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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 : Calculating The Length Of The Diagonal Of A Quadrilateral

Rhombus RHOM has area 56.

Which of the following could be true about the values of and ?

Possible Answers:

RO = 4, HM = 14

RO = 14, HM = 14

RO = 16, HM = 7

None of the other responses gives a correct answer.

RO = 28, HM = 28

Correct answer:

RO = 16, HM = 7

Explanation:

The area of a rhombus is half the product of the lengths of its diagonals, which here are $\overline{RO}$ and $\overline{HM}$ . This means

$\frac{1}{2} \cdot RO \cdot HM = 56$

Therefore, we need to test each of the choices to find the pair of diagonal lengths for which this holds.

RO = 28, HM = 28 :

Area: $\frac{1}{2} \cdot RO \cdot HM = \frac{1}{2} \cdot 28 \times 28 = 392$

RO = 14, HM = 14

Area: $\frac{1}{2} \cdot RO \cdot HM = \frac{1}{2} \cdot 14 \times 14 =98$

RO = 4, HM = 14

Area: $\frac{1}{2} \cdot RO \cdot HM = \frac{1}{2} \cdot 4 \times 14 = 28$

RO = 16, HM = 7

Area: $\frac{1}{2} \cdot RO \cdot HM = \frac{1}{2} \cdot16 \times 7 =56$

RO = 16, HM = 7 is the correct choice.

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Example Question #51 : Rhombuses

Rhombus RHOM has perimeter 64; $m \angle R = 60 ^{\circ }$ . What is the length of $\overline{HM}$ ?

Possible Answers:

$16\sqrt{2}$

$16\sqrt{3}$

$8\sqrt{3}$

$8\sqrt{2}$

Correct answer:

Explanation:

The sides of a rhombus are all congruent; since the perimeter of Rhombus RHOM is 64, each side measures one fourth of this, or 16.

The referenced rhombus, along with diagonal $\overline{HM}$ , is below:

Rhombus

Since consecutive angles of a rhombus, as with any other parallelogram, are supplementary, $\angle RHO$ and $\angle RMO$ have measure $120^{\circ }$ ; $\overline{HM}$ bisects both into $60 ^{\circ }$ angles, making $\bigtriangleup RHM$ equilangular and, as a consequence, equilateral. Therefore, HM = RH = 16 .

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Example Question #2 : Calculating The Length Of The Diagonal Of A Quadrilateral

Rhombus RHOM has perimeter 48; $m \angle R = 60 ^{\circ }$ . What is the length of $\overline{RO}$ ?

Possible Answers:

$6\sqrt{2}$

$6\sqrt{3}$

$12\sqrt{3}$

Correct answer:

$12\sqrt{3}$

Explanation:

The referenced rhombus, along with diagonals $\overline{RO}$ and $\overline{HM}$ , is below.

Rhombus

The four sides of a rhombus have equal measure, so each side has measure one fourth of the perimeter of 48, which is 12.

Since consecutive angles of a rhombus, as with any other parallelogram, are suplementary, $\angle RHO$ and $\angle RMO$ have measure $120^{\circ }$ ; the diagonals bisect $\angle MRH$ and $\angle RHO$ into $30 ^{\circ }$ and $60 ^{\circ }$ angles, respectively, to form four 30-60-90 triangles. $\bigtriangleup HXR$ is one of them; by the 30-60-90 Triangle Theorem, $HX = \frac{1}{2} \cdot RH = \frac{1}{2} \cdot 12 =6$ ,

and

$RX = HX \cdot \sqrt{3} = 6 \sqrt{3}$ .

Since the diagonals of a rhombus bisect each other, $RO = 2 \cdot RX = 2 \cdot 6 \sqrt{3} = 12\sqrt{3}$ .

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Example Question #51 : Rhombuses

If the area of a rhombus is $60\:cm^2$ , and the length of one of its diagonals is $2\:cm$ , what must be the length of the other diagonal?

