Quadrilaterals - Geometry

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What is the area of the following kite?

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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$$

Which of the following shapes is a kite?

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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.

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 .

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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}$=h$

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

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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$

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

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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$

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

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

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

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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$$

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

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

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

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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}$$

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

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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}$$

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

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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$

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

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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.

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

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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}$$d_{2}$$

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

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

$A=\frac{\sqrt{8}$\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.

The rectangle area is 220. What is the area of the inscribed

kite ?

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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 .

.

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

The area of the rectangle is , what is the area of the kite?

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The area of a kite is half the product of the diagonals.

$A=\frac{p \cdot q}{2}$$

The diagonals of the kite are the height and width of the rectangle it is superimposed in, and we know that because the area of a rectangle is base times height.

Therefore our equation becomes:

$A=\frac{l \cdot w}{2}$$ .

We also know the area of the rectangle is $A=l \cdot w=80$ . Substituting this value in we get the following:

$A=\frac{l \cdot w}{2}$=\frac{80}{2}$=40$

Thus,, the area of the kite is .

Given: Quadrilateral such that $\overline{KI}$ \cong $\overline{KE}$ , $\overline{TI}$ \cong $\overline{TE}$ , $m \angle K = $60^{\circ }$$ , $\angle T$ is a right angle, and diagonal $\overline{I E}$ has length 24.

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

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The Quadrilateral is shown below with its diagonals $\overline{I E}$ and $\overline{KT}$ .

. We call the point of intersection :

The diagonals of a quadrilateral with two pairs of adjacent congruent sides - a kite - are perpendicular; also, $\overline{KT}$ bisects the $60 ^{\circ }$ and angles of the kite. Consequently, $\bigtriangleup KXI$ is a 30-60-90 triangle and $\bigtriangleup TXI$ is a 45-45-90 triangle. Also, the diagonal that connects the common vertices of the pairs of adjacent sides bisects the other diagonal, making the midpoint of $\overline{IE}$ . Therefore,

$IX = $\frac{1}{2}$ \cdot IE = $\frac{1}{2}$ \cdot 24 = 12$ .

By the 30-60-90 Theorem, since $\overline{IX}$ and $\overline{KX}$ are the short and long legs of $\bigtriangleup KXI$ ,

$KX = IX \cdot $\sqrt{3}$= 12 $\sqrt{3}$$

By the 45-45-90 Theorem, since $\overline{IX}$ and $\overline{XT}$ are the legs of a 45-45-90 Theorem,

.

The diagonal $\overline{KT}$ has length

$KT =XT + KX= 12+ 12$\sqrt{3}$$ .

If the area of a kite is square units, and one diagonal is units longer than the other, what is the length of the shorter diagonal?

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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 .

$198=\frac{x(x+4)}{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.

A kite has two perpendicular interior diagonals. One diagonal is twice the length of the other diagonal. The total area of the kite is $196\textup{ $units}^{2}$$ . Find the length of each interior diagonal.

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To solve this problem, apply the formula for finding the area of a kite:

$Area=\frac{diagonalA\times diagonalB}{2}$$

However, in this problem the question only provides information regarding the exact area. The lengths of the diagonals are represented as a ratio, where

Therefore, it is necessary to plug the provided information into the area formula. Diagonal is represented by and diagonal .

The solution is:

$196=\frac{x\times 2x}{2}$$

$196\times2=x\times 2x$

$x=$\sqrt{196}$=14$

Thus, if , then diagonal must equal

A kite has two perpendicular interior diagonals. One diagonal has a measurement of and the area of the kite is $60\textup{ $units}^{2}$$ . Find the length of the other interior diagonal.

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This problem can be solved by applying the area formula:

Since this question provides the area of the kite and length of one diagonal, plug that information into the equation to solve for the missing diagonal.

Thus the solution is:

$60=\frac{8\times diagonal B}{2}$$

$60\times2=8\times diagonalB$

$diagonal B=\frac{120}{8}$=15$