Plane Geometry - Advanced Geometry

Card 0 of 20

Question

What is the area of the following kite?

Answer

The formula for the area of a kite:

$A = \frac{1}{2} (a)(b)$ ,

where represents the length of one diagonal and represents the length of the other diagonal.

Plugging in our values, we get:

$A = \frac{1}{2} (4: m)(5: m)$

$A = \frac{1}{2} (20: m^2) = 10: m^2$

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Question

Which of the following shapes is a kite?

Answer

A kite is a four-sided shape with straight sides that has two pairs of sides. Each pair of adjacent sides are equal in length. A square is also considered a kite.

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Question

Two congruent equilateral triangles with sides of length are connected so that they share a side. Each triangle has a height of . Express the area of the shape in terms of .

Answer

The shape being described is a rhombus with side lengths 1. Since they are equilateral triangles connected by one side, that side becomes the lesser diagonal, so .

The greater diagonal is twice the height of the equaliteral triangles, .

The area of a rhombus is half the product of the diagonals, so:

$A=\frac{pq}{2}=\frac{1 \cdot 2h}{2}=h$

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Question

The diagonal lengths of a kite are and . What is the area?

Answer

The area of a kite is given below. Substitute the given diagonals to find the area.

$A= \frac{d1\cdot d2}{2} = \frac{6\cdot 7}{2}=21$

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Question

Find the area of a kite if the length of the diagonals are and .

Answer

Substitute the given diagonals into the area formula for kites. Solve and simplify.

$A=\frac{d1\cdot d2}{2} = \frac{3\cdot 10}{2}=15$

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Question

Find the area of a kite if the diagonal dimensions are and .

Answer

The area of the kite is given below. The FOIL method will need to be used to simplify the binomial.

$A=\frac{d1\cdot d2}{2}=\frac{(x+3)(x-6)}{2}= \frac{x^2-6x+3x-18}{2}$

$A= \frac{x^2-3x-18}{2}$

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Question

Find the area of a kite if one diagonal is long, and the other diagonal is long.

Answer

The formula for the area of a kite is

$A=\frac{d_1\cdot d_2}{2}$

Plug in the values for each of the diagonals and solve.

$A = \frac{2\cdot 10}{2}=\frac{20}{2}=10:cm^2$

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Question

The diagonals of a kite are $\sqrt10:cm$ and $\sqrt5:cm$ . Find the area.

Answer

The formula for the area for a kite is

$A=\frac{d_1\cdot d_2}{2}$ , where and are the lengths of the kite's two diagonals. We are given the length of these diagonals in the problem, so we can substitute them into the formula and solve for the area:

$A = \frac{d1\cdot d2}{2} = \frac{\sqrt10 \cdot \sqrt5}{2} = \frac{\sqrt50}{2} = \frac{5\sqrt2}{2}:cm^2$

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Question

The diagonals of a kite are and $\frac{1}{x}:units$ . Express the kite's area in simplified form.

Answer

Write the formula for the area of a kite.

$A=\frac{d_1\cdot d_2}{2}$

Substitute the diagonals and reduce.

$A=\frac{d_1\cdot d_2}{2} = \frac{(2x+1)(\frac{1}{x})}{2}$

Multiply the parenthetical elements together by distributing the $\frac{1}{x}$ :

$A= \frac{(2x\cdot \frac{1}{x})+(1\cdot \frac{1}{x})}{2}$

$A=\frac{\frac{2x}{x}+\frac{1}{x}}{2}$

$A=\frac{2+\frac{1}{x}}{2}$

You can consider the outermost fraction with in the denominator as multiplying everything in the numerator by $\frac{1}{2}$ :

$A=(\frac{1}{2})(2+\frac{1}{x})$

$A=1+\frac{1}{2x}$

Change the added to $\frac{2x}{2x}$ to create a common denominator and add the fractions to arrive at the correct answer:

$A=\frac{2x}{2x} + \frac{1}{2x} = \frac{2x+1}{2x}$

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Question

If the diagonals of a kite are $\frac{1}{6x}:units$ and , express the area in simplified form.

Answer

Write the formula for the area of a kite:

$A=\frac{d_1 \cdot d_2}{2}$

Substitute the given diagonals:

$A = \frac{(\frac{1}{6x})(2x+5)}{2}$

Distribute the $\frac{1}{6x}$ :

$A= \frac{\frac{1}{3}+\frac{5}{6x}}{2}$

Create a common denominator for the two fractions in the numerator:

$A= \frac{\frac{2x}{6x}+\frac{5}{6x}}{2}$

$A= \frac{\frac{2x+5}{6x}}{2}$

You can consider the outermost division by as multiplying everything in the numerator by $\frac{1}{2}$ :

$A=(\frac{1}{2})(\frac{2x+5}{6x})$

Multiply across to arrive at the correct answer:

$A=\frac{2x+5}{12x}$

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Question

Find the area of a kite with the diagonal lengths of and .

Answer

Write the formula to find the area of a kite. Substitute the diagonals and solve.

$A=\frac{d1 \cdot d2}{2} = \frac{2a \cdot 2b}{2} =2ab$

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Question

Find the area of a kite with diagonal lengths of and .

Answer

Write the formula for the area of a kite.

$A= \frac{d1\cdot d2}{2}$

Plug in the given diagonals.

