A regular dodecagon is inscribed in a circle with radius $1.$ Compute \[(P_1 P_2)^2 + (P_1 P_3)^2 + \dots + (P_{11} P_{12})^2.\](The sum includes all terms of the form $(P_i P_j)^2,$ where $1 \le i < j \le 12.$)

Sorry if it's in latex.

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Step-by-step explanation:

Problem

A regular dodecagon ($12$ sides) is inscribed in a circle with radius $r$ inches. The area of the dodecagon, in square inches, is:

$\textbf{(A)}\ 3r^2\qquad\textbf{(B)}\ 2r^2\qquad\textbf{(C)}\ \frac{3r^2\sqrt{3}}{4}\qquad\textbf{(D)}\ r^2\sqrt{3}\qquad\textbf{(E)}\ 3r^2\sqrt{3}$

Solution

The formula for the area of a regular dodecagon is $3r^2$. The answer is $\boxed{\textbf{(A)}}$. (If you don't know this formula, it's pretty easy to figure out that the area of a square inscribed in a circle is $2r^2$, and all the choices except $3r^2$ are less than $2r^2$. Remember, the more sides a regular polygon has, the closer its area gets to $\pi r^2$.)

See Also

1962 AHSC (Problems • Answer Key • Resources)

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Problem 17Followed by

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0

step-by-step explanation:

you can not divide by 0 so 4 ÷ 0 would be 0

0 × 9999 =0

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$32,660

step-by-step explanation: