A geometry problem by abeel muhammad

Answer For N Vertices is nC2 - n

Derived formula for N vertices to get number of diagonal is N*(N-3)/2

every vertices has 7 diagonals. so, answer is ( 7*10 )/2 = 35

we can notice the following :

for 4 vertices >> we can draw 2 diagonals

for 5 vertices >> we can draw 5 diagonals " 2+3 "

for 6 vertices >> we can draw 9 diagonals " 2+3+4 "

so it is a sum of sequence { 2 ,3,4,5,6,7,8 ....... }

the first term of sequence : $2$

the final term is : $n-2$

the number of terms is : \(n-3)

so, the sum of terms = the number of diagonals of a shape with n vertices = \(\frac{(n-2)+2}{2} \times {(n-3)} \) $=$ $\frac{n(n-3)}{2}$

for $n = 10$ >>> the number of diagonals $= \frac{10 x 3}{2} = 35$

no of diagonals =(10*7)/2=35

can easily be done using combination.You will need 2 vertices for drawing each diagonal. For this the total number should be nC2 for a n sided diagram. But the vertices forming each side can never make diagonal. That is why the number of diagonal is = nC2-n=10C2-10=35.

(n^2-3n)/2 is the formula for finding no. of diagonals of a polygon if n is the no. of sides. thus,(100-30)/2=70/2=35

no of diagonals = n(n-3)/2 =10 *7/2 = 35

Start with a smaller polygon: A triangle has no diagonals A square has 2 diagonals A pentagon has 5 diagonals A hexagon has 9 diagonals A septagon has 14 diagonals

Look for a pattern:

0, 2, 5, 9, 14, ....

They are all increasing by 1 more than the previous one (with the exception of the very first two, which increase by 2)

So to continue the pattern: 0, 2, 5, 9, 14, 20, 27, 35,...

So a regular decagon would have a total of 35 diagonals.