AMC 8 · 2010 · #23

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference

area-circlespythagorean-theoremcoordinate-geometryratio-proportion identify-subproblemscoordinate-geometry ↑ Prerequisites: area-circlespythagorean-theorem

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Semicircles $POQ$ and $ROS$ pass through the center $O$ . What is the ratio of the combined areas of the two semicircles to the area of circle $O$ ?

Pick an answer.

(A)

$\frac{\sqrt{2}}{4}$

(B)

$\frac{1}{2}$

(C)

$\frac{2}{\pi}$

(D)

$\frac{2}{3}$

(E)

$\frac{\sqrt{2}}{2}$

View mode:

Toolkit + CCSS Solution

Understand

Restated: A circle $O$ is centered at the origin and passes through the four points $P(-1,1)$, $Q(1,1)$, $R(-1,-1)$, and $S(1,-1)$. Two semicircles are drawn: one on chord $PQ$ as its diameter (passing through center $O$) and one on chord $RS$ as its diameter (also passing through $O$). What is the ratio of the combined area of the two semicircles to the area of circle $O$?

Givens: Circle $O$ has center at the origin and passes through $P(-1,1)$, $Q(1,1)$, $R(-1,-1)$, $S(1,-1)$; Semicircle $POQ$ has $PQ$ as its diameter and passes through $O$; Semicircle $ROS$ has $RS$ as its diameter and passes through $O$; Answer choices: (A) $\tfrac{\sqrt{2}}{4}$, (B) $\tfrac{1}{2}$, (C) $\tfrac{2}{\pi}$, (D) $\tfrac{2}{3}$, (E) $\tfrac{\sqrt{2}}{2}$

Unknowns: The ratio (combined area of the two semicircles) : (area of circle $O$)

Understand

Plan

Primary tool: #11 Use Coordinates

Secondary: #7 Identify Subproblems

The figure is already drawn on the coordinate plane with the four corner points labeled, so Tool #11 (Use Coordinates) turns every length into a distance calculation. Tool #7 (Identify Subproblems) splits the question into two clean pieces — find the big circle's radius from $O$ to $P$, and find each semicircle's radius from the chord length $PQ$ — so the area ratio at the end is just $\dfrac{2 \cdot \tfrac{1}{2}\pi r_{\text{small}}^2}{\pi r_{\text{big}}^2}$.

Execute — Answer: B

#11 Use Coordinates 8.G.B.8 Step 1

Find the radius of circle $O$.
Since $O = (0,0)$ and $P = (-1, 1)$ lies on the circle, the radius is the distance $OP$.
Apply the distance formula (Pythagorean Theorem on the legs $1$ and $1$).

$$r_{\text{big}} = \sqrt{(-1-0)^2 + (1-0)^2} = \sqrt{1 + 1} = \sqrt{2}$$

💡 Distance between two points on the coordinate plane via the Pythagorean Theorem is the Grade 8 distance formula.

#7 Identify Subproblems 7.G.B.4 Step 2

Compute the area of circle $O$ using $A = \pi r^2$ with $r = \sqrt{2}$.

$$A_{\text{big}} = \pi (\sqrt{2})^2 = 2\pi$$

💡 Applying $A = \pi r^2$ to a circle is the standard Grade 7 area-of-a-circle formula.

#11 Use Coordinates 6.NS.C.8 Step 3

Find the diameter of each small semicircle.
The chord $PQ$ runs from $(-1, 1)$ to $(1, 1)$ — both points share $y = 1$, so the chord is horizontal and its length is just the difference of $x$-coordinates.
The diameter equals $PQ$, so each semicircle's radius is half of that.

$$PQ = 1 - (-1) = 2 \;\Rightarrow\; r_{\text{small}} = \tfrac{PQ}{2} = 1$$

💡 Reading horizontal distance off the coordinate plane as a difference of $x$-coordinates is a Grade 6 coordinate-plane skill.

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

Compute the combined area of the two semicircles.
Each has radius $1$, so each has area $\tfrac{1}{2}\pi (1)^2 = \tfrac{\pi}{2}$.
Two of them together make a full unit circle's worth of area.

$$A_{\text{semis}} = 2 \cdot \tfrac{1}{2}\pi (1)^2 = \pi$$

💡 Splitting the combined area into "two halves of a unit circle" is the Tool #7 subproblems move.

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

Take the ratio of the combined semicircle area to the area of circle $O$.

$$\dfrac{A_{\text{semis}}}{A_{\text{big}}} = \dfrac{\pi}{2\pi} = \dfrac{1}{2} \;\Rightarrow\; \textbf{(B)}$$

💡 Expressing one quantity as a fraction of another is the Grade 6 ratio concept.

[1] #11 8.G.B.8 Find the radius of circle $O$. Since $O = (0,0)$ and $P = (-1, 1)$ lies on the c

[2] #7 7.G.B.4 Compute the area of circle $O$ using $A = \pi r^2$ with $r = \sqrt{2}$.

[3] #11 6.NS.C.8 Find the diameter of each small semicircle. The chord $PQ$ runs from $(-1, 1)$ t

[4] #7 7.G.B.4 Compute the combined area of the two semicircles. Each has radius $1$, so each h

[5] #7 6.RP.A.1 Take the ratio of the combined semicircle area to the area of circle $O$.

Review

Reasonableness: The big circle has radius $\sqrt{2} \approx 1.41$, so its area $2\pi$ is exactly twice the area of a unit circle. The two semicircles glue together (in area) into one unit circle of area $\pi$. So the ratio must be $\pi : 2\pi = 1 : 2$, matching answer (B). The other choices fail simple sanity checks: $\tfrac{2}{\pi} \approx 0.637$ would be a dimensional mismatch (no stray $\pi$ survives the ratio), and $\tfrac{\sqrt{2}}{4}, \tfrac{\sqrt{2}}{2}$ would require some length, not area, in the ratio.

Alternative: Tool #2 (Find a Pattern) via scaling: any circle is similar to any other circle, so areas scale by the square of the radius. The small semicircles have radius $1$ and the big circle has radius $\sqrt{2}$, so $\dfrac{r_{\text{small}}^2}{r_{\text{big}}^2} = \dfrac{1}{2}$. Two semicircles equal one full small circle, so the combined area equals one small circle. The ratio of (one small circle) : (big circle) is $\dfrac{1}{2}$ — answer (B), with no $\pi$ ever needing to be computed.

CCSS standards used (min grade 8)

6.NS.C.8 Solve real-world and mathematical problems by graphing points in the coordinate plane (Reading the horizontal length $PQ = 2$ directly from the $x$-coordinates of $P(-1,1)$ and $Q(1,1)$.)
6.RP.A.1 Understand the concept of a ratio (Expressing the final answer as the ratio $\dfrac{\pi}{2\pi} = \dfrac{1}{2}$ of the combined semicircle area to the big-circle area.)
7.G.B.4 Know the formulas for the area and circumference of a circle and use them to solve problems (Applying $A = \pi r^2$ to get the big circle's area $2\pi$ and each small semicircle's area $\tfrac{\pi}{2}$.)
8.G.B.8 Apply the Pythagorean Theorem to find the distance between two points in a coordinate system (Computing $OP = \sqrt{1^2 + 1^2} = \sqrt{2}$ as the radius of circle $O$.)

⭐ This AMC 8 geometry problem only needs the Grade 8 distance formula and the Grade 7 circle-area formula that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 8)

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