Other Polygons - PSAT Math

Card 0 of 20

Question

A regular polygon has a measure of $140^\circ$ for each of its internal angles. How many sides does it have?

Answer

To determine the measure of the angles of a regular polygon use:

Angle = (n – 2) x 180° / n

Thus, (n – 2) x 180° / n = 140°

180° n - 360° = 140° n

40° n = 360°

n = 360° / 40° = 9

Compare your answer with the correct one above

Question

A regular seven sided polygon has a side length of 14”. What is the measurement of one of the interior angles of the polygon?

Answer

The formula for of interior angles based on a polygon with a number of side n is:

Each Interior Angle = (n-2)180/n

= (7-2)180/7 = 128.57 degrees

Compare your answer with the correct one above

Question

If angle A and angle C are complementary angles and B and D are supplementary angles, which of the following must be true?

Answer

This question is very misleading, because while each answer COULD be true, none of them MUST be true. Between angle A and C, onne of the angles could be very small (0.001 degrees) and the other one could be very large. For instance, if A = 89.9999 and C = 0.0001, AC = 0.009. On the other hand, the two angles could be very siimilar. If B = 90 and D = 90 then BD = 8100 and BD > AC. If we use these same values we disprove AD = BC as 8100 ≠ .009. Finally, if B is a very small value, then B/C will be very small and smaller than A/D.

Compare your answer with the correct one above

Question

In isosceles triangle ABC, the measure of angle A is 50 degrees. Which is NOT a possible measure for angle B?

Answer

If angle A is one of the base angles, then the other base angle must measure 50 degrees. Since 50 + 50 + x = 180 means x = 80, the vertex angle must measure 80 degrees.

If angle A is the vertex angle, the two base angles must be equal. Since 50 + x + x = 180 means x = 65, the two base angles must measure 65 degrees.

The only number given that is not possible is 95 degrees.

Compare your answer with the correct one above

Question

In triangle ABC, the measure of angle A = 70 degrees, the measure of angle B = x degrees, and the measure of angle C = y degrees. What is the value of y in terms of x?

Answer

Since the three angles of a triangle sum to 180, we know that 70 + x + y = 180. Subtract 70 from both sides and see that x + y = 110. Subtract x from both sides and see that y = 110 – x.

Compare your answer with the correct one above

Question

What is the measure, in degrees, of each interior angle of a regular convex polygon that has twelve sides?

Answer

The sum of the interior angles, in degrees, of a regular polygon is given by the formula 180(n – 2), where n is the number of sides. The problem concerns a polygon with twelve sides, so we will let n = 12. The sum of the interior angles in this polygon would be 180(12 – 2) = 180(10) = 1800.

Because the polygon is regular (meaning its sides are all congruent), all of the angles have the same measure. Thus, if we divide the sum of the measures of the angles by the number of sides, we will have the measure of each interior angle. In short, we need to divide 1800 by 12, which gives us 150.

The answer is 150.

Compare your answer with the correct one above

Question

In the figure above, polygon ABDFHGEC is a regular octagon. What is the measure, in degrees, of angle FHI?

Answer

Angle FHI is the supplement of angle FHG, which is an interior angle in the octagon. When two angles are supplementary, their sum is equal to 180 degrees. If we can find the measure of each interior angle in the octagon, then we can find the supplement of angle FHG, which will give us the measure of angle FHI.

The sum of the interior angles in a regular polygon is given by the formula 180(n – 2), where n is the number of sides in the polygon. An octagon has eight sides, so the sum of the angles of the octagon is 180(8 – 2) = 180(6) = 1080 degrees. Because the octagon is regular, all of its sides and angles are congruent. Thus, the measure of each angle is equal to the sum of its angles divided by 8. Therefore, each angle in the polygon has a measure of 1080/8 = 135 degrees. This means that angle FHG has a measure of 135 degrees.

Now that we know the measure of angle FHG, we can find the measure of FHI. The sum of the measures of FHG and FHI must be 180 degrees, because the two angles form a line and are supplementary. We can write the following equation:

Measure of FHG + measure of FHI = 180

135 + measure of FHI = 180

Subtract 135 from both sides.

