Advanced Geometry : How to find the length of the diagonal of a rhombus

Study concepts, example questions & explanations for Advanced Geometry

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Example Questions

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

Assume quadrilateral \(\displaystyle \small EFGH\) is a rhombus. If the perimeter of \(\displaystyle \small EFGH\) is \(\displaystyle \small 24\) and the length of diagonal \(\displaystyle \small \overline{EG}=10\), what is the length of diagonal \(\displaystyle \small \overline{FH}\)?

Possible Answers:

\(\displaystyle \small 2\sqrt{5}\)

\(\displaystyle \small 2\sqrt{11}\)

\(\displaystyle \small \sqrt{11}\)

\(\displaystyle \small 5\)

\(\displaystyle \small 11\)

Correct answer:

\(\displaystyle \small 2\sqrt{11}\)

Explanation:

To find the value of diagonal \(\displaystyle \small \overline{FH}\), we must first recognize some important properties of rhombuses. Since the perimeter is of \(\displaystyle \small EFGH\) is \(\displaystyle \small 24\), and by definition a rhombus has four sides of equal length, each side length of the rhombus is equal to \(\displaystyle \small 6\). The diagonals of rhombuses also form four right triangles, with hypotenuses equal to the side length of the rhombus and legs equal to one-half the lengths of the diagonals. We can therefore use the Pythagorean Theorem to solve for one-half of the unknown diagonal:

\(\displaystyle \small 6^2=5^2+x^2\), where \(\displaystyle \small 6\) is the rhombus side length, \(\displaystyle \small 5\) is one-half of the known diagonal, and \(\displaystyle \small x\) is one-half of the unknown diagonal. We can therefore solve for \(\displaystyle \small x\):

\(\displaystyle \small x^2=6^2-5^2=36-25=11\)

\(\displaystyle \small x\) is therefore equal to \(\displaystyle \small \sqrt{11}\). Since \(\displaystyle \small x\) represents one-half of the unknown diagonal, we need to multiply by \(\displaystyle \small 2\) to find the full length of diagonal \(\displaystyle \small \overline{FH}\).

The length of diagonal \(\displaystyle \small \overline{FH}\) is therefore \(\displaystyle \small 2\sqrt{11}\)

Example Question #2 : How To Find The Length Of The Diagonal Of A Rhombus

Assume quadrilateral \(\displaystyle \small JKMN\) is a rhombus. If the area of \(\displaystyle \small JKMN\) is \(\displaystyle \small 273\) square units, and the length of diagonal \(\displaystyle \small \overline{JM}\) is \(\displaystyle \small 13\) units, what is the length of diagonal \(\displaystyle \small \overline{KN}\)?

Possible Answers:

\(\displaystyle \small 42\)

\(\displaystyle \small 21\)

\(\displaystyle \small 1775\)

\(\displaystyle \small 26\)

\(\displaystyle \small 13\)

Correct answer:

\(\displaystyle \small 42\)

Explanation:

This problem relies on the knowledge of the equation for the area of a rhombus, \(\displaystyle \small \small A=\frac{pq}{2}\), where \(\displaystyle \small A\) is the area, and \(\displaystyle \small p\) and \(\displaystyle \small q\) are the lengths of the individual diagonals. We can substitute the values that we know into the equation to obtain:

\(\displaystyle \small 273=\frac{(13)q}{2}\)

\(\displaystyle \small q=\frac{2 \cdot 273}{13}\)

\(\displaystyle \small q=42\)

Therefore, our final answer is that the diagonal \(\displaystyle \small \overline{KN}=42\)

Example Question #3 : How To Find The Length Of The Diagonal Of A Rhombus

If the area of a rhombus is \(\displaystyle 12\), and one of the diagonal lengths is \(\displaystyle 6\), what is the length of the other diagonal?

Possible Answers:

\(\displaystyle 4\sqrt2\)

\(\displaystyle 2\sqrt2\)

\(\displaystyle \sqrt2\)

\(\displaystyle 2\)

\(\displaystyle 4\)

Correct answer:

\(\displaystyle 4\)

Explanation:

The area of a rhombus is given below.

\(\displaystyle A=\frac{d1\cdot d2}{2}\)

Substitute the given area and a diagonal.  Solve for the other diagonal.