Possible Answers:

$30\:cm$

$60\:cm$

$90\:cm$

$120\:cm$

$45\:cm$

Correct answer:

$60\:cm$

Explanation:

Write the formula for the area of a rhombus.

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

Plug in the given area and diagonal length. Solve for the other diagonal.

$60\:cm^2=\frac{2\:cm\cdot d_2}{2}$

$2(60\:cm^2)=2(\frac{2\:cm\cdot d_2}{2})$

$120\:cm^2=2\:cm\cdot d_2$

$\frac{120\:cm}{2\:cm}=\frac{2\:cm\cdot d_2}{2\:cm}$

$d_2 =60\:cm$

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Example Question #141 : Quadrilaterals

ABCD is a rhombus. Find $m(\overline{BD})$ .

Varsity3

Possible Answers:

$m(\overline{BD}) = 36$

$m(\overline{BD}) = 6$

$m(\overline{BD}) \approx 9.83$

$m(\overline{BD}) = 35$

$m(\overline{BD}) = 12$

Correct answer:

$m(\overline{BD}) \approx 9.83$

Explanation:

Using the Law of Sines,

$\frac{sin(35^{o})}{6} = \frac{sin(110^{o})}{BD}$

$BD = \frac{6}{sin(35^{o})}\cdot sin(110^{o})$

$BD = m(\overline{BD})$

$m(\overline{BD})\approx 9.83$

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Example Question #142 : Quadrilaterals

Find the lengths of the two diagonals, the longer diagonal is $d_{1}$ , the shorter diagonal is $d_{2}$ .

Varsity1

Possible Answers:

$d_{1} = 13$

$d_{2} = 5$

$d_{1} = 6$

$d_{2} = 12$

$d_{1} = 2$

$d_{2} = 6$

$d_{1} = 12$

$d_{2} = 6$

$d_{1} = 6\cdot \sqrt{3}$

$d_{2} = 6$

Correct answer:

$d_{1} = 6\cdot \sqrt{3}$

$d_{2} = 6$

Explanation:

1) All sides of a rhombus are congruent.

2) Because all sides of a rhombus are congruent, the expressions of the side lengths can be set equal to each other. The resulting equation is then solved,

2x - 4 = 3x - 9

9 - 4 = 3x - 2x

5 = x

x = 5

3) Because the sides of a rhombus are congruent, x = 5 can be substituted into either 2x - 4 or 3x - 9 to find the length of a side,

2(5) - 4 = 6 , or, 3(5) - 9 = 6 .

4) Each of the composing triangles are right triangles, so then is the length of the hypotenuse for each triangle.

5) 6 = 3(2) .

6) The standard 30 - 60 - 90 right triangle has a hypotenuse length equal to .

7) The hypotenuse of a standard 30 - 60 - 90 right triangle is being multiplied by .

The result is , so then is the scale factor for the triangle side lengths.

8) For the standard 30 - 60 - 90 right triangle, the other two side lengths are and $\sqrt{3}$ , so then the height of the triangle from step 7) has a height of 3(1) , and the base length is $3(\sqrt{3})$ .

9) The base of the triangle from step 7) is

$\frac{1}{2}d_{1} = 3\cdot(\sqrt{3})$ ,

and the height is

$\frac{1}{2}d_{2} = 3(1)$ .

10) Diagonal

$d_2=2\cdot \frac{1}{2}d_{2} = 2\cdot3 = 6$ ,

and diagonal

$d_{1} = 2\cdot \frac{1}{2}d_{1} = 2\cdot3(\sqrt{3}) = 6(\sqrt{3})$ .

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Example Question #141 : Quadrilaterals

Screen_shot_2015-03-06_at_3.16.41_pm

What is the second diagonal for the above rhombus?

Possible Answers:

Correct answer:

Explanation:

Because a rhombus has vertical and horizontal symmetry, it can be broken into four congruent triangles, each with a hypotenuse of 13 and a base of 5 (half the given diagonal).

The Pythagorean Theorem

a^2+b^2=c^2 will yield,

$\sqrt{169-25}=12$ for the height of the triangles.