$A= \frac{(a+b)\cdot (2a-2b)}{2}$

Pull out a common factor of two in and simplify.

$A= \frac{(a+b)\cdot2(a-b)}{2}$

Use the FOIL method to simplify.

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Question

Show algebraically how the formula of the area of a kite is developed.

Answer

The given kite is divded into two congruent triangles.

Each triangle has a height $h = \frac{1}{2}d_{1}$ and a base $b = d_{2}$ .

The area of each triangle is $A = \frac{1}{2}bh$ .

The areas of the two triangles are added together,

$= \frac{1}{2}(\frac{1}{2}d_{1})(d_{2}) + \frac{1}{2}(\frac{1}{2}d_{1})(d_{2})$

$= \frac{1}{4}d_{1}d_{2} +\frac{1}{4}d_{1}d_{2}$

$=\frac{1}{2}d_{1}d_{2}$

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Question

Find the area of a kite, if the diagonals of the kite are $\sqrt{8}, \sqrt{7}$ .

Answer

You find the area of a kite by using the lengths of the diagonals.

$A=\frac{d_{1}\times d_{2}}{2}$

$A=\frac{\sqrt{8}\times \sqrt{7}}{2}$ , which is equal to $\frac{\sqrt{56}}{2}$ .

You can reduce this to,

$\frac{\sqrt{2\cdot 4\cdot 7}}{2}=\frac{2\sqrt{2\cdot 7}}{2}=\sqrt{2\cdot 7}=\sqrt{14}$ for the final answer.

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Question

The rectangle area $A_{R}$ is 220. What is the area $A_{K}$ of the inscribed

kite ?

Answer

The measures of the kite diagonals $\overline{GH}$ and $\overline{BE}$ have to be found.

Using the circumscribed rectangle, , and .

has to be found to find $m(\overline{GH})$ .

The rectangle area $A_{R} = 220$ .

$A_{R} = bh$ .

From step 1) and step 2), using substitution, .

Solving the equation for x,

$\frac{(19 + x)(10)}{10} = \frac{220}{10}$

$m(\overline{GH}) = 19 + 3 = 22$

Kite area $A_{K} = \frac{1}{2} \cdot 22 \cdot 10 = 110$

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Question

A rhombus has a side length of 5. Which of the following is NOT a possible value for its area?

Answer

The area of a rhombus will vary as the angles made by its sides change. The "flatter" the rhombus is (with two very small angles and two very large angles, say 2, 178, 2, and 178 degrees), the smaller the area is. There is, of course, a lower bound of zero for the area, but the area can get arbitrarily small. This implies that the correct answer would be the largest choice. In fact, the largest area of a rhombus occurs when all four angles are equal, i.e. when the rhombus is a square. The area of a square of side length 5 is 25, so any value bigger than 25 is impossible to acheive.

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Question

Which of the following shapes is a rhombus?

Answer

A rhombus is a four-sided figure where all sides are straight and equal in length. All opposite sides are parallel. A square is considered to be a rhombus.

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Question

Assume quadrilateral $\small ABCD$ is a rhombus. The perimeter of $\small ABCD$ is , and the length of one of its diagonals is $\small 12$ . What is the area of $\small ABCD$ ?

Answer

To solve for the area of the rhombus $\small ABCD$ , we must use the equation $\small A=\frac{pq}{2}$ , where $\small p$ and $\small q$ are the diagonals of the rhombus. Since the perimeter of the rhombus is $\small 40$ , and by definition all 4 sides of a rhombus have the same length, we know that the length of each side is $\small 10$ . We can find the length of the other diagonal if we recognize that the two diagonals combined with a side edge form a right triangle. The length of the hypotenuse is $\small 10$ , and each leg of the triangle is equal to one-half of each diagonal. We can therefore set up an equation involving Pythagorean's Theorem as follows:

$\small 10^2=x^2+6^2$ , where $\small x$ is equal to one-half the length of the unknown diagonal.

We can therefore solve for $\small x$ as follows:

$\small x^2=10^2-6^2=100-36=64$

$\small x$ is therefore equal to 8, and our other diagonal is 16. We can now use both diagonals to solve for the area of the rhombus:

$\small A=\frac{(16)(12)}{2}=\frac{192}{2}=96$

The area of rhombus $\small ABCD$ is therefore equal to $\small 96$

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Question

Assume quadrilateral is a rhombus. If diagonal $\overline{MO}=29$ and diagonal $\overline{NP}=14$ , what is the area of rhombus

Answer

Solving for the area of rhombus requires knowledge of the equation for finding the area of a rhombus. The equation is $\small A=\frac{pq}{2}$ , where and $\small q$ are the two diagonals of the rhombus. Since both of these values are given to us in the original problem, we merely need to substitute these values into the equation to obtain:

$\small A=\frac{(29)(14)}{2}=\frac{406}{2}=203$

The area of rhombus $\small MNOP$ is therefore $\small 203$ square units.

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Question

What is the area of the rhombus above?

Answer

The formula for the area of a rhombus from the diagonals is half the product of the diagonals, or in mathematical terms:

$A=\frac{p\cdot q}{2}$ where and are the lengths of the diagonals.

Substituting our values yields,

$A=\frac{8*12}{2}=\frac{96}{2}=48$

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