Other Shapes - GED Math

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Question

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

Answer

The sum of the degree measures of the angles of a polygon with sides is . Since an octagon has eight sides, substitute to get:

$180(8-2 ) = 180 \times6 = 1,080$

Each angle of a regular polygon has equal measure, so divide this by 8 to get the measure of one angle:

$1,080 \div 8 = 135$ , the degree measure of one angle.

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Question

Give the number of sides of a regular polygon whose interior angles have measure $170^{\circ }$ .

Answer

The easiest way to solve this is to look at the exterior angles, each of which have measure $(180-170)^{\circ } = 10^{\circ }$ . Since each exterior angle of a regular polygon with sides is $\frac{360}{N}$ , we solve for in the following equation:

$\frac{360}{N} = 10$

$\frac{360}{N} \cdot \frac{N }{10} = 10 \cdot \frac{N }{10}$

The polygon has 36 sides.

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Question

Give the measure of each interior angle of a regular 72-sided polygon.

Answer

A regular polygon with sides has interior angles of measure $\frac{(n-2)180^{\circ }}{n}$ each. Substitute 72 for .

$\frac{(72-2)180^{\circ }}{72}= \frac{70 \times 180^{\circ }}{72} = 175^{\circ }$

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Question

Note: Figure NOT drawn to scale.

The above hexagon is regular. What is ?

Answer

Two of the angles of the quadrilateral formed are angles of a regular hexagon, so each measures

$\frac{180^{\circ } (6-2)}{6}= 120^{\circ }$ .

The four angles of the quadrilateral are $120 ^{\circ }, 120^{\circ }, 45^{\circ }, x^{\circ }$ . Their sum is $360^{\circ }$ , so we can set up an equation and solve for :

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Question

The above octagon is regular. What is ?

Answer

Three of the angles of the pentagon formed are angles of a regular octagon, so each measures

$\frac{180^{\circ } (8-2)}{8}= 135^{\circ }$ .

The five angles of the pentagon are $135^{\circ }, 135^{\circ }, 135^{\circ }, 80^{\circ }, y^{\circ }$ . Their sum is $180^{\circ } (5-3) = 540^{\circ }$ , so we can set up an equation and solve for :

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Question

Refer to the above diagram.

Which of these is a valid alternative name for $\angle FEC$ ?

Answer

When naming an angle after three points, the middle letter must be its vertex, or the point at which its sides meet - this is . The other two letters must refer to points on its two sides. Therefore, $\angle FEC$ includes on one side, making one of its sides $\overrightarrow{EF}$ , and on the other, making the other side $\overrightarrow{EC}$ .

An alternative name for this angle must be one of two things:

It can be named only after its vertex - that is, $\angle E$ - but only if there is no ambiguity as to which angle is being named. Since more than one angle in the diagram has vertex , $\angle E$ is not a correct choice.

It can be named after three points. Again, the middle letter must be vertex , so we can throw out $\angle GCE$ and $\angle GFE$ .

The only possible choice is $\angle CEA$ .

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Question

Refer to the above figure. $\Delta OKR$ is equilateral, and Quadrilateral is a square.

Evaluate $m \angle ROP$ .

Answer

By angle addition,

$m \angle ROP = m \angle ROK + m \angle KOP$ .

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

$\angle KOP$ , as an angle of a square, has measure $90 ^{\circ }$ .

Therefore,

$m \angle ROP = 60 ^{\circ } + 90^{\circ } = 150 ^{\circ }$ .

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Question

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

Evaluate $m \angle NOQ$ .

Answer

By angle addition,

$m \angle NOQ = m \angle NOK + m \angle KOQ$ .

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

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

$m \angle NOQ = 108^{\circ } + 45^{\circ } = 153^{\circ }$

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Question

Note: Figure NOT drawn to scale.

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

Evaluate $m \angle NRO$ .

Answer

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

$\angle ROK$ , 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:

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

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

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

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

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

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Question

Three consecutive even angles add up to . What must be the value of the second largest angle?

Answer

Let be an even angle. The next consecutive even values are .

Set up an equation such that all angles added equal to 180.

Divide by three on both sides.

$\frac{3x}{3}=\frac{174}{3}$

The second largest angle is .

Substitute the value of in to the expression.

The answer is:

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. . $\angle C$ is a right angle. What percent of $\Delta ACE$ has been shaded in?

Answer

$\Delta BCD$ is a right triangle with legs ; its area is half the product of its legs, which is

$Area= \frac{1}{2} \times 3 \times 4 = 6$

$\Delta ACE$ is a right triangle with legs

and

;

its area is half the product of its legs, which is

$Area= \frac{1}{2} \times 5 \times 12 = 30$

The shaded region is the former triangle removed from the latter triangle; its area is the difference of the two: .

This region is therefore

$\frac{24}{30 } \times 100 % = 80 %$ of $\Delta ACE$ .

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Question

Note: Figure NOT drawn to scale.

