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Advanced Geometry : Rhombuses

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Example Question #11 : Finding Sides

Rhombus_1

ABCD is a rhombus. $\overline{AB} = 17\:cm$ $\overline{AB} = 17$ and $\overline{AC} = 30\:cm$ . Find $\overline{BD}$ .

Possible Answers:

$16\:cm$

$13\:cm$

$8\:cm$

$\sqrt{13}\sqrt{47}\:cm$

$\frac{17}{\sqrt{2}}\:cm$

Correct answer:

$16\:cm$

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 $\overline{BD}$ . From the problem, we are given that the sides are $17\:cm$ and $\overline{AC} = 30\:cm$ . Because the diagonals bisect each other, we know:

$\overline{AD} = 17\:cm$

$\overline{AE} = 15\:cm$

Using the Pythagorean Theorem,

a^2 + b^2 = c^2

$(\overline{ED})^2 + (15\:cm)^2 = (17\:cm)^2$

$(\overline{ED})^2 + 225\:cm^2 = 289\:cm^2$

$(\overline{ED})^2 = 64\:cm^2$

$\overline{ED} = 8\:cm$

$\overline{BD} = 16\:cm$

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Example Question #1 : How To Find The Length Of The Diagonal Of A Rhombus

Rhombus_1

ABCD is a rhombus. $\overline{BC} = 20$ and $\overline{BD} = 20$ . Find $\overline{AC}$ .

Possible Answers:

$10\sqrt{3}$

$20\sqrt{3}$

$10\sqrt{2}$

$20\sqrt{2}$

Correct answer:

$20\sqrt{3}$

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 problem, we are given that the sides are and $\overline{BD} = 20$ . Because the diagonals bisect each other, we know:

$\overline{AD} = 20$

$\overline{ED} = 10$

Using the Pythagorean Theorem,

a^2 + b^2 = c^2

$\overline{AE}^2 + 10^2 = 20^2$

$\overline{AE}^2 = 300$

$\overline{AE} = 10\sqrt{3}$

$\overline{AC} = 20\sqrt{3}$

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Example Question #1 : How To Find The Length Of The Diagonal Of A Rhombus

Rhombus_1

ABCD is a rhombus. $\overline{AB} = 4x + 9$ , $\overline{BD} = 2x + 6$ , and $\overline{AC} = 12x$ . Find .

Possible Answers:

1.5

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 and use the Pythagorean Theorem to solve for . From the problem:

$\overline{AD} = 4x + 9$

$\overline{BD} = 2x + 6$

$\overline{AC} = 12x$

Because the diagonals bisect each other, we know:

$\overline{ED} = x + 3$

$\overline{AE} = 6x$

Using the Pythagorean Theorem,

a^2 + b^2 = c^2

$\overline{AE}^2 + \overline{ED}^2 = \overline{AD}^2$

$\left ( 6x\right )^2 + \left ( x + 3\right )^2 = \left ( 4x + 9\right )^2$

36x^2 + x^2 + 6x + 9 = 16x^2 + 72x + 81

21x^2 - 66x - 72 = 0

Using the quadratic formula,

$x = \frac{b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{66 \pm \sqrt{\left (-66 \right )^2 - 4\left (21 \right )\left (-72 \right )}}{2\left (21 \right )}$

With this equation, we get two solutions:

x = 4, -0.857

Only the positive solution is valid for this problem.

x = 4

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Example Question #1 : How To Find The Length Of The Diagonal Of A Rhombus

Rhombus_1

ABCD is a rhombus. $\overline{AB} = 3x - 2$ , $\overline{AC} = 4x + 4$ , and $\overline{BD} = 2x$ . Find .

Possible Answers:

1.5

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 and use the Pythagorean Theorem to solve for . From the problem:

$\overline{AD} = 3x-2$

$\overline{BD} = 2x$

$\overline{AC} = 4x + 4$

Because the diagonals bisect each other, we know:

$\overline{ED} = x$

$\overline{AE} = 2x + 2$

Using the Pythagorean Theorem,

a^2 + b^2 = c^2

$\overline{AE}^2 + \overline{ED}^2 = \overline{AD}^2$

$\left ( 2x + 2\right )^2 + \left ( x \right )^2 = \left ( 3x - 2\right )^2$

4x^2 + 8x + 4 + x^2 = 9x^2 - 12x + 4

0 = 4x^2 - 20x

Factoring,

0 = x and 0 = 4x - 20

The first solution is nonsensical for this problem.

0 = 4x - 20

x = 5

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

Rhombus_1

ABCD is a rhombus. $\overline{BD} = 18$ and $\overline{AC} = 80$ . Find the length of the sides.

Possible Answers:

$18\sqrt{2}$

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 side $\overline{AD}$ . From the problem, we are given $\overline{BD} = 18$ and $\overline{AC} = 80$ . Because the diagonals bisect each other, we know:

$\overline{AE} = 40$

$\overline{ED} = 9$

Using the Pythagorean Theorem,

a^2 + b^2 = c^2

$40^2 + 9^2 = \overline{AD}^2$

$1681 = \overline{AD}^2$

$41 = \overline{AD}$

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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 = 28, HM = 28

RO = 4, HM = 14

RO = 16, HM = 7

None of the other responses gives a correct answer.

RO = 14, HM = 14

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 #1 : Calculating The Length Of The Diagonal Of A Quadrilateral

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

Possible Answers:

$8\sqrt{2}$

$16\sqrt{3}$

$16\sqrt{2}$

$8\sqrt{3}$

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 #14 : How To Find The Length Of The Diagonal Of A Rhombus

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

Possible Answers:

$6\sqrt{2}$

$12\sqrt{3}$

$6\sqrt{3}$

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 #15 : How To Find The Length Of The Diagonal Of A Rhombus

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:

$90\:cm$

$45\:cm$

$120\:cm$

$60\:cm$

$30\:cm$

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 #51 : Rhombuses

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

Varsity3

Possible Answers:

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

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

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

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

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

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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Advanced Geometry : Rhombuses

Study concepts, example questions & explanations for Advanced Geometry

All Advanced Geometry Resources

Example Questions

Example Question #11 : Finding Sides

Example Question #1 : How To Find The Length Of The Diagonal Of A Rhombus

Example Question #1 : How To Find The Length Of The Diagonal Of A Rhombus

Example Question #1 : How To Find The Length Of The Diagonal Of A Rhombus

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

Example Question #1 : Calculating The Length Of The Diagonal Of A Quadrilateral

Example Question #1 : Calculating The Length Of The Diagonal Of A Quadrilateral

Example Question #14 : How To Find The Length Of The Diagonal Of A Rhombus

Example Question #15 : How To Find The Length Of The Diagonal Of A Rhombus

Example Question #51 : Rhombuses

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