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Intermediate Geometry : Quadrilaterals

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Example Questions

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

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

Possible Answers:

30.5in $\displaystyle 30.5in$

30in $\displaystyle 30in$

16in $\displaystyle 16in$

22.5in $\displaystyle 22.5in$

Correct answer:

22.5in $\displaystyle 22.5in$

Explanation:

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

P=4S $\displaystyle P=4S$, where $\displaystyle S$ is the side length.

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

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

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Example Question #231 : Intermediate Geometry

Given that a rhombus has an area of $\displaystyle 70$ square units, and a height of $\displaystyle 5$. What is the length of one side of the rhombus?

Possible Answers:

12.5 $\displaystyle 12.5$

$\displaystyle 15$

$\displaystyle 14$

$\displaystyle 13$

Correct answer:

$\displaystyle 14$

Explanation:

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

$Area=(base\times height)$ $\displaystyle Area=(base\times height)$

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

$70=(base\times height)$ $\displaystyle 70=(base\times height)$

$70=(base\times 5)$ $\displaystyle 70=(base\times 5)$

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

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Example Question #61 : Rhombuses

Find the length of a side of a rhombus if it has diagonals possessing the following lengths: $\displaystyle 2$ and $\displaystyle 4$.

Possible Answers:

$\sqrt2$ $\displaystyle \sqrt2$

Cannot be determined

$\sqrt5$ $\displaystyle \sqrt5$

$\sqrt6$ $\displaystyle \sqrt6$

Correct answer:

$\sqrt5$ $\displaystyle \sqrt5$

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.

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

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

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

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

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

$\text{side length}=\sqrt5$ $\displaystyle \text{side length}=\sqrt5$

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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: $\displaystyle 4$ and $\displaystyle 10$.

Possible Answers:

$\sqrt{31}$ $\displaystyle \sqrt{31}$

$\sqrt{35}$ $\displaystyle \sqrt{35}$

$3\sqrt3$ $\displaystyle 3\sqrt3$

$\sqrt{29}$ $\displaystyle \sqrt{29}$

Correct answer:

$\sqrt{29}$ $\displaystyle \sqrt{29}$

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.

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

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

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

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

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

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

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Example Question #61 : Rhombuses

Find the length of a side of a rhombus if it has diagonals possessing the following lengths: $\displaystyle 12$ and $\displaystyle 20$.

Possible Answers:

$6\sqrt{10}$ $\displaystyle 6\sqrt{10}$

$12\sqrt2$ $\displaystyle 12\sqrt2$

$2\sqrt{34}$ $\displaystyle 2\sqrt{34}$

$3\sqrt{15}$ $\displaystyle 3\sqrt{15}$

Correct answer:

$2\sqrt{34}$ $\displaystyle 2\sqrt{34}$

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.

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

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

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

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

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

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

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Example Question #62 : Rhombuses

Find the length of a side of a rhombus if it has diagonals possessing the following lengths: $\displaystyle 14$ and $\displaystyle 22$.

Possible Answers:

$\displaystyle 8$

$19\sqrt{3}$ $\displaystyle 19\sqrt{3}$

$2\sqrt{77}$ $\displaystyle 2\sqrt{77}$

$\sqrt{170}$ $\displaystyle \sqrt{170}$

Correct answer:

$\sqrt{170}$ $\displaystyle \sqrt{170}$

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.

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

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

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

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

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

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

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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: $\displaystyle 16$ and $\displaystyle 18$.

Possible Answers:

$3\sqrt{15}$ $\displaystyle 3\sqrt{15}$

$12\sqrt2$ $\displaystyle 12\sqrt2$

$7\sqrt3$ $\displaystyle 7\sqrt3$

$\sqrt{145}$ $\displaystyle \sqrt{145}$

Correct answer:

$\sqrt{145}$ $\displaystyle \sqrt{145}$

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.

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

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

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

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

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

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

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Example Question #62 : Rhombuses

Find the length of a side of a rhombus if it has diagonals possessing the following lengths: $\displaystyle 6$ and $\displaystyle 8$.

Possible Answers:

$5\sqrt3$ $\displaystyle 5\sqrt3$

$\displaystyle 4$

$\displaystyle 5$

$2\sqrt3$ $\displaystyle 2\sqrt3$

Correct answer:

$\displaystyle 5$

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.

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

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

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

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

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

$\text{side length}=5$ $\displaystyle \text{side length}=5$

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Example Question #61 : Rhombuses

Find the length of a side of a rhombus if it has diagonals possessing the following lengths: $\displaystyle 10$ and $\displaystyle 24$.

Possible Answers:

$\sqrt{39}$ $\displaystyle \sqrt{39}$

$\sqrt{26}$ $\displaystyle \sqrt{26}$

$3\sqrt3$ $\displaystyle 3\sqrt3$

$\displaystyle 13$

Correct answer:

$\displaystyle 13$

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.

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

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

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

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

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

$\text{side length}=13$ $\displaystyle \text{side length}=13$

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Example Question #61 : Quadrilaterals

Find the length of a side of a rhombus if it has diagonals possessing the following lengths: $\displaystyle 12$ and $\displaystyle 16$.

Possible Answers:

$11\sqrt2$ $\displaystyle 11\sqrt2$

$\displaystyle 12$

$8\sqrt3$ $\displaystyle 8\sqrt3$

$\displaystyle 10$

Correct answer:

$\displaystyle 10$

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.

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

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

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

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

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

$\text{side length}=10$ $\displaystyle \text{side length}=10$

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Intermediate Geometry : Quadrilaterals

Study concepts, example questions & explanations for Intermediate Geometry

All Intermediate Geometry Resources

Example Questions

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

Example Question #231 : Intermediate Geometry

Example Question #61 : Rhombuses

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

Example Question #61 : Rhombuses

Example Question #62 : Rhombuses

Example Question #236 : Intermediate Geometry

Example Question #62 : Rhombuses

Example Question #61 : Rhombuses

Example Question #61 : Quadrilaterals

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