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 $\overset{\frown}{AB}$ ?

A circular segment is the region between a chord (a segment joining two points on a circle) and an arc. Find its area by subtracting the triangle formed by the chord and two radii from the matching sector. Bisect the chord to uncover a special right triangle, then use the central angle to find the sector’s share of the circle. The minor arc calls for the smaller sector.

Hints

Hint 1
The radii from $O$ to $A$ and $B$ have the same length. Draw a line from $O$ to the midpoint of the chord. It makes two right triangles. How long is each half of the chord?
Hint 2
A 30-60-90 triangle has side lengths in the ratio $1:\sqrt{3}:2$ . Match the radius to $2$ and the half-chord to $\sqrt{3}$ . What angle does each right triangle have at $O$ ?
Hint 3
A sector is the slice between two radii and an arc. Use the central angle to find its fraction of the circle, then subtract the triangle. The chord and minor arc enclose less than the sector.

Step-by-step

Subtract a triangle from the minor sector
Step 1
Split the chord into two right triangles
The radii $OA$ and $OB$ both have length $10$ , so $OAB$ is an isosceles triangle, with two equal sides. Join $O$ to the midpoint $M$ of $AB$ . In an isosceles triangle, that line is perpendicular to the base, so it makes two matching right triangles. Each half of the chord is $AM=MB=\frac{10\sqrt{3}}{2}=5\sqrt{3}$ .
Step 2
Find the central angle and triangle height
In right triangle $OAM$ , $OA=10$ is the hypotenuse, the side across from the right angle. Its length and $AM=5\sqrt{3}$ match the $30^\circ$ - $60^\circ$ - $90^\circ$ side ratio $2:\sqrt{3}:1$ . So the remaining side is $OM=5$ , and $\angle AOM=60^\circ$ because $AM$ is opposite that angle. The other right triangle matches, so $m\angle AOB=60^\circ+60^\circ=120^\circ$ . That is the angle for the minor arc, not the rest of the circle.
Step 3
Find the minor sector’s area
A sector is the slice between two radii and their arc. Its $120^\circ$ angle takes $\frac{120}{360}$ of the circle, so its area is $\frac{120}{360}\pi(10)^2$ . Type $\frac{120}{360}\cdot10^2$ in Desmos for the coefficient of $\pi$ . It shows $33.333\ldots$ ; tap the fraction button to see $\frac{100}{3}$ . The minor sector’s area is $\frac{100\pi}{3}$ .
Step 4
Find the triangle’s area
Triangle $OAB$ has base $AB=10\sqrt{3}$ and height $OM=5$ , which is perpendicular to the base. Its area is $\frac12(10\sqrt{3})(5)$ . Type $\frac12\cdot10\cdot5$ in Desmos; it shows $25$ , so the area of the whole triangle is $25\sqrt{3}$ .
Step 5
Subtract the triangle from the sector
The circular segment bounded by the chord and minor arc is the minor sector with triangle $OAB$ removed. Subtract it once: $A_{\text{segment}}=\frac{100\pi}{3}-25\sqrt{3}$ . That is the requested area in square units. Choice A.

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

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