This problem requires us to find the number of diagonals in a regular polygon given a relationship between its interior and exterior angle sums.
Let $n$ be the number of sides of the regular polygon.
The problem states that the sum of the interior angles is four times the sum of its exterior angles:
$ (n-2) \times 180^\circ = 4 \times 360^\circ $Simplify the equation:
$ (n-2) \times 180 = 1440 $Divide both sides by 180:
$ n-2 = \frac{1440}{180} $ $ n-2 = 8 $Add 2 to both sides:
$ n = 8 + 2 $ $ n = 10 $The polygon has 10 sides.
The formula for the number of diagonals ($D$) in an $n$-sided polygon is:
$ D = \frac{n(n-3)}{2} $Substitute $n=10$ into the formula:
$ D = \frac{10(10-3)}{2} $ $ D = \frac{10 \times 7}{2} $ $ D = \frac{70}{2} $ $ D = 35 $Therefore, the polygon has 35 diagonals.
ABCDEF is a regular hexagon. Side of the hexagon is 36 cm. What is the area of the triangle AOB ?
If ∆ABC ~ ∆DEF, and BC = 4 cm, EF = 5 cm and the area of triangle ABC = 80 cm 2, then the area of the triangle DEF is:
If Δ ABC is right angled at B, AB = 12 cm and ∠CAB = 60°, determine the length of BC.
If ΔABC and ΔDEF are congruent triangles, then which of the following is FALSE?
D and E are points on the sides AB and AC, respectively, of ΔABC such that DE is parallel to BC and AD ∶ DB = 7 ∶ 9. If CD and BE intersect each other at F. then find the ratio of areas of ΔDEF and ΔCBF.