Question 146·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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A circle has a diameter of 16. A square and an equilateral triangle are each inscribed in the circle (so all of their vertices lie on the circle).

Which choice gives the difference between the area of the square and the area of the equilateral triangle?

For inscribed polygons, every vertex lies on the circle, but a circle’s diameter is not necessarily a polygon side. The square’s diagonal is the diameter; for the equilateral triangle, a diameter through one vertex reveals a $30$ - $60$ - $90$ triangle. Find each area, then subtract in the order asked. Watch for the tempting mistake of using the radius as a side length.

Hints

Hint 1
A square’s diagonal joins opposite vertices. For a square inscribed in a circle, that diagonal spans the diameter. How can the Pythagorean theorem give you the square’s area without first finding its side?
Hint 2
Draw a diameter through one vertex of the equilateral triangle. An inscribed angle facing that diameter is a right angle, and the diameter splits the triangle’s vertex angle in half. What special right triangle does that make?
Hint 3
In a $30$ - $60$ - $90$ triangle, the side lengths have the ratio $1:\sqrt3:2$ . Use the circle’s diameter as the longest side to find a side of the equilateral triangle, then find its perpendicular height.

Step-by-step

Use the diameter to find both areas
Step 1
Relate the square’s diagonal to its area
Opposite vertices of the inscribed square are opposite ends of a diameter, so its diagonal is $16$ . Let $s$ be its side. The diagonal makes a right triangle with two legs $s$ , so the Pythagorean theorem gives:
$s^2+s^2=16^2.$
Combine the equal terms:
$2s^2=16^2.$
Divide by $2$ ; $s^2$ is the square’s area:
$A_{\square}=s^2=\frac{16^2}{2}.$
Step 2
Calculate the square’s area
Type $\frac{16^2}{2}$ in Desmos. It shows $128$ , the square’s area. The diameter is the square’s diagonal, not its side, so squaring $16$ alone would give the wrong area.
Step 3
Reveal a special right triangle
Label the equilateral triangle’s vertices $A$ , $B$ , and $C$ . Draw a diameter from $A$ to a point $D$ on the opposite side of the circle. Because $B$ lies on the circle and faces diameter $AD$ , $\angle ABD$ is a right angle. The diameter through $A$ follows the triangle’s line of symmetry, splitting its $60^\circ$ angle at $A$ in half. So $\angle BAD=30^\circ$ .
Step 4
Find the triangle’s side
Triangle $ABD$ is a $30$ - $60$ - $90$ triangle. Its hypotenuse, the diameter $AD$ , is $16$ . The side $AB$ is the long leg, which is half the hypotenuse times $\sqrt3$ :
$AB=\frac{16}{2}\sqrt3=8\sqrt3.$
So $8\sqrt3$ is a full side of the equilateral triangle, not its radius.
Step 5
Find the triangle’s height
The altitude, or perpendicular height, of an equilateral triangle splits it into two $30$ - $60$ - $90$ triangles. Its height is $\frac{\sqrt3}{2}$ of its side. Type $\frac{\sqrt3}{2}(8\sqrt3)$ in Desmos; it shows $12$ , the height to use in the area formula.
Step 6
Calculate the triangle’s area
Use the whole side $8\sqrt3$ as the base and $12$ as the perpendicular height. Type $\frac12(8\sqrt3)(12)$ in Desmos; it shows about $83.14$ . Keep the area exact for the answer choices. First halve the base:
$A_{\triangle}=\frac12(8\sqrt3)(12)=4\sqrt3(12).$
Then multiply:
$A_{\triangle}=48\sqrt3.$
Step 7
Subtract in the requested order
The square’s area is $128$ , and the triangle’s area is $48\sqrt3$ . Subtract the triangle’s area from the square’s:
$A_{\square}-A_{\triangle}=128-48\sqrt3.$
Type $128-48\sqrt3$ in Desmos; it shows about $44.86$ , a positive difference. The difference is $128-48\sqrt3$ square units. Choice D.

Question 146·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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