Advanced Geometry : Kites

Study concepts, example questions & explanations for Advanced Geometry

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Example Questions

Example Question #41 : Kites

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

Possible Answers:

\displaystyle \sqrt{14}

\displaystyle \sqrt{56}

\displaystyle 2\sqrt{14}

\displaystyle \sqrt{23}

\displaystyle 28

Correct answer:

\displaystyle \sqrt{14}

Explanation:

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

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

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

You can reduce this to,

 \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. 

Example Question #41 : Plane Geometry

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

Varsity5

 

Possible Answers:

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

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

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

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

\displaystyle =\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})

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

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

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

 \displaystyle =\frac{1}{2}d_{1}d_{2}

\displaystyle (\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_{2})

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

Correct answer:

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

\displaystyle = \frac{1}{4}d_{1}d_{2} +\frac{1}{4}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 \displaystyle h = \frac{1}{2}d_{1} and a base \displaystyle b = d_{2}.

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

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

\displaystyle A + A

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

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

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

Example Question #41 : Kites

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

kite \displaystyle GBHE?

 Varsity4

Possible Answers:

\displaystyle 60

\displaystyle 110

\displaystyle 190

\displaystyle 47.5

\displaystyle 95

Correct answer:

\displaystyle 110

Explanation:

1) The measures of the kite diagonals \displaystyle \overline{GH} and \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 \displaystyle m(\overline{GH}).

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

5) \displaystyle A_{R} = bh.

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

7) Solving the equation for x,

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

\displaystyle 19 + x = 22

\displaystyle x = 22 - 19

\displaystyle x = 3

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

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

 

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