How to find the area of a kite

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Math › How to find the area of a kite

Questions 1 - 10

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 .

$\frac{h}{2}$

Explanation

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:

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

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

Explanation

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.

Find the area of the following kite:

Kite

Explanation

The formula for the area of a kite is:

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

Where $d_{1}$ is the length of one diagonal and $d_{2}$ is the length of the other diagonal

Plugging in our values, we get:

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

$A = \frac{1}{2}(6m\cdot 13m) = 39m^2$

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

$\frac{5\sqrt2}{2}:cm^2$

$\frac{2\sqrt5}{2}:cm^2$

$5\sqrt2:cm^2$

$\sqrt5:cm^2$

Explanation

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:

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

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

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

$(x^2-3x-18)\sqrt2$

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

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

Explanation

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 the following kite:

Screen_shot_2014-03-01_at_9.16.34_pm

$18\sqrt{3}+18m^2$

$18\sqrt{3}m^2$

$18\sqrt{3}-18m^2$

Explanation

The formula for the area of a kite is:

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

where is the length of one diagonal and is the length of another diagonal.

Use the formulas for a triangle and a triangle to find the lengths of the diagonals. The formula for a triangle is $a-a-a \sqrt{2}$ and the formula for a triangle is $a-a \sqrt{3}-2a$ .

Our triangle is: $3 \sqrt{2}m-3 \sqrt{2}m-6m$

Our triangle is: $3 \sqrt{2}m-3 \sqrt{6}m-6\sqrt{2}m$

Plugging in our values, we get:

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

$A = \frac{1}{2} (6\sqrt{2}m) (3\sqrt{6}m + 3\sqrt{2}m)$

$A = \frac{1}{2} (18\sqrt{12}+18\sqrt{4}) = 18\sqrt{3}+18m^2$

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

kite ?

Varsity4

Explanation

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,