\(\displaystyle 12=\frac{6\cdot d2}{2}\)

\(\displaystyle 24=6\cdot d2\)

\(\displaystyle 4 =d2\)

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

If the area of a rhombus is \(\displaystyle 2\), and a diagonal has a length of \(\displaystyle 1\), what is the length of the other diagonal?

Possible Answers:

\(\displaystyle 3\)

\(\displaystyle \frac{1}{4}\)

\(\displaystyle 4\)

\(\displaystyle 2\)

\(\displaystyle \frac{\sqrt2}{2}\)

Correct answer:

\(\displaystyle 4\)

Explanation:

The area of a rhombus is given below. Plug in the area and the given diagonal. Solve for the other diagonal.

\(\displaystyle A=\frac{d1 \cdot d2}{2}\)

\(\displaystyle 2=\frac{1 \cdot d2}{2}\)

\(\displaystyle d2=4\)

Example Question #5 : How To Find The Length Of The Diagonal Of A Rhombus

The area of a rhombus is \(\displaystyle 5\). The length of a diagonal is twice as long as the other diagonal. What is the length of the shorter diagonal?

Possible Answers:

\(\displaystyle \sqrt5\)

\(\displaystyle 5\sqrt2\)

\(\displaystyle 2\sqrt5\)

\(\displaystyle \frac{5}{2}\)

\(\displaystyle 5\)

Correct answer:

\(\displaystyle \sqrt5\)

Explanation:

Let the shorter diagonal be \(\displaystyle d1\), and the longer diagonal be \(\displaystyle d2\).  The longer dimension is twice as long as the other diagonal.  Write an expression for this.

\(\displaystyle d2=2\cdot d1\)

Write the area of the rhombus.

\(\displaystyle A=\frac{d1 \cdot d2}{2}\)

Since we are solving for the shorter diagonal, it's best to setup the equation in terms \(\displaystyle d1\), so that we can solve for the shorter diagonal.  Plug in the area and expression to solve for \(\displaystyle d1\).

\(\displaystyle 5=\frac{d1 \cdot (2\cdot d1)}{2}\)

\(\displaystyle 5=d1^2\)

\(\displaystyle \sqrt5=d1\)

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

Rhombus_1

\(\displaystyle ABCD\) is a rhombus with side length \(\displaystyle 10\:cm\). Diagonal \(\displaystyle \overline{BD}\) has a length of \(\displaystyle 12\:cm\). Find the length of diagonal \(\displaystyle \overline{AC}\).

Possible Answers:

\(\displaystyle 2\sqrt{34}\:cm\)

\(\displaystyle 2\sqrt{61}\:cm\)

\(\displaystyle 8\:cm\)

\(\displaystyle 10\sqrt{2}\:cm\)

\(\displaystyle 16\:cm\)

Correct answer:

\(\displaystyle 16\:cm\)

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 \(\displaystyle AED\) to find the length of diagonal \(\displaystyle \overline{AC}\). From the problem, we are given that the sides are \(\displaystyle 10\:cm\) and \(\displaystyle \overline{BD} = 12\:cm\). Because the diagonals bisect each other, we know:

\(\displaystyle \overline{AD} = 10\:cm\)

\(\displaystyle \overline{ED} = 6\:cm\)

Using the Pythagorean Theorem,

\(\displaystyle a^2 + b^2 = c^2\)

\(\displaystyle (\overline{AE})^2 + (6\:cm)^2 = (10\:cm)^2\)

\(\displaystyle (\overline{AE})^2 + 36\:cm^2 = 100\:cm^2\)

\(\displaystyle (\overline{AE})^2 = 64\:cm^2\)

\(\displaystyle \overline{AE} = 8\:cm\)

\(\displaystyle \overline{AC} = 16\:cm\)

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

Rhombus_1

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

Possible Answers:

\(\displaystyle \frac{17}{\sqrt{2}}\:cm\)

\(\displaystyle 8\:cm\)

\(\displaystyle 13\:cm\)

\(\displaystyle 16\:cm\)

\(\displaystyle \sqrt{13}\sqrt{47}\:cm\)

Correct answer:

\(\displaystyle 16\:cm\)

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 \(\displaystyle AED\) to find the length of diagonal \(\displaystyle \overline{BD}\). From the problem, we are given that the sides are \(\displaystyle 17\:cm\) and \(\displaystyle \overline{AC} = 30\:cm\). Because the diagonals bisect each other, we know:

\(\displaystyle \overline{AD} = 17\:cm\)