Measure of FHI = 45 degrees.

The answer is 45.

Compare your answer with the correct one above

Question

What is the measure of each angle in a regular octagon?

Answer

An octagon contains six triangles, or 1080 degrees. This means with 8 angles, each angle is 135 degrees.

Compare your answer with the correct one above

Question

What is the measure of each central angle of an octagon?

Answer

There are 360 degrees and 8 angles, so dividing leaves 45 degrees per angle.

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

Refer to the above figure. $\Delta OKR$ is equilateral and Pentagon is regular.

Evaluate $m \angle MNR$ .

Answer

By angle addition,

$m \angle MNR + m \angle RNO = m \angle MNO$

$\angle MNO$ is an angle of a reguar pentagon, so its measure is $\frac{180^{\circ}\times (5-2)}{5} = 108 ^{\circ}$ .

To find $m \angle RNO$ , first we find $m \angle NOR$ .

By angle addition,

$m \angle NOR + m \angle ROK = m \angle NOK$

$\angle NOK$ is an angle of a regular pentagon and has measure $108 ^{\circ}$ .

$\angle ROK$ , as an angle of an equilateral triangle, has measure $60 ^{\circ }$ .

$m \angle NOR + 60 ^{\circ }= 108 ^{\circ }$

$m \angle NOR = 48 ^{\circ }$

$\Delta OKR$ is equilateral, so $\overline{OR} \cong \overline{OK}$ ; Pentagon is regular, so $\overline{OK} \cong \overline{NO}$ . Therefore, $\overline{OR} \cong \overline{ON}$ , and by the Isosceles Triangle Theorem, $m \angle NRO = m \angle RNO$ .

The degree measures of three angles of a triangle total $180 ^{\circ }$ , so:

$m \angle NRO + m \angle RNO + m \angle NOR = 180^{\circ }$

$m \angle RNO + m \angle RNO + 48^{\circ } = 180^{\circ }$

$2 m \angle RNO + 48^{\circ } = 180^{\circ }$

$2 m \angle RNO = 132^{\circ }$

$m \angle RNO = 66^{\circ }$

Since

$m \angle MNR + m \angle RNO = m \angle MNO$

we have

$m \angle MNR + 66 ^{\circ } = 108^{\circ }$

$m \angle MNR = 42^{\circ }$

Compare your answer with the correct one above

Question

Pentagon is regular. If diagonal $\overline{BE}$ is drawn, which of the following describes Quadrilateral ?

Answer

The figure described is below.

Each of the angles of the pentagon has measure $\frac{180^{\circ } (5-2)}{5} = 108^{\circ }$

$\Delta ABE$ is an isosceles triangle, and $m\angle A = 108^{\circ }$ , so

$m \angle ABE = \frac{1}{2} \left (180^{\circ } - 108^{\circ } \right ) = 36^{\circ }$

and

$m \angle CBE = m \angle CBA - m \angle ABE = 108^{\circ } - 36^{\circ } = 72^{\circ }$

Since

$m \angle C =108^{\circ }$ ,

$m \angle C + m\angle CBE = 108^{\circ } + 72^{\circ } = 180^{\circ }$

and by the parallel postulate,

$\overline{CD} || \overline{BE}$

Quadrilateral has exactly one pair of parallel sides, so it is a trapezoid.

Compare your answer with the correct one above

Question

If square A has a side of length 5 inches, how many times bigger is the area of square B if it has a side of length 25 inches?

Answer

First find the area of both squares using the formula .

For square A, s = 5.

For square B, s = 25.

The question is asking for the ratio of these two areas, which will tell us how many times bigger square B is. Divide the area of square B by the area of square A to find the answer.

$\frac{A_B}{A_A}=\frac{625}{25}=25$

Compare your answer with the correct one above

Question

A square has an area of 36 cm2. A circle is inscribed and cut out. What is the area of the remaining shape? Use 3.14 to approximate π.

Answer

We need to find the area of both the square and the circle and then subtract the two. Inscribed means draw within a figure so as to touch in as many places as possible. So the circle is drawn inside the square. The opposite is circumscribed, meaning drawn outside.

