Pentagons - PSAT Math

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Question

What is the measure, in degrees, of one interior angle of a regular pentagon?

Answer

The formula for the sum of the interior angles of any regular polygon is as follows:

$\small = 180(n-2)$

where $\small n$ is equal to the number of sides of the regular polygon.

Therefore, the sum of the interior angles for a regular pentagon is:

$\small =180(5-2)$

$\small =180(3)$

$\small =540$

To find the measure of one interior angle of a regular pentagon, simply divide by the number of sides (or number of interior angles):

$\small =\frac{540}{5}$

$\small =108$

The measure of one interior angle of a regular pentagon is 108 degrees.

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Question

Refer to the above figure, which shows Square and regular Pentagon .

Evaluate $m \angle LOQ$ .

Answer

By angle addition,

$m \angle LOQ = m \angle LOK + m \angle KOQ$

$\angle KOQ$ is one of two acute angles of isosceles right triangle $\Delta KOQ$ , so $m \angle KOQ = 45^{\circ }$ .

To find $m \angle LOK$ we examine $\Delta LOK$ .

$\angle LKO$ is an angle of a regular pentagon and has measure $\frac{180^{\circ}\times (5-2)}{5} = 108 ^{\circ}$ .

Also, since, in $\Delta LOK$ , sides $\overline{KL} \cong \overline{KO}$ , by the Isosceles Triangle Theorem, $m \angle LOK = m \angle KLO$ .

Since the angles of a triangle must total $180^{\circ }$ in measure,

$m \angle LOK + m \angle KLO + m \angle LKO = 180^{\circ }$

$m \angle LOK + m \angle LOK + 108 ^{\circ } = 180^{\circ }$

$2 m \angle LOK + 108 ^{\circ } = 180^{\circ }$

$2 m \angle LOK = 72^{\circ }$

$m \angle LOK = 36^{\circ }$

$m \angle LOQ = m \angle LOK + m \angle KOQ$

$m \angle LOQ = 36^{\circ } + 45 ^{\circ } = 81^{\circ }$

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