Geometry - How to find the length of the diagonal of a rhombus

If the area of a rhombus is , and one of the diagonal lengths is , what is the length of the other diagonal?

$2\sqrt2$

$4\sqrt2$

$\sqrt2$

Explanation

The area of a rhombus is given below.

$A=\frac{d1\cdot d2}{2}$

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

$12=\frac{6\cdot d2}{2}$

$24=6\cdot d2$

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

$\small 2\sqrt{11}$

$\small 11$

$\small 5$

$\small \sqrt{11}$

$\small 2\sqrt{5}$

Explanation

To find the value of diagonal $\small \overline{FH}$ , we must first recognize some important properties of rhombuses. Since the perimeter is of $\small EFGH$ is $\small 24$ , and by definition a rhombus has four sides of equal length, each side length of the rhombus is equal to $\small 6$ . 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:

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

$\small x^2=6^2-5^2=36-25=11$

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

The length of diagonal $\small \overline{FH}$ is therefore $\small 2\sqrt{11}$

Screen_shot_2015-03-06_at_3.16.41_pm

What is the second diagonal for the above rhombus?

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

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:

$d=12 \cdot 2$

The area of a rhombus is . The length of a diagonal is twice as long as the other diagonal. What is the length of the shorter diagonal?

$\sqrt5$

$\