Asquare = s2 = 36 cm2 so the side is 6 cm

6 cm is also the diameter of the circle and thus the radius is 3 cm

A circle = πr2 = 3.14 * 32 = 28.28 cm2

The resulting difference is 7.74 cm2

Compare your answer with the correct one above

Question

In the square above, the radius of each half-circle is 6 inches. What is the area of the shaded region?

Answer

We can find the area of the shaded region by subtracting the area of the semicircles, which is much easier to find. Two semi-circles are equivalent to one full circle. Thus we can just use the area formula, where r = 6:

π(62) → 36π

Now we must subtract the area of the semi-circles from the total area of the square. Since we know that the radius also covers half of a side, 6(2) = 12 is the full length of a side of the square. Squaring this, 122 = 144. Subtracting the area of the circles, we get our final terms,

= 144 – 36π

Compare your answer with the correct one above

Question

If Bailey paints the wall shaped like above and uses one bucket per 5 square units, how many buckets does Bailey need?

Answer

To solve, we will need to find the area of the wall. We can do this by finding the areas of each section and adding them together. Break the area into a rectange and two triangles.

The area of the rectangle will be equal to the base times the height. The area of each triangle will be one half its given base times its height.

$A_{rec}=bh$

$A_{tri}=\frac{1}{2}bh$

For the rectangle, the base is 12 and the height is 4 (both given in the figure).

$A_{rec}=(12)(4)=48$

The triangle to the right has a given base of 6, but we need to solve for its height. The height will be equal to the difference between the total height (6) and the height of the rectangle (4).

$h_{tri}=6-4=2$

We now have the base and height of the triangle to the right, allowing us to calculate its area.

$A_{tri_r}=\frac{1}{2}(6)(2)=6$

Now we need to solve the triangle to the left. We solved for its height (2), but we still need to solve for its base. The total base of the rectangle is 12. Subtract the base of the right-side triangle (6) and the small segment at the top of the rectangle (3) from this total length to solve for the base of the left triangle.

The left-side triangle has a base of 3 and a height of 2, allowing us to calculate its area.

$A_{tri_l}=\frac{1}{2}(3)(2)=3$

Add together the two triangles and the rectangle to find the total area.

We know that each bucket of paint will cover 5 square units, and we have 57 square units total. Divide to find how many buckets are required.

$\frac{57}{5}=11.4$

We will need 11 full buckets and part of a twelfth bucket to cover the wall, meaning that we will need 12 buckets total.

Compare your answer with the correct one above

Question

A square is inscribed within a circle with a radius $\small 3\sqrt{2}$ . Find the area of the circle that is not covered by the square.

Answer

First, find the area of the circle.

$A=\pi(3\sqrt{2})^2$

$\small A=18\pi$

Next, find the length of 1 side of the square using the Pythagorean Theorem. Two radii from the center of the circle to adjacent corners of the square will create a right angle at the center of the circle. The radii will be the legs of the triangle and the side of the square will be the hypotenuse.

$\small (3\sqrt{2})^2 +(3\sqrt{2})^2 =c^2$

$\small 18+18=36=c^2$

$\small c=6$

Find the area of the square.

$\small A=c^2=6^2=36$

Subtract the area of the square from the area of the circle.

$\small 18\pi-36$

Compare your answer with the correct one above

Question

Refer to the above figure. Quadrilateral is a square. Give the area of Polygon in terms of .

Answer

$\overline{BE }$ is both one side of Square and the hypotenuse of $\Delta ABE$ ; its hypotenuse can be calculated from the lengths of the legs using the Pythagorean Theorem:

$BE = \sqrt{(AB)^{2}+ \left ( AE\right )^{2}}$

$BE = \sqrt{x^{2}+ \left ( 2x\right )^{2}} = \sqrt{x^{2}+ 4x ^{2}} = \sqrt{5x ^{2}} = x\sqrt{5}$

Polygon is the composite of $\Delta ABE$ and Square , so its area is the sum of those of the two figures.

