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.

i did my best to undestand your question i hope this helped u and u undestand but i cant anwser it cus ? queston is? sorry u have u write ur queston more batter if u need more help call me  at 4076321760

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

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40

the answer is here: : )

: