Let p be the number of all triangles that can be formed by joining the vertices of a regul…

Question

Let $p$ be the number of all triangles that can be formed by joining the vertices of a regular polygon $P$ of $n$ sides, and let $q$ be the number of all quadrilaterals that can be formed by joining the vertices of $P$. If $p+q=126$, then the eccentricity of the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{n}=1$ is:

Accepted Answer

$$\begin{aligned}
& \text { Total trangles }=\Rightarrow={ }^{\mathrm{h}} \mathrm{C}_3 \
& \text { Total auadrilaterals }={ }^{\mathrm{h}} \mathrm{C}_4=\mathrm{q} \
& { }^{\mathrm{n}} \mathrm{C}_3+{ }^{\mathrm{n}} \mathrm{C}_4=126 \Rightarrow{ }^{\mathrm{n}+1} \mathrm{C}_4=126 \
& \Rightarrow \mathrm{n}+1=9 \Rightarrow \mathrm{n}=8 \
& \frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{\mathrm{n}}=1 \Rightarrow \frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{8}=1 \
& \mathrm{e}=\sqrt{1-\frac{8}{16}}=\sqrt{\frac{8}{16}}=\frac{1}{\sqrt{2}}
\end{aligned}$$

Solutions · 1