The area of $\Delta ABE$ is half the product of its legs:

$A= \frac{1}{2} \cdot x \cdot 2x = x^{2}$

The area of Square is the square of the length of a side:

$A = \left ( x\sqrt{5} \right )^{2} = 5x^{2}$

Add the areas:

$x^{2}+ 5x^{2} = 6x^{2}$

This is the area of the polygon.

Compare your answer with the correct one above

Question

Note: Figure NOT drawn to scale.

The above figure shows Rectangle ; is the midpoint of $\overline{CA}$ ; $EC = 2 \cdot RE$ ; $RC = 2 \cdot RN$ .

What percent of Rectangle is shaded?

Answer

The area of , the shaded region in question, is that of the rectangle minus those of $\Delta ERN$ and $\Delta ECT$ . We look at both.

The answer is independent of the sidelengths of the rectangle, so to ease calculations - this will become more apparent later - we will arbitrarily assign to the rectangle the dimensions

and, subsequently,

and, since is the midpoint of $\overline{CA}$ ,

.

The area of Rectangle is equal to $RN \cdot RC = 6 \cdot 12 = 72$ .

Since $EC = 2 \cdot RE$ , and ,

and

The area of $\Delta ERN$ is equal to $\frac{1}{2}(RE)(RN)= \frac{1}{2} \cdot 4 \cdot 6 = 12$ .

The area of $\Delta ECT$ is equal to $\frac{1}{2}(EC)(CT)= \frac{1}{2} \cdot 8 \cdot 3 = 12$

The area of the shaded region is therefore , which is

$\frac{48}{72} \times 100 % = 66 \frac{2}{3} %$ of the rectangle.

Compare your answer with the correct one above

Question

Regular Octagon has sidelength 1.

Give the length of diagonal $\overline{AD}$ .

Answer

The trick is to construct segments perpendicular to $\overline{AD}$ from and , calling the points of intersection and respectively.

Each interior angle of a regular octagon measures

$\frac{\left ( 8 - 2\right ) 180^{\circ }}{8} = 135^{\circ }$ ,

and by symmetry, $m \angle HAD = m \angle ADE = 90^{\circ }$ ,

so $m \angle BAX = m \angle CDY= 45^{\circ }$ .

This makes $\Delta BAX$ and $\Delta CDY$ $45^{\circ } -45^{\circ } -90^{\circ }$ triangles.

Since their hypotenuses are sides of the octagon with length 1, then their legs - in particular, $\overline{AX}$ and $\overline{YD}$ - have length $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ .

Also, since a rectangle was formed when the perpendiculars were drawn, .

The length of diagonal $\overline{AD}$ is

$AD = AX + XY + YD = \frac{\sqrt{2}}{2}+ 1 + \frac{\sqrt{2}}{2} = 1 + \sqrt{2}$ .

Compare your answer with the correct one above

Question

Regular Polygon (a twelve-sided polygon, or dodecagon) has sidelength 1.

Give the length of diagonal $\overline{AD}$ to the nearest tenth.

Answer

The trick is to construct segments perpendicular to $\overline{AD}$ from and , calling the points of intersection and respectively.

Each interior angle of a regular dodecagon measures

$\frac{\left ( 12 - 2\right ) 180^{\circ }}{12} = 150^{\circ }$ .

Since $\overline{BX}$ and $\overline{CY}$ are perpendicular to $\overline{AD}$ , it can be shown via symmetry that they are also perpendicular to $\overline{BC}$ . Therefore,

$\angle ABX$ and $\angle DCY$ both measure $60^{\circ }$

and $\Delta ABX$ and $\Delta DCY$ are $30^{\circ}-60^{\circ}-90^{\circ}$ triangles with long legs $\overline{AX}$ and $\overline{YD}$ . Since their hypotenuses are sides of the dodecagon and therefore have length 1,

$AX = YD = \frac{\sqrt{3}}{2}$ .

Also, since Quadrilateral is a rectangle, .

The length of diagonal $\overline{AD}$ is $AD = AX + XY + YD = \frac{\sqrt{3}}{2}+ 1 + \frac{\sqrt{3}}{2} = 1 + \sqrt{3}$ .

Compare your answer with the correct one above

Tap the card to reveal the answer