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)

Preceded by

Problem 17Followed by

Problem 19

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6,250

step-by-step explanation:

to get the area, multiply 360 by 160.(57,600)

to find out how many bags of fertilizer it will take, first figure out how many square feet one bag can hold. divide 1000 by 8(125) to get the unit rate(how many square feet one pound of fertilizer can hold)and multiply it by 50(6250). if i haven't made a mistake, that is the answer.