\(\displaystyle \overline{AE} = 15\:cm\)

Using the Pythagorean Theorem,

\(\displaystyle a^2 + b^2 = c^2\)

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

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

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

\(\displaystyle \overline{ED} = 8\:cm\)

\(\displaystyle \overline{BD} = 16\:cm\)

Example Question #7 : How To Find The Length Of The Diagonal Of A Rhombus

Rhombus_1

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

Possible Answers:

\(\displaystyle 10\sqrt{2}\)

\(\displaystyle 20\sqrt{3}\)

\(\displaystyle 10\sqrt{3}\)

\(\displaystyle 20\)

\(\displaystyle 20\sqrt{2}\)

Correct answer:

\(\displaystyle 20\sqrt{3}\)

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 \(\displaystyle AED\) to find the length of diagonal \(\displaystyle \overline{AC}\). From the problem, we are given that the sides are \(\displaystyle 20\) and \(\displaystyle \overline{BD} = 20\). Because the diagonals bisect each other, we know:

\(\displaystyle \overline{AD} = 20\)

\(\displaystyle \overline{ED} = 10\)

Using the Pythagorean Theorem,

\(\displaystyle a^2 + b^2 = c^2\)

\(\displaystyle \overline{AE}^2 + 10^2 = 20^2\)

\(\displaystyle \overline{AE}^2 = 300\)

\(\displaystyle \overline{AE} = 10\sqrt{3}\)

\(\displaystyle \overline{AC} = 20\sqrt{3}\)

Example Question #2 : How To Find The Length Of The Diagonal Of A Rhombus

 

 

Rhombus_1

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

Possible Answers:

\(\displaystyle 8\)

\(\displaystyle 1.5\)

\(\displaystyle 5\)

\(\displaystyle 4\)

\(\displaystyle 2\)

Correct answer:

\(\displaystyle 4\)

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 \(\displaystyle AED\) and use the Pythagorean Theorem to solve for \(\displaystyle x\). From the problem:

\(\displaystyle \overline{AD} = 4x + 9\)

\(\displaystyle \overline{BD} = 2x + 6\)

\(\displaystyle \overline{AC} = 12x\)

Because the diagonals bisect each other, we know:

\(\displaystyle \overline{ED} = x + 3\)

\(\displaystyle \overline{AE} = 6x\)

Using the Pythagorean Theorem,

\(\displaystyle a^2 + b^2 = c^2\)

\(\displaystyle \overline{AE}^2 + \overline{ED}^2 = \overline{AD}^2\)

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

\(\displaystyle 36x^2 + x^2 + 6x + 9 = 16x^2 + 72x + 81\)

\(\displaystyle 21x^2 - 66x - 72 = 0\)

Using the quadratic formula,

\(\displaystyle x = \frac{b \pm \sqrt{b^2 - 4ac}}{2a}\)

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

With this equation, we get two solutions:

\(\displaystyle x = 4, -0.857\)

Only the positive solution is valid for this problem.

\(\displaystyle x = 4\)

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

Rhombus_1

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

Possible Answers:

\(\displaystyle 1.5\)

\(\displaystyle 3\)

\(\displaystyle 5\)

\(\displaystyle 4\)

\(\displaystyle 2\)

Correct answer:

\(\displaystyle 5\)

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 \(\displaystyle AED\) and use the Pythagorean Theorem to solve for \(\displaystyle x\). From the problem:

\(\displaystyle \overline{AD} = 3x-2\)

\(\displaystyle \overline{BD} = 2x\)

\(\displaystyle \overline{AC} = 4x + 4\)

Because the diagonals bisect each other, we know:

\(\displaystyle \overline{ED} = x\)

\(\displaystyle \overline{AE} = 2x + 2\)

Using the Pythagorean Theorem,

\(\displaystyle a^2 + b^2 = c^2\)

\(\displaystyle \overline{AE}^2 + \overline{ED}^2 = \overline{AD}^2\)

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

\(\displaystyle 4x^2 + 8x + 4 + x^2 = 9x^2 - 12x + 4\)

\(\displaystyle 0 = 4x^2 - 20x\)

Factoring,

\(\displaystyle 0 = x\) and \(\displaystyle 0 = 4x - 20\)

The first solution is nonsensical for this problem. 

\(\displaystyle 0 = 4x - 20\)

\(\displaystyle x = 5\)

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