Refer to the above diagram. . $\angle C$ is a right angle. What is the area of the shaded region?

Answer

$\Delta BCD$ is a right triangle with legs ; its area is half the product of its legs, which is

$Area= \frac{1}{2} \times 8 \times 8 = 32$

$\Delta ACE$ is a right triangle with legs

and

;

its area is half the product of its legs, which is

$Area= \frac{1}{2} \times 12 \times 24= 144$

The shaded region is the former triangle removed from the latter triangle; its area is the difference of the two: .

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Question

The above hexagon is regular. Give its area.

Answer

A regular hexagon can be divided into six triangles, each of which can be easily proved equilateral, as seen in the diagram below:

All segments shown are congruent, and, since the diameter shown in the original diagram is 4, each sidelength is half this, or 2.

Each equilateral triangle has area

$A = \frac{s^{2} \sqrt{3}}{4} = \frac{2^{2} \sqrt{3}}{4} = \frac{4\sqrt{3}}{4} = \sqrt{3}$ .

There are six such triangles, so the total area of the hexagon is six times this, or $6\sqrt{3}$ .

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Question

Give the area of a regular hexagon with perimeter 36.

Answer

A hexagon has six sides; a regular hexagon with perimeter 36 has sidelength

$36 \div 6 = 6$ .

A regular hexagon can be divided into six triangles, each of which can be easily proved equilateral, as seen in the diagram below:

Each equilateral triangle has sidelength 6, so each has area

$A = \frac{s^{2} \sqrt{3}}{4} = \frac{6^{2} \sqrt{3}}{4} = \frac{36 \sqrt{3}}{4} = 9 \sqrt{3}$ .

The total area of the hexagon is the area of six such triangles:

$6 \cdot 9 \sqrt{3} = 54 \sqrt{3}$

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Question

Determine the area of a square with a side length of .

Answer

Write the area of a square.

Substitute the side into the formula.

$A=(2x^5)^2 = (2x^5)(2x^5) = 4x^{10}$

The answer is: $4x^{10}$

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Question

Figure NOT drawn to scale.

Refer to the above figure. Every angle shown is a right angle.

Give its area.

Answer

Examine the bottom figure, in which the bottom two sides have been connected. Note that the figure is now a rectangle cut out of a rectangle, and, since the opposite sides of a rectangle have the same length, we can fill in some of the side lengths as shown:

The figure is a 60-by-40 rectangle cut from a 100-by-100 square, so, to get the area of the figure, subtract the area of the former from that of the latter. The area of a rectangle is equal to the product of its dimensions, so the areas of the rectangle and the square are, respectively,

$40 \times 60 = 2,400$

and

$100 \times 100 = 10,000$ ,

making the area of the figure

.

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Question

A circle is inscribed in square that has a side length of , as shown by the figure below.

Find the area of the shaded region. Use $\pi=3.14$ .

Answer

Since the circle is inscribed in the square, the diameter of the circle is the same length as the length of a square.

Start by finding the area of the square.

$\text{Area of Square}=\text{side}^2$

For the given square,

$\text{Area}=144$

Now, because the diameter of the circle is the same as the length of a side of the square, we now also know that the radius of the circle must be . Next recall how to find the area of a circle.

$\text{Area of Circle}=\pi r^2$

Plug in the found radius to find the area of the circle.

$\text{Area of Circle}=\pi(6)^2=36\pi=113.04$

Now, the shaded area is the area left over when the area of the circle is subtracted from the area of the square. Thus, we can write the following equation to find the area of the shaded region.

$\text{Area of Shaded Region}=144-113.04=30.96$

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Question

Josh wants to build a circular pool in his square yard that measures feet on each side. He wants to build the pool as big as possible, then pave the rest of his yard in tile. In square feet, what is the area of the yard that will be tiled? Round your answer to the nearest tenths place.

Answer

Start by drawing out the square yard and the circular pool in a way that maximizes the area of the pool.

Notice that the diameter of the pool will be the same length as the side of the square.

Since the question asks about the area that is left over after the pool is built, we can find that area by subtracting the area of the pool from the area of the square.

Start by finding the area of the square.

$\text{Area of Square}=50^2=2500$

Next, find the area of the circular pool.

Since the diameter of the pool is , the radius of the pool must be . Recall how to find the area of a circle:

$\text{Area}=\pi r^2$

Plug in the radius of the circle.

$\text{Area}=\pi (25)^2=625\pi$

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

$\text{Tiled Area}=2500-625\pi=536.5$

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Question

The above figure is a regular octagon. Give its perimeter in yards.

Answer

A regular octagon has eight sides of equal length, so multiply the length of one side by eight:

$6 \frac{1}{4} \times 8 = \frac{25}{4} \times \frac{8}{1} = \frac{200}{4} = 50$ feet.

Divide by three to get the equivalent in yards:

$50 \div 3 = 16 \frac{2}{3}$ yards.

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Question

Identify the above polygon.

Answer

A polygon with eight sides is called an octagon.

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