$, (E) $4\pi$
Unknowns: The area of $\triangle ABC$
Understand
Restated: Isosceles right triangle $ABC$ has its right angle at $C$ and encloses a semicircle whose area is $2\pi$. The center $O$ of the semicircle lies on the hypotenuse $\overline{AB}$, and the semicircle is tangent to the two legs $\overline{AC}$ and $\overline{BC}$. Find the area of triangle $ABC$.
Givens: $\triangle ABC$ is an isosceles right triangle with the right angle at $C$, so $\angle A = \angle B = 45^\circ$ and $AC = BC$; A semicircle of area $2\pi$ sits inside the triangle; The semicircle's center $O$ lies on the hypotenuse $\overline{AB}$; The semicircle is tangent to both legs $\overline{AC}$ and $\overline{BC}$; Answer choices: (A) $6$, (B) $8$, (C) $3\pi$, (D) $10$, (E) $4\pi$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #13 Use Symmetry
The figure already shows the triangle and semicircle; the missing ingredient is the two radii from $O$ to the points of tangency on $\overline{AC}$ and $\overline{BC}$. Tool #1 (Draw a Diagram) means adding those two radii: each is perpendicular to its leg, so together with the right angle at $C$ they form a small square inside the triangle. Once the square appears, the radius (found from the area $2\pi$) and a $45$-$45$-$90$ triangle at corner $A$ together give the leg length. Tool #13 (Use Symmetry) is the alternative angle: reflect the triangle across hypotenuse $\overline{AB}$ and the half-disk becomes a full circle inside a square. The triangle is half that square, so the triangle's area is half the square's area.
Execute — Answer: B
7.G.B.4 Step 1 - Find the radius from the semicircle's area.
- The area of a semicircle of radius $r$ is half the area of the full circle, so $\tfrac{1}{2}\pi r^2 = 2\pi$.
- Multiply both sides by $2$ to clear the half, then divide by $\pi$.
💡 The Grade 7 "area of a circle" formula in reverse: given the area, solve for the radius. The semicircle is half the disk, so the equation has a $\tfrac{1}{2}$ in front.
7.G.B.5 Step 2 - Add the two radii to the diagram.
- Let $D$ be where the semicircle touches leg $\overline{AC}$ and $E$ where it touches leg $\overline{BC}$.
- Draw $\overline{OD}$ and $\overline{OE}$.
- Each radius is perpendicular to its tangent leg, so $\angle ODC = \angle OEC = 90^\circ$.
- With $\angle DCE = 90^\circ$ already (the triangle's right angle), the quadrilateral $ODCE$ has three right angles, so the fourth is $90^\circ$ as well.
- Both $OD$ and $OE$ are radii, hence equal, so $ODCE$ is a square of side $r = 2$.
- In particular $CD = CE = 2$.
💡 Drawing the radii to the tangent points is the standard Grade 7 "tangent meets radius at a right angle" move. Three right angles in a quadrilateral force the fourth, and equal sides upgrade the rectangle to a square.
8.G.A.5 Step 3 - Use the $45$-$45$-$90$ corner at $A$ to find $AD$.
- In $\triangle ADO$, $\angle ODA = 90^\circ$ (radius perpendicular to leg) and $\angle DAO = 45^\circ$ (a base angle of the isosceles right triangle).
- The remaining angle $\angle AOD$ is therefore $45^\circ$, so $\triangle ADO$ is a $45$-$45$-$90$ isosceles right triangle with legs $AD$ and $OD$.
- The legs are equal, so $AD = OD = 2$.
💡 Grade 8 "angle sum of a triangle." The two given angles force the third, which makes the small triangle isosceles, which forces $AD = OD$.
6.G.A.1 Step 4 - Assemble the leg and compute the area.
- Leg $AC$ splits as $AC = AD + DC = 2 + 2 = 4$.
- Because the triangle is isosceles with legs $AC$ and $BC$, we also have $BC = 4$.
- The two legs meet at the right angle $C$, so they serve as base and height of the right triangle, and the area is half their product.
💡 Grade 6 "area of a right triangle is half the product of its legs." Once both legs are known, the area is one multiplication and a halving.
7.G.B.4 Find the radius from the semicircle's area. The area of a semicircle of radius $ 7.G.B.5 Add the two radii to the diagram. Let $D$ be where the semicircle touches leg $\ 8.G.A.5 Use the $45$-$45$-$90$ corner at $A$ to find $AD$. In $\triangle ADO$, $\angle O 6.G.A.1 Assemble the leg and compute the area. Leg $AC$ splits as $AC = AD + DC = 2 + 2 Review
Reasonableness: Sanity check the size. The semicircle has area $2\pi \approx 6.28$, and it fits inside the triangle, so the triangle's area must be larger than $6.28$. That rules out (A) $6$. Choice (B) $8$ is just a bit larger than $6.28$, which fits the picture: the half-disk fills most of the triangle but leaves two small corner regions near $A$ and $B$. Choices (D) $10$ and (E) $4\pi \approx 12.57$ would leave too much empty room; (C) $3\pi \approx 9.42$ has no reason to involve $\pi$ once the semicircle's area cancels. A direct check: with leg $4$, the leg-to-tangent-point distance is $AD = 2$, the radius $OD = 2$, and the right triangle's area $\tfrac{1}{2}\cdot 4 \cdot 4 = 8$ matches answer (B) exactly.
Alternative: Tool #13 (Use Symmetry): reflect $\triangle ABC$ across hypotenuse $\overline{AB}$. The semicircle's flat side lies on $\overline{AB}$, so the reflected half completes the full circle. The two reflected triangles form a square whose side equals leg $AC$, and the inscribed circle has diameter equal to that side. The circle's radius is $r = 2$, so the square's side is $2r = 4$ and its area is $16$. The triangle is half the square, giving area $8$ — answer (B).
CCSS standards used (min grade 8)
6.G.A.1Find the area of right triangles, other triangles, special quadrilaterals, and polygons (Computing the area of $\triangle ABC$ as half the product of its two legs once both legs are known.)7.G.B.4Know the formulas for the area and circumference of a circle and use them to solve problems (Solving $\tfrac{1}{2}\pi r^2 = 2\pi$ for the radius $r = 2$ from the given semicircle area.)7.G.B.5Use facts about supplementary, complementary, vertical, and adjacent angles in a multi-step problem to write and solve simple equations for an unknown angle in a figure (Using the right angles where the radii meet the tangent legs (and the triangle's right angle at $C$) to identify $ODCE$ as a square of side $2$.)8.G.A.5Use informal arguments to establish facts about the angle sum and exterior angle of triangles (Combining $\angle ODA = 90^\circ$ and $\angle DAO = 45^\circ$ to conclude $\triangle ADO$ is a $45$-$45$-$90$ triangle, so $AD = OD = 2$.)
⭐ When a circle is tangent to lines, the first move is almost always to draw the radii to the touch points. Here those two radii build a square inside the triangle, and the $45$-$45$-$90$ corners fill in the rest of each leg — giving leg $4$ and area $8$.
⭐ When a circle is tangent to lines, the first move is almost always to draw the radii to the touch points. Here those two radii build a square inside the triangle, and the $45$-$45$-$90$ corners fill in the rest of each leg — giving leg $4$ and area $8$.
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