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Advanced Geometry : Kites

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Example Questions

Example Question #41 : Kites

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

Possible Answers:

$\sqrt{14}$ $\displaystyle \sqrt{14}$

$\sqrt{56}$ $\displaystyle \sqrt{56}$

$2\sqrt{14}$ $\displaystyle 2\sqrt{14}$

$\sqrt{23}$ $\displaystyle \sqrt{23}$

$\displaystyle 28$

Correct answer:

$\sqrt{14}$ $\displaystyle \sqrt{14}$

Explanation:

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

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

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

You can reduce this to,

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

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Example Question #41 : Plane Geometry

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

Varsity5

Possible Answers:

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

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

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

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

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

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

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

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

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

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

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

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

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

Correct answer:

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

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

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

Explanation:

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

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

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

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

A + A $\displaystyle A + A$

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

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

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

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Example Question #41 : Kites

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

kite GBHE $\displaystyle GBHE$ ?

Varsity4

Possible Answers:

$\displaystyle 60$

110 $\displaystyle 110$

190 $\displaystyle 190$

47.5 $\displaystyle 47.5$

$\displaystyle 95$

Correct answer:

110 $\displaystyle 110$

Explanation:

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

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

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

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

5) $A_{R} = bh$ $\displaystyle A_{R} = bh$ .

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

7) Solving the equation for x,

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

$\displaystyle 19 + x = 22$

$\displaystyle x = 22 - 19$

x = 3 $\displaystyle x = 3$

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

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

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Advanced Geometry : Kites

Study concepts, example questions & explanations for Advanced Geometry

All Advanced Geometry Resources

Example Questions

Example Question #41 : Kites

Example Question #41 : Plane Geometry

Example Question #41 : Kites

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