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Advanced Geometry : Quadrilaterals

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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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Example Question #1 : Trapezoids

Which of the following shapes is a trapezoid?

Shapes

Possible Answers:

C, D

B, D

None

B, C, D

Correct answer:

B, D

Explanation:

A trapezoid is a four-sided shape with straight sides that has a pair of opposite parallel sides. The other sides may or may not be parallel. A square and a rectangle are both considered trapezoids.

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Example Question #1 : Trapezoids

What is the area of the following trapezoid?

Trapezoid

Possible Answers:

36m^2

24m^2

32m^2

40m^2

80m^2

Correct answer:

36m^2

Explanation:

The formula for the area of a trapezoid is:

$A = \frac{1}{2} (b_{1} + b_{2})(h)$ ,

where $b_{1}$ is the value of the top base, $b_{2}$ is value of the bottom base, and is the value of the height.

Plugging in our values, we get:

$A = \frac{1}{2} (8\: m + 10\: m)(4\: m)$

$A = \frac{1}{2} (18\: m)(4\: m)$

$A = (9\: m)(4\: m) = 36\: m^2$

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Example Question #1 : Trapezoids

Screen_shot_2015-03-06_at_2.30.09_pm

What is the area of the trapezoid pictured above in square units?

Possible Answers:

238

119

168

Correct answer:

119

Explanation:

The formula for the area of a trapezoid is the average of the bases times the height,

$A=\frac{b_{1}+b_{2}}{2}h$ .

Looking at this problem and when the appropriate values are plugged in, the formula yields:

$A=(\frac{10+24}{2})7$

$A=\frac{34}{2}\cdot 7$

$A=\frac{238}{2}=119$

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Example Question #1 : How To Find The Area Of A Trapezoid

Screen_shot_2015-03-06_at_2.44.49_pm

What is the height of the trapezoid pictured above?

Possible Answers:

Correct answer:

Explanation:

To find the height, we must introduce two variables, $b_{1}, b_{2}$ , each representing the bases of the triangles on the outside, so that $b_{1}+b_{2}+11=25, b_{1}+b_{2}=14$ . (Equation 1)

The next step is to set up two Pythagorean Theorems,

$13^2=b_{1}^2+h^2, 15^2=b_{2}^2+h^2$ (Equation 2, 3)

The next step is a substitution from the first equation,

$b_{1}=14-b_{2}, b_{1}^2=196-28b_{2}+b_{2}^2$ (Equation 4)

and plugging it in to the second equation, yielding

$13^2=196-28b_{2}+b_{2}^2+h^2$ (Equation 5)

The next step is to substitute from Equation 3 into equation 5,

$h^2=15^2-b_{2}^2$ , $169=196-28b_{2}+b_{2}^2+225-b_{2}^2$ which simplifies to

$-252=-28b_{2}, b_{2}=9$

Once we have one of the bases, just plug into the Pythagorean Theorem, 225=81+h^2, 144=h^2, h=12

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Example Question #1 : Trapezoids

A isosceles trapezoid with sides , , , and has a height of , what is the area?

Possible Answers:

180

194

138

160

Correct answer:

180

Explanation:

An isosceles trapezoid has two sides that are the same length and those are not the bases, so the bases are 10 and 20.

The area of the trapezoid then is:

$A=\frac{b_{1}+b_{2}}{2}h=\frac{10+20}{2}12=15 \cdot 12=180$

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Example Question #1 : How To Find The Area Of A Trapezoid

If the height of a trapezoid is , bottom base is , and the top base is , what is the area?

Possible Answers:

150

300

200

Correct answer:

150

Explanation:

The formula for finding the area of a trapezoid is:

$A=\frac{1}{2}(b1+b2)h$

Substitute the given values to find the area.

$A=\frac{1}{2}(20+10)(10)$

$A= \frac{1}{2}(30)(10)$

$A=\frac{1}{2}(300)$

A=150

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Example Question #1 : Trapezoids

Find the area of a trapezoid with bases of length $12\:cm$ and $15\:cm$ and a height of $7\:cm$ .

Possible Answers:

$94.5\:cm^2$

$105\:cm^2$

$52.5\:cm^2$

$84\:cm^2$

$189\:cm^2$

Correct answer:

$94.5\:cm^2$

Explanation:

The formula for the area of a trapezoid is:

$A = \frac{a + b}{2}h$

Where and are the bases and is the height. Using this formula and the given values, we get:

$A = \frac{12\:cm + 15\:cm}{2}\left ( 7\:cm) = 94.5\:cm^2$

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Advanced Geometry : Quadrilaterals

Study concepts, example questions & explanations for Advanced Geometry

All Advanced Geometry Resources

Example Questions

Example Question #41 : Kites

Example Question #41 : Quadrilaterals

Example Question #41 : Kites

Example Question #1 : Trapezoids

Example Question #1 : Trapezoids

Example Question #1 : Trapezoids

Example Question #1 : How To Find The Area Of A Trapezoid

Example Question #1 : Trapezoids

Example Question #1 : How To Find The Area Of A Trapezoid

Example Question #1 : Trapezoids

All Advanced Geometry Resources

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