Super-Hard SAT Math Question 181 | Free SAT Prep | Aniko

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Question 181·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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In a circle with center $O$ and radius $10$ , chord $\overline{AB}$ has length $10\sqrt{3}$ . Which choice gives the area, in square units, of the region inside the circle that is bounded by chord $\overline{AB}$ and the minor arc $\overset{\frown}{AB}$ ? Аniкo Queѕtіon Bank

For chord-and-arc area problems, identify the region as a circular segment and immediately plan to do “sector minus triangle.” The key move is to convert the chord length into a central angle using $c=2R\sin(\theta/2)$ , being careful to choose the angle that matches minor versus major arc. Once $\theta$ is known, the sector is a simple fraction of $\pi R^2$ , and the triangle area comes quickly from $\tfrac12 ab\sin(C)$ with $a=b=R$ . Thіѕ question iѕ ‌from ‍Аn⁠ікo

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Relate the chord to the central angle

Use $c=2R\sin\left(\frac{\theta}{2}\right)$ with $c=10\sqrt{3}$ and $R=10$ to find $\theta$ . Wrіttеn by A⁠nікo

Use the minor arc information

There are two possible angles that give the same sine value; choose the one that corresponds to the minor arc (the smaller central angle).

Segment = sector − triangle

The area you want is not the whole sector. Subtract the area of triangle formed by the two radii and the chord.

Desmos Guide

Find the central angle numerically

Enter the chord formula in degrees by graphing

$y=20\sin(x/2)$
$y=10\sqrt{3}$

Then add a degree symbol to the trig input if needed by editing to $y=20\sin((x/2)^\circ)$ so $x$ is interpreted in degrees.

Use the intersection for the minor angle

Click the intersection point and take the smaller $x$ -value (this corresponds to the minor arc’s central angle).

Compute sector minus triangle

In new lines, define

$R=10$
$\theta=$ (the minor-angle value you found)
$S=(\theta/360)\pi R^2$
$T=\frac12 R^2\sin(\theta^\circ)$

Then compute $S-T$ and match it to the listed choices. © Aniko

Step-by-step Explanation

Find the central angle using the chord

Let $\theta$ be the measure (in degrees) of central angle $\angle AOB$ . A chord of length $c$ in a circle of radius $R$ satisfies

c = 2R\sin\left(\frac{\theta}{2}\right)

Here, $c=10\sqrt{3}$ and $R=10$ , so

10\sqrt{3}=2(10)\sin\left(\frac{\theta}{2}\right) \Rightarrow \sin\left(\frac{\theta}{2}\right)=\frac{\sqrt{3}}{2}

Since the problem asks for the minor arc, we use the acute value $\frac{\theta}{2}=60^\circ$ , giving $\theta=120^\circ$ .

Compute the sector area

The area of a sector with central angle $\theta$ (in degrees) is

\frac{\theta}{360}\cdot \pi R^2

With $\theta=120$ and $R=10$ :

\text{sector area}= \frac{120}{360}\cdot \pi(10)^2 = \frac{1}{3}\cdot 100\pi = \frac{100\pi}{3}

Compute the triangle area

Triangle $AOB$ has sides $OA=OB=10$ with included angle $120^\circ$ . Its area is anіko.ai

\frac{1}{2}ab\sin(C)=\frac{1}{2}(10)(10)\sin(120^\circ)

And $\sin(120^\circ)=\sin(60^\circ)=\frac{\sqrt{3}}{2}$ , so

\text{triangle area}=50\cdot\frac{\sqrt{3}}{2}=25\sqrt{3}

Subtract to get the segment area

The region bounded by the chord and the minor arc is the sector minus the triangle:

\text{segment area}= \frac{100\pi}{3}-25\sqrt{3}

So the correct choice is $\frac{100\pi}{3}-25\sqrt{3}$ .

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation