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Advanced Geometry : Advanced Geometry

Study concepts, example questions & explanations for Advanced Geometry

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6 Diagnostic Tests 57 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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

The diagonal lengths of a kite are and . What is the area?

Possible Answers:

$\frac{169}{4}$

$42\sqrt2$

$21\sqrt2$

Correct answer:

Explanation:

The area of a kite is given below. Substitute the given diagonals to find the area.

$A= \frac{d1\cdot d2}{2} = \frac{6\cdot 7}{2}=21$

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Example Question #2 : How To Find The Area Of A Kite

Find the area of a kite if the length of the diagonals are and .

Possible Answers:

$30\sqrt3$

$15\sqrt2$

$30\sqrt2$

Correct answer:

Explanation:

Substitute the given diagonals into the area formula for kites. Solve and simplify.

$A=\frac{d1\cdot d2}{2} = \frac{3\cdot 10}{2}=15$

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Example Question #3 : How To Find The Area Of A Kite

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

Possible Answers:

$5\:cm^2$

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

$5\sqrt2\:cm^2$

$\sqrt5\:cm^2$

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

Correct answer:

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

Explanation:

The formula for the area for a kite is

$A=\frac{d_1\cdot d_2}{2}$ , where d_1 and d_2 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:

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

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Example Question #4 : How To Find The Area Of A Kite

The diagonals of a kite are $(2x+1)\:units$ and $\frac{1}{x}\:units$ . Express the kite's area in simplified form.

Possible Answers:

$\frac{x+1}{2x}\:units^2$

$\frac{2x+1}{2x}\:units^2$

$\frac{2+x}{x}\:units^2$

$\frac{2}{x}\:units^2$

$\frac{2x+1}{4x}\:units^2$

Correct answer:

$\frac{2x+1}{2x}\:units^2$

Explanation:

Write the formula for the area of a kite.

$A=\frac{d_1\cdot d_2}{2}$

Substitute the diagonals and reduce.

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

Multiply the parenthetical elements together by distributing the $\frac{1}{x}$ :

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

$A=\frac{\frac{2x}{x}+\frac{1}{x}}{2}$

$A=\frac{2+\frac{1}{x}}{2}$

You can consider the outermost fraction with in the denominator as multiplying everything in the numerator by $\frac{1}{2}$ :

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

$A=1+\frac{1}{2x}$

Change the added to $\frac{2x}{2x}$ to create a common denominator and add the fractions to arrive at the correct answer:

$A=\frac{2x}{2x} + \frac{1}{2x} = \frac{2x+1}{2x}$

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Example Question #31 : Kites

If the diagonals of a kite are $\frac{1}{6x}\:units$ and $2x+5\:units$ , express the area in simplified form.

Possible Answers:

$\frac{2x+5}{12x}\:units^2$

$\frac{x+5}{6}\:units^2$

$\frac{2x+5}{3x}\:units^2$

$\frac{2x+5}{6x}\:units^2$

$\frac{x+5}{12}\:units^2$

Correct answer:

$\frac{2x+5}{12x}\:units^2$

Explanation:

Write the formula for the area of a kite:

$A=\frac{d_1 \cdot d_2}{2}$

Substitute the given diagonals:

$A = \frac{(\frac{1}{6x})(2x+5)}{2}$

Distribute the $\frac{1}{6x}$ :

$A= \frac{\frac{1}{3}+\frac{5}{6x}}{2}$

Create a common denominator for the two fractions in the numerator:

$A= \frac{\frac{2x}{6x}+\frac{5}{6x}}{2}$

$A= \frac{\frac{2x+5}{6x}}{2}$

You can consider the outermost division by as multiplying everything in the numerator by $\frac{1}{2}$ :

$A=(\frac{1}{2})(\frac{2x+5}{6x})$

Multiply across to arrive at the correct answer:

$A=\frac{2x+5}{12x}$

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

Find the area of a kite if one diagonal is $2\:cm$ long, and the other diagonal is $10\:cm$ long.

Possible Answers:

$3\sqrt{10}$ cm^2

$20\:cm^2$

$10\:cm^2$

$10\sqrt2\:cm^2$

$10\sqrt3\:cm^2$

Correct answer:

$10\:cm^2$

Explanation:

The formula for the area of a kite is

$A=\frac{d_1\cdot d_2}{2}$

Plug in the values for each of the diagonals and solve.

$A = \frac{2\cdot 10}{2}=\frac{20}{2}=10\:cm^2$

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Example Question #31 : Kites

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

Possible Answers:

4a+4b

2ab

2a+2b

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 #1 : How To Find The Area Of A Kite

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

Possible Answers:

a^2-b^2

2a^2+2b^2

2a^2-2b^2

a^2-ab-b^2

a^2+b^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 #331 : Advanced Geometry

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

Possible Answers:

$\sqrt{14}$

$\sqrt{23}$

$\sqrt{56}$

$2\sqrt{14}$

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 #42 : Kites

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

Varsity5

Possible Answers:

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

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

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

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

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

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

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

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

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

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

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

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

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

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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Advanced Geometry : Advanced Geometry

Study concepts, example questions & explanations for Advanced Geometry

All Advanced Geometry Resources

Example Questions

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

Example Question #2 : How To Find The Area Of A Kite

Example Question #3 : How To Find The Area Of A Kite

Example Question #4 : How To Find The Area Of A Kite

Example Question #31 : Kites

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

Example Question #31 : Kites

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

Example Question #331 : Advanced Geometry

Example Question #42 : Kites

All Advanced Geometry Resources

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