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Intermediate Geometry : Quadrilaterals

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Example Questions

Example Question #221 : Intermediate Geometry

Given that a rhombus has a perimeter of 90in , find the length of a side of the rhombus.

Possible Answers:

16in

30.5in

22.5in

30in

Correct answer:

22.5in

Explanation:

To find the length of one side of the rhombus, apply the formula:

P=4S , where is the side length.

Since we are given the perimeter, we plug that value into the equation and solve for .

90=4(S)
$S=\frac{90}{4}=22.5$

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Example Question #231 : Intermediate Geometry

Given that a rhombus has an area of square units, and a height of . What is the length of one side of the rhombus?

Possible Answers:

12.5

Correct answer:

Explanation:

To find the length of a side of the rhombus, work backwards using the area formula:

$Area=(base\times height)$

Since we are given the area and the height, we plug these values into the equation and solve for the base.

$70=(base\times height)$

$70=(base\times 5)$

$base=\frac{70}{5}=14$

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Example Question #232 : Intermediate Geometry

Find the length of a side of a rhombus if it has diagonals possessing the following lengths: and .

Possible Answers:

$\sqrt5$

$\sqrt6$

$\sqrt2$

Cannot be determined

Correct answer:

$\sqrt5$

Explanation:

Recall that the diagonals of a rhombus bisect each other and are perpendicular to one another.

In order to find the length of the side, we can use Pythagorean's theorem. The lengths of half of the diagonals are the legs, and the length of the side of the rhombus is the hypotenuse.

First, find the lengths of half of the diagonals.

$\frac{\text{Diagonal 1}}{2}=\frac{2}{2}=1$

$\frac{\text{Diagonal 2}}{2}=\frac{4}{2}=2$

Next, substitute these half diagonals as the legs in a right triangle using the Pythagorean theorem.

$1^2+2^2=\text{side length}^2$

$\text{side length}^2=1+4=5$

$\text{side length}=\sqrt5$

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

Find the length of a side of a rhombus if it has diagonals possessing the following lengths: and .

Possible Answers:

$\sqrt{29}$

$\sqrt{35}$

$3\sqrt3$

$\sqrt{31}$

Correct answer:

$\sqrt{29}$

Explanation:

Recall that the diagonals of a rhombus bisect each other and are perpendicular to one another.

In order to find the length of the side, we can use Pythagorean's theorem. The lengths of half of the diagonals are the legs, and the length of the side of the rhombus is the hypotenuse.

First, find the lengths of half of the diagonals.

$\frac{\text{Diagonal 1}}{2}=\frac{4}{2}=2$

$\frac{\text{Diagonal 2}}{2}=\frac{10}{2}=5$

Next, substitute these half diagonals as the legs in a right triangle using the Pythagorean theorem.

$2^2+5^2=\text{side length}^2$

$\text{side length}^2=4+25=29$

$\text{side length}=\sqrt{29}$

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Example Question #234 : Intermediate Geometry

Find the length of a side of a rhombus if it has diagonals possessing the following lengths: and .

Possible Answers:

$6\sqrt{10}$

$12\sqrt2$

$2\sqrt{34}$

$3\sqrt{15}$

Correct answer:

$2\sqrt{34}$

Explanation:

Recall that the diagonals of a rhombus bisect each other and are perpendicular to one another.

In order to find the length of the side, we can use Pythagorean's theorem. The lengths of half of the diagonals are the legs, and the length of the side of the rhombus is the hypotenuse.

First, find the lengths of half of the diagonals.

$\frac{\text{Diagonal 1}}{2}=\frac{12}{2}=6$

$\frac{\text{Diagonal 2}}{2}=\frac{20}{2}=10$

Next, substitute these half diagonals as the legs in a right triangle using the Pythagorean theorem.

$6^2+10^2=\text{side length}^2$

$\text{side length}^2=36+100=136$

$\text{side length}=2\sqrt{34}$

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Example Question #235 : Intermediate Geometry

Find the length of a side of a rhombus if it has diagonals possessing the following lengths: and .

Possible Answers:

$2\sqrt{77}$

$\sqrt{170}$

$19\sqrt{3}$

Correct answer:

$\sqrt{170}$

Explanation:

Recall that the diagonals of a rhombus bisect each other and are perpendicular to one another.

In order to find the length of the side, we can use Pythagorean's theorem. The lengths of half of the diagonals are the legs, and the length of the side of the rhombus is the hypotenuse.

First, find the lengths of half of the diagonals.

