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PSAT Math : How to find an angle in a pentagon

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Example Question #1 : How To Find An Angle In A Pentagon

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

Possible Answers:

144

108

180

120

Correct answer:

108

Explanation:

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

$\small = 180(n-2)$ $\displaystyle \small = 180(n-2)$

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

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

$\small =180(5-2)$ $\displaystyle \small =180(5-2)$

$\small =180(3)$ $\displaystyle \small =180(3)$

$\small =540$ $\displaystyle \small =540$

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

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

$\small =108$ $\displaystyle \small =108$

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

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Example Question #1 : How To Find An Angle In A Pentagon

Pentagon_1

Refer to the above figure, which shows Square OPQK $\displaystyle OPQK$ and regular Pentagon KLMNO $\displaystyle KLMNO$.

Evaluate $m \angle LOQ$ $\displaystyle m \angle LOQ$.

Possible Answers:

$75 ^{\circ }$ $\displaystyle 75 ^{\circ }$

$90 ^{\circ }$ $\displaystyle 90 ^{\circ }$

$72 ^{\circ }$ $\displaystyle 72 ^{\circ }$

$67\frac{1}{2} ^{\circ }$ $\displaystyle 67\frac{1}{2} ^{\circ }$

$81^{\circ }$ $\displaystyle 81^{\circ }$

Correct answer:

$81^{\circ }$ $\displaystyle 81^{\circ }$

Explanation:

By angle addition,

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

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

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

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

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

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

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

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

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

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

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

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

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

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PSAT Math : How to find an angle in a pentagon

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Example Question #1 : How To Find An Angle In A Pentagon

Example Question #1 : How To Find An Angle In A Pentagon

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