The sum of the interior angles of a polygon with n sides is given by the formula:
$ \text{Sum} = (n-2) \times 180^\circ $
For a pentagon, n = 5. Therefore, the sum of its interior angles is:
$ (5-2) \times 180^\circ = 3 \times 180^\circ = 540^\circ $
The interior angles are in the ratio 1 : 3 : 5 : 7 : 11. Let the angles be represented as $1x$, $3x$, $5x$, $7x$, and $11x$, where x is a common multiplier.
The sum of these angles is:
$ 1x + 3x + 5x + 7x + 11x = 27x $
Equating the sum of the ratio parts to the total sum of the interior angles of the pentagon:
$ 27x = 540^\circ $
Now, solve for x:
$ x = \frac{540^\circ}{27} $
$ x = 20^\circ $
The smallest interior angle corresponds to the smallest part of the ratio, which is $1x$. Therefore, the smallest angle is:
$ 1x = 1 \times 20^\circ = 20^\circ $
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.