questionr
Early access
Exams
Joint Entrance Examination
Mathematics
Ellipse
Question 15
Question 15 of 183
Medium
MCQ
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 q be the number of all quadrilaterals that can be formed by joining the vertices of P. If p + q = 126, then the eccentrici...
A
1
2
\frac{1}{\sqrt{\sqrt{2}}}
2
1
B
1
2
\frac{1}{2}
2
1
C
7
4
\frac{\sqrt{\sqrt{7}}}{4}
4
7
D
3
4
\frac{3}{4}
4
3
Check answer
Press A-D to select · Enter to check
Solutions · 1
Post your solution
Official solution
=
⇒
=
h
C
3
=
h
C
4
=
q
n
C
3
+
n
C
4
=
126
⇒
n
+
1
C
4
=
126
⇒
n
+
1
=
9
⇒
n
=
8
x
2
16
+
y
2
n
=
1
⇒
x
2
16
+
y
2
8
=
1
e
=
1
−
8
16
=
8
16
=
1
2
\begin{gathered} =\Rightarrow ={}^{h}{C}_{3} \\ ={}^{h}{C}_{4}=q \\ {}^{n}{C}_{3}+{}^{n}{C}_{4}=126\Rightarrow {}^{n+1}{C}_{4}=126 \\ \Rightarrow n+1=9\Rightarrow n=8 \\ \frac{{x}^{2}}{16}+\frac{{y}^{2}}{n}=1\Rightarrow \frac{{x}^{2}}{16}+\frac{{y}^{2}}{8}=1 \\ e=\sqrt{1-\frac{8}{16}}=\sqrt{\frac{8}{16}}=\frac{1}{\sqrt{2}} \end{gathered}
=⇒=
h
C
3
=
h
C
4
=
q
n
C
3
+
n
C
4
=
126
⇒
n
+
1
C
4
=
126
⇒
n
+
1
=
9
⇒
n
=
8
16
x
2
+
n
y
2
=
1
⇒
16
x
2
+
8
y
2
=
1
e
=
1
−
16
8
=
16
8
=
2
1
Answer the question to unlock
Previous
Next