The greater diagonal is twice the height of the triangles therefore, the greater diagonal becomes:

$d=12 \cdot 2$

d=24

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Example Question #11 : How To Find The Length Of The Diagonal Of A Rhombus

ABCD is rhombus with side lengths in meters. $\overline{AB} = 13$ and $\overline{BD} = 10$ . What is the length, in meters, of $\overline{AC}$ ?

Rhombus_1

Possible Answers:

Correct answer:

Explanation:

A rhombus is a quadrilateral with four sides of equal length. Rhombuses have diagonals that bisect each other at right angles.

Rhombus_2

Thus, we can consider the right triangle AED to find the length of diagonal $\overline{AC}$ . From the given information, each of the sides of the rhombus measures meters and $\overline{BD} = 10$ .

Because the diagonals bisect each other, we know:

$\overline{AD} = 13$

$\overline{ED} = 5$

Using the Pythagorean theorem,

a^2 + b^2 = c^2

$\overline{AE}^2 + 5^2 = 13^2$

$\overline{AE}^2 = 144$

$\overline{AE} = 12$

$\overline{AC} = 24$

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Example Question #441 : Advanced Geometry

Let $\small \small f(x)=x^4-5x^3+2x^2+9$ . If $\small g(x)$ is equal to $\small f(x)$ when flipped across the x-axis, what is the equation for $\small g(x)$ ?

Possible Answers:

$\small x^4+5x^3+2x^2+9$

$\small x^4+5x^3+2x^2-9$

$\small -x^4-5x^3-2x^2-9$

$\small x^4-5x^3+2x^2+9$

$\small -x^4+5x^3-2x^2-9$

Correct answer:

$\small -x^4+5x^3-2x^2-9$

Explanation:

When a function $\small f(x)$ is flipped across the x-axis, the new function $\small g(x)$ is equal to $\small -f(x)$ . Therefore, our function $\small g(x)$ is equal to:

$\small g(x)=-f(x)=-(x^4-5x^3+2x^2+9)=-x^4+5x^3-2x^2-9$

Our final answer is therefore $\small g(x)=-x^4+5x^3-2x^2-9$

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Example Question #442 : Advanced Geometry

Let $\small f(x)=x^5-2x^4-6x^3+x^2+5x$ . If we let $\small g(x)$ equal $\small f(x)$ when it is flipped across the y-axis, what is the equation for $\small \small g(x)$ ?

Possible Answers:

$\small x^5-2x^4+6x^3-x^2+5x$

$\small x^5+2x^4-6x^3-x^2+5x$

$\small x^5+2x^4+6x^3+x^2+5x$

$\small -x^5+2x^4+6x^3-x^2-5x$

$\small -x^5-2x^4+6x^3+x^2-5x$

Correct answer:

$\small -x^5-2x^4+6x^3+x^2-5x$

Explanation:

When a function $\small f(x)$ is flipped across the y-axis, the resulting function $\small g(x)$ is equal to $\small f(-x)$ . Therefore, to find our $\small g(x)$ , we must substitute in $\small -x$ for every $\small x$ is our equation:

$\small g(x)=(-x)^5-2(-x)^4-6(-x)^3+(-x)^2+5(-x)=-x^5-2x^4+6x^3+x^2-5x$

Our final answer is therefore $\small g(x)=-x^5-2x^4+6x^3+x^2-5x$

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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 : Calculating The Length Of The Diagonal Of A Quadrilateral

Example Question #51 : Rhombuses

Example Question #2 : Calculating The Length Of The Diagonal Of A Quadrilateral

Example Question #51 : Rhombuses

Example Question #141 : Quadrilaterals

Example Question #142 : Quadrilaterals

Example Question #141 : Quadrilaterals

Example Question #11 : How To Find The Length Of The Diagonal Of A Rhombus

Example Question #441 : Advanced Geometry

Example Question #442 : Advanced Geometry

All Advanced Geometry Resources

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