Plane Geometry

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Questions 1 - 10

A kite has an area of square units, and one diagonal is units longer than the other. In unites, what is the length of the shorter diagonal?

Explanation

Let be the length of the shorter diagonal. Then the length of the longer diagonal can be represented by .

Recall how to find the area of a kite:

$\text{Area}=\frac{(\text{Diagonal 1})(\text{Diagonal 2})}{2}$

Plug in the given area and solve for .

$119=\frac{x(x+3)}{2}$

Since we are dealing with geometric shapes, the answer must be a positive value. Thus, .

The length of the shorter diagonal is units long.

Find the area of the rhombus.

Explanation

Recall that one of the ways to find the area of a rhombus is with the following formula:

$\text{Area}=\text{base}\times\text{height}$

Now, since all four sides in the rhombus are the same, we know from the given side value what the length of the base will be. In order to find the length of the height, we will need to use sine.

$\sin x=\frac{\text{height}}{\text{side}}$ , where is the given angle.

$\text{height}=(\text{side})(\sin x)$

Now, plug this into the equation for the area to get the following equation:

$\text{Area}=(side)^2 \sin x$

Plug in the given side length and angle values to find the area.

$\text{Area}=(5)^2\sin 33=13.62$

Make sure to round to places after the decimal.

$\bigtriangleup ABC$ is an equilateral triangle; $\bigtriangleup DEF$ is an equiangular triangle.

True or false: From the given information, it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ .

True

False

Explanation

As we are establishing whether or not $\bigtriangleup ABC \sim \bigtriangleup DEF$ , then , , and correspond respectively to , , and .

A triangle is equilateral (having three sides of the same length) if and only if it is also equiangular (having three angles of the same measure, each of which is $60^{\circ }$ ). It follows that all angles of both triangles measure $60^{\circ }$ .

Specifically, $\angle A \cong \angle D$ and $\angle B \cong \angle E$ , making two pairs of corresponding angles congruent. By the Angle-Angle Similarity Postulate, it follows that $\bigtriangleup ABC \sim \bigtriangleup DEF$ , making the correct answer "true".

An isosceles trapezoid has base measurements of and . The perimeter of the trapezoid is . Find the length for one of the two remaining sides.

Explanation

To solve this problem, first note that an isosceles trapezoid has two parallel bases that are nonequivalent in length. Additionally, an isosceles trapezoid must have two nonparallel sides that have equivalent lengths. Since this problem provides the length for both of the bases as well as the total perimeter, the missing sides can be found using the following formula: Perimeter= Base one Base two (leg), where the length of "leg" is one of the two equivalent nonparallel sides.

Thus, the solution is:

$leg=\frac{10}{2}=5$

Check the solution by plugging in the answer:

Find the length of a side of a rhombus that has diagonal lengths of and .

Explanation

Recall that in a rhombus, the diagonals are not only perpendicular to each other, but also bisect one another.

Thus, we can find the lengths of half of each diagonal and use that in the Pythagorean Theorem to find the length of the side of the rhombus.

First, find the lengths of half of each diagonal.

$\text{Half Diagonal 1}=\frac{12}{2}=6$

$\text{Half Diagonal 2}=\frac{14}{2}=7$

Now, use these half diagonals as the legs of a right triangle that has the side of the rhombus as its hypotenuse.

$\text{Side of Rhombus}=\sqrt{(\text{Half Diagonal 1})^2+(\text{Half Diagonal 2})^2}$

Plug in the lengths of the half diagonals to find the length of the rhombus.

$\text{Side of Rhombus}=\sqrt{6^2+7^2}=\sqrt{85}=9.22$

Make sure to round to places after the decimal.

Find the length of the hypotenuse.

$\sqrt{14}$

$4\sqrt2$

$\sqrt{15}$

$3\sqrt3$

Explanation

Recall how to find the length of the hypotenuse, , of a right triangle by using the Pythagorean Theorem.

Substitute in the given values.

$c^2=(\sqrt6)^2+(2\sqrt2)^2$

Simplify.

Solve.

Now, because we want to solve for just , take the square root of the value you found above.

$\sqrt{c^2}=c=\sqrt{14}$

What is the hypotenuse of a right triangle with side lengths and ?

Explanation

The Pythagorean Theorem states that . This question gives us the values of and , and asks us to solve for .

Take and and plug them into the equation as and :

$12^{2}+16^{2}=c^{2}$

Now we can start solving for :

$144+256=c^{2}$

$400=c^{2}$

$\sqrt{400}=\sqrt{c^{2}}$

The length of the hypotenuse is .

Kite_series_2

Using the kite shown above, find the length of side

Explanation

A kite is a geometric shape that has two sets of equivalent adjacent sides.

Thus, the length of side .

Since, , must equal .

Find the value of if the area of this trapezoid is .

Explanation

The formula to find the area of a trapezoid is

$\text{Area}=\frac{1}{2}(base_1+base_2)(height)$ .

Substitute in the values for the area, a base, and the height. Then solve for .

$90=\frac{1}{2}(14+a)6$

Given Trapezoid with bases $\overline{TR}$ and $\overline{AP}$ . The trapezoid has midsegment $\overline{MN}$ , where and are the midpoints of $\overline{TP}$ and $\overline{RA}$ , respectively.

True, false, or undetermined: Trapezoid $TRNM \sim$ Trapezoid .

False

True

Undetermined

Explanation

The fact that the trapezoid is isosceles is actually irrelevant. Since and are the midpoints of legs $\overline{TP}$ and $\overline{RA}$ , it holds by definition that

$\overline{TM} \cong \overline{MP}$

and

$\overline{RN} \cong \overline{NA}$

It follows that $\frac{TM}{MP} = \frac{RN}{NA} = 1$

However, the bases of each trapezoid are noncongruent, by definition, so, in particular,

$TR \ne AP$

Assume that $\overline{AP}$ is the longer base - this argument works symmetrically if the opposite is true. Let - equivalently, .

Since the bases are of unequal length, $x \ne 0$ . The length of the midsegment $\overline{MN}$ is the arithmetic mean of the lengths of these two bases, so