ISEE Upper Level Quantitative Reasoning - How to find an angle in other polygons

Thingy

Note: Figure NOT drawn to scale.

Refer to the above diagram. Pentagon is regular. What is the measure of $\angle 1$ ?

$144^{\circ}$

$128^{\circ}$

$120^{\circ}$

$136^{\circ }$

$108^{\circ}$

Explanation

The answer can be more clearly seen by extending $\overline{PA}$ to a ray $\overrightarrow{PA}$ :

Thingy

Note that angles have been newly numbered.

$\angle 2$ and $\angle 4$ are exterior angles of a (five-sided) regular pentagon in relation to two parallel lines, so each has a measure of $\left (360 \div 5 \right )^{\circ}= 72^{\circ}$ . $\angle 3$ is a corresponding angle to $\angle 4$ , so its measure is also $72^{\circ}$ .

By angle addition,

$m \angle 1 = m\angle 2 + m\angle 3 = 72^{\circ}+ 72^{\circ} = 144^{\circ}$

Thingy

In the above figure, the seven-side polygon, or heptagon, shown is regular. What is the measure of $\angle 1$ ?

$102\frac{6}{7}^{\circ}^{\circ}$

$128\frac{4}{7}^{\circ}^{\circ}$

$141\frac{3}{7}^{\circ }$

$154\frac{2}{7}^{\circ}$

The correct answer is not given among the other responses.

Explanation

The answer can be more clearly seen by extending the lower right side of the heptagon to a ray, as shown:

Thingy

Note that angles have been newly numbered.

$\angle 2$ and $\angle 4$ are exterior angles of a (seven-sided) regular heptagon, so each has a measure of $\left (360 \div 7 \right )^{\circ}= 51\frac{3}{7}^{\circ}$ . $\angle 3$ is a corresponding angle to $\angle 4$ in relation to two parallel lines, so its measure is also $51\frac{3}{7}^{\circ}$ .

By angle addition,

$m \angle 1 = m\angle 2 + m\angle 3 = 51\frac{3}{7}^{\circ}^{\circ}+ 51\frac{3}{7}^{\circ}^{\circ} = 102\frac{6}{7}^{\circ}^{\circ}$

Heptagon

The seven-sided polygon - or heptagon - in the above diagram is regular. What is the measure of $\angle 1$ ?

$51\frac{3}{7}^{\circ }$

$77\frac{1}{7}^{\circ }$

$38\frac{4}{7}^{\circ }$

$70^{\circ}$

$60 ^{\circ}$

Explanation

In the diagram below, some other angles have been numbered for the sake of convenience.

Heptagon

An interior angle of a regular heptagon has measure

$\frac{(7-2)180^{\circ }}{7}= 128\frac{4}{7}^{\circ }$ .

This is the measure of $\angle 2$ .

As a result of the Isosceles Triangle Theorem, $\angle 3 \cong \angle 4$ , so

$m \angle 3 = \frac{180^{\circ} - m \angle 2}{2} = \frac{180^{\circ} - 128\frac{4}{7}^{\circ }}{2}= 25\frac{5}{7}^{\circ }$ .

This is also the measure of $\angle 5$ .

By angle addition,

$m\angle 6 = 128\frac{4}{7}^{\circ } - 2 \cdot 25\frac{5}{7}^{\circ } = 77\frac{1}{7}^{\circ }$

Again, as a result of the Isosceles Triangle Theorem, $\angle 1 \cong \angle 7$ , so

$m \angle 1 = \frac{180^{\circ} - m \angle 6}{2} = \frac{180^{\circ} - 77\frac{1}{7}^{\circ }^{\circ }}{2}= 51\frac{3}{7}^{\circ }$

The measures of the angles of a ten-sided polygon, or decagon, form an arithmetic sequence. The least of the ten degree measures is $100 ^{\circ }$ . What is the greatest of the ten degree measures?

This polygon cannot exist.

$116^{\circ}$

$140^{\circ}$

$170^{\circ}$

$160^{\circ}$

Explanation

The total of the degree measures of any ten-sided polygon is

$\left (10-2 \right )180^{\circ} =1,440^{\circ}$ .

In an arithmetic sequence, the terms are separated by a common difference, which we will call . Since the least of the degree measures is $100^{\circ }$ , the measures of the angles are

Their sum is

$100+ \left (100+d \right )+\left (100+2d \right ) ...\left (100+9d \right )=1,440$

$d = 9\frac{7}{9}$

The greatest of the angle measures is

$100+9d = 100 + 9\left ( 9\frac{7}{9}\right )= 100+88 = 188$

However, an angle measure cannot exceed $180^{\circ}$ . The correct choice is that this polygon cannot exist.

