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Advanced Geometry : How to find the area of a kite

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Example Question #31 : Quadrilaterals

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

Possible Answers:

2ab

2a+2b

4a+4b

4ab

Correct answer:

2ab

Explanation:

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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Example Question #11 : Geometry

Find the area of a kite with diagonal lengths of a+b and 2a-2b .

Possible Answers:

a^2+b^2

a^2-ab-b^2

2a^2+2b^2

a^2-b^2

2a^2-2b^2

Correct answer:

a^2-b^2

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 2a-2b and simplify.

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

A=(a+b)(a-b)

Use the FOIL method to simplify.

A=(a+b)(a-b)= (a)(a)+(a)(-b)+(b)(a)+(b)(-b)

A= a^2-b^2

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Example Question #41 : Kites

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

Possible Answers:

$\sqrt{56}$

$\sqrt{23}$

$\sqrt{14}$

$2\sqrt{14}$

Correct answer:

$\sqrt{14}$

Explanation:

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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Example Question #41 : Quadrilaterals

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

Varsity5

Possible Answers:

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

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

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

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

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

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

$\frac{1}{4}d_{1}\cdot \frac{1}{4}d_{2}$

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

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

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

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

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

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

Correct answer:

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

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

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

Explanation:

1) The given kite is divded into two congruent triangles.

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

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

4) The areas of the two triangles are added together,

A + A

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

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

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

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Example Question #41 : Kites

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

kite GBHE ?

Varsity4

Possible Answers:

110

190

47.5

Correct answer:

110

Explanation:

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

2) Using the circumscribed rectangle, GH = 19 + x , and BE = 5 + 5 = 10 .

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

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

5) $A_{R} = bh$ .

6) From step 1) and step 2), using substitution, (19 + x)(10) = 220 .

7) Solving the equation for x,

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

19 + x = 22

x = 22 - 19

x = 3

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

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

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Advanced Geometry : How to find the area of a kite

Study concepts, example questions & explanations for Advanced Geometry

All Advanced Geometry Resources

Example Questions

Example Question #31 : Quadrilaterals

Example Question #11 : Geometry

Example Question #41 : Kites

Example Question #41 : Quadrilaterals

Example Question #41 : Kites

All Advanced Geometry Resources

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