$\frac{\text{Diagonal 1}}{2}=\frac{14}{2}=7$

$\frac{\text{Diagonal 2}}{2}=\frac{22}{2}=11$

Next, substitute these half diagonals as the legs in a right triangle using the Pythagorean theorem.

$7^2+11^2=\text{side length}^2$

$\text{side length}^2=49+121=170$

$\text{side length}=\sqrt{170}$

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Example Question #236 : Intermediate Geometry

Find the length of a side of a rhombus if it has diagonals possessing the following lengths: and .

Possible Answers:

$3\sqrt{15}$

$12\sqrt2$

$7\sqrt3$

$\sqrt{145}$

Correct answer:

$\sqrt{145}$

Explanation:

Recall that the diagonals of a rhombus bisect each other and are perpendicular to one another.

In order to find the length of the side, we can use Pythagorean's theorem. The lengths of half of the diagonals are the legs, and the length of the side of the rhombus is the hypotenuse.

First, find the lengths of half of the diagonals.

$\frac{\text{Diagonal 1}}{2}=\frac{16}{2}=8$

$\frac{\text{Diagonal 2}}{2}=\frac{18}{2}=9$

Next, substitute these half diagonals as the legs in a right triangle using the Pythagorean theorem.

$8^2+9^2=\text{side length}^2$

$\text{side length}^2=64+81=145$

$\text{side length}=\sqrt{145}$

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Example Question #237 : Intermediate Geometry

Find the length of a side of a rhombus if it has diagonals possessing the following lengths: and .

Possible Answers:

$5\sqrt3$

$2\sqrt3$

Correct answer:

Explanation:

Recall that the diagonals of a rhombus bisect each other and are perpendicular to one another.

In order to find the length of the side, we can use Pythagorean's theorem. The lengths of half of the diagonals are the legs, and the length of the side of the rhombus is the hypotenuse.

First, find the lengths of half of the diagonals.

$\frac{\text{Diagonal 1}}{2}=\frac{6}{2}=3$

$\frac{\text{Diagonal 2}}{2}=\frac{8}{2}=4$

Next, substitute these half diagonals as the legs in a right triangle using the Pythagorean theorem.

$3^2+4^2=\text{side length}^2$

$\text{side length}^2=9+16=25$

$\text{side length}=5$

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Example Question #238 : Intermediate Geometry

Find the length of a side of a rhombus if it has diagonals possessing the following lengths: and .

Possible Answers:

$\sqrt{39}$

$\sqrt{26}$

$3\sqrt3$

Correct answer:

Explanation:

Recall that the diagonals of a rhombus bisect each other and are perpendicular to one another.

In order to find the length of the side, we can use Pythagorean's theorem. The lengths of half of the diagonals are the legs, and the length of the side of the rhombus is the hypotenuse.

First, find the lengths of half of the diagonals.

$\frac{\text{Diagonal 1}}{2}=\frac{10}{2}=5$

$\frac{\text{Diagonal 2}}{2}=\frac{24}{2}=12$

Next, substitute these half diagonals as the legs in a right triangle using the Pythagorean theorem.

$5^2+12^2=\text{side length}^2$

$\text{side length}^2=25+144=169$

$\text{side length}=13$

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Example Question #239 : Intermediate Geometry

Find the length of a side of a rhombus if it has diagonals possessing the following lengths: and .

Possible Answers:

$8\sqrt3$

$11\sqrt2$

Correct answer:

Explanation:

Recall that the diagonals of a rhombus bisect each other and are perpendicular to one another.

In order to find the length of the side, we can use Pythagorean's theorem. The lengths of half of the diagonals are the legs, and the length of the side of the rhombus is the hypotenuse.

First, find the lengths of half of the diagonals.

$\frac{\text{Diagonal 1}}{12}=\frac{2}{2}=6$

$\frac{\text{Diagonal 2}}{2}=\frac{16}{2}=8$

Next, substitute these half diagonals as the legs in a right triangle using the Pythagorean theorem.

$6^2+8^2=\text{side length}^2$

$\text{side length}^2=36+64=100$

$\text{side length}=10$

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Intermediate Geometry : Quadrilaterals

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

Example Question #221 : Intermediate Geometry

Example Question #231 : Intermediate Geometry

Example Question #232 : Intermediate Geometry

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

Example Question #234 : Intermediate Geometry

Example Question #235 : Intermediate Geometry

Example Question #236 : Intermediate Geometry

Example Question #237 : Intermediate Geometry

Example Question #238 : Intermediate Geometry

Example Question #239 : Intermediate Geometry

All Intermediate Geometry Resources

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