The measures of the angles of an octagon form an arithmetic sequence. The greatest of the eight degree measures is $166^{\circ }$ . What is the least of the eight degree measures?

$104^{\circ}$

$74^{\circ}$

$144^{\circ}$

$114^{\circ}$

This octagon cannot exist.

Explanation

The total of the degree measures of any eight-sided polygon is

$\left (8-2 \right )180^{\circ} =1,080^{\circ}$ .

In an arithmetic sequence, the terms are separated by a common difference, which we will call . Since the greatest of the degree measures is $166^{\circ }$ , the measures of the angles are

Their sum is

$\left (166-7d \right )+\left ( 166-6d \right ) +...+ \left (166-d \right )+166 = 1,080$

$d= 8\frac{6}{7}$

The least of the angle measures is

$166-7d = 166 - 7 \cdot 8\frac{6}{7} = 166 - 62 = 104$

The correct choice is $104^{\circ}$ .

Thingy

In the above figure, the seven-side polygon, or heptagon, shown is regular. What is the measure of $\angle 1$ ?

$154\frac{2}{7}^{\circ}$

$102\frac{6}{7}^{\circ}^{\circ}$

$128\frac{4}{7}^{\circ}^{\circ}$

$141\frac{3}{7}^{\circ }$

The correct answer is not given among the other responses.

Explanation

The answer can be more clearly obtained by extending the top of the two parallel lines as follows:

Note that two angles have been newly labeled.

Thingy

$\angle 2$ is an interior angle of a regular heptagon and therefore has measure

$\frac{\left ( 7-2\right )180^{\circ }}{7}= 128\frac{4}{7}^{\circ }$

By the Isosceles Triangle Theorem, since the two sides of the heptagon that help form the triangle are congruent, so are the two acute angles, and

$m \angle 3 =\frac{1}{2}\left ( 180 -128\frac{4}{7} \right ) ^{\circ }= 25\frac{5}{7}^{\circ }$

$\angle 1$ is supplementary to $\angle 3$ , so

$m \angle 1 = 180^{\circ } - m \angle 3 = 180^{\circ } -25\frac{5}{7}^{\circ }= 154\frac{2}{7}^{\circ}$

Pentagon

Note: Figure NOT drawn to scale.

In the above figure, Pentagon is regular. Give the measure of $\angle XZY$ .

$41 ^{\circ}$

$29 ^{\circ}$

$59^{\circ}$

$51 ^{\circ}$

$39 ^{\circ}$

Explanation

The sum of the degree measures of the angles of Quadrilateral is 360, so

$m \angle XZY+ m \angle EXZ + m \angle E + m \angle EYZ = 360^{\circ}$

Each interior angle of a regular pentagon measures

$\frac{(5-2)180^{\circ}}{5} = 108^{\circ}$ ,

which is therefore the measure of $\angle E$ .

It is also given that $m \angle EXZ = 90^{\circ}$ and $m \angle EYZ = 121^{\circ}$ , so substitute and solve:

$m \angle XZY+ m \angle EXZ + m \angle E + m \angle EYZ = 360^{\circ}$

$m \angle XZY+ 90^{\circ} +108^{\circ} +121^{\circ} = 360^{\circ}$

$m \angle XZY+ 319^{\circ} = 360^{\circ}$

$m \angle XZY=41 ^{\circ}$

The measures of the angles of a nine-sided polygon, or nonagon, form an arithmetic sequence. The least of the nine degree measures is $127 ^{\circ }$ . What is the greatest of the nine degree measures?

$153^{\circ}$

$167\frac{6}{11} ^{\circ }$

$147^{\circ}$

$15 2\frac{5}{7}^{\circ}$

$143^{\circ}$

Explanation

The total of the degree measures of any nine-sided polygon is

$\left (9-2 \right )180^{\circ} = 1,260^{\circ}$ .

In an arithmetic sequence, the terms are separated by a common difference, which we will call . Since the least of the degree measures is $127 ^{\circ }$ , the measures of the angles are

Their sum is

$127+\left (127+d \right )+\left ( 127+2d \right )+...+\left (127+8d \right )= 1,260$

$d = 3\frac{1}{4}$

The greatest of the angle measures, in degrees, is

$127+8d = 127+8 \left ( 3\frac{1}{4}\right )= 127 +26 = 153$

$153^{\circ}$ is the correct choice.

What is the sum of all the interior angles of a decagon (a polygon with ten sides)?

Explanation

The sum of the angles in a polygon can be found using the equation below, in which t is equal to the total sum of the angles, and n is equal to the number of sides.