Super-Hard SAT Math Question 146 | Free SAT Prep | Aniko

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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? A‌nikо АІ‌ Т‍utor

When multiple shapes are inscribed in the same circle, translate “inscribed” into a specific length relationship with the circle: an inscribed square has diagonal equal to the diameter, and an inscribed equilateral triangle has side length $R\sqrt{3}$ because $R=\frac{s}{\sqrt{3}}$ . Compute each area from its side length, then subtract in the order the question specifies. Co‌ntеnt by Ani‌кo‍.‍a⁠і

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Start with what an inscribed square tells you

In a circle, the diagonal of an inscribed square is the same as the circle’s diameter.

Connect the equilateral triangle to the radius

For an equilateral triangle, the circle through its vertices is its circumcircle. Use the relationship between circumradius and side length.

Compute each area separately, then subtract

Find the square’s area and the triangle’s area from their side lengths. The question asks for square area minus triangle area. Thіѕ quеst‌iοn is ‍frоm Anікο

Desmos Guide

Define the radius

Type R=8.

Compute square area from the diagonal

Since the diagonal is 16, type Asq=(16/sqrt(2))^2.

Compute triangle area using the circumradius relationship

Type s=R*sqrt(3) and then type Atri=(sqrt(3)/4)*s^2. From⁠ a‌nіkо.а⁠i

Form the requested difference

Type D=Asq-Atri. Compare the displayed value of D to the answer choices; select the one that matches.

Step-by-step Explanation

Use the diameter to get the radius

The circle’s diameter is 16, so its radius is аniко.ai

R=\frac{16}{2}=8.

Find the area of the inscribed square

For a square inscribed in a circle, the square’s diagonal equals the circle’s diameter.

So the diagonal is 16, and if $s$ is the side length,

\begin{aligned} 16 &= s\sqrt{2} \\ s &= \frac{16}{\sqrt{2}}=8\sqrt{2}. \end{aligned}

Then the square’s area is

A_{\text{sq}}=s^2=(8\sqrt{2})^2=128.

Find the area of the inscribed equilateral triangle

For an equilateral triangle with side length $s$ , the circumradius is $R=\frac{s}{\sqrt{3}}$ , so $s=R\sqrt{3}$ .

With $R=8$ :

s=8\sqrt{3}.

The area of an equilateral triangle is $A_{\triangle}=\frac{\sqrt{3}}{4}s^2$ , so

\begin{aligned} A_{\triangle} &= \frac{\sqrt{3}}{4}(8\sqrt{3})^2 \\ &= \frac{\sqrt{3}}{4}\cdot 64\cdot 3 \\ &= 48\sqrt{3}. \end{aligned}

Subtract and match a choice

The requested difference is

A_{\text{sq}}-A_{\triangle}=128-48\sqrt{3}.

So the correct choice is $128-48\sqrt{3}$ .

Strategy

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Step-by-step Explanation

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Step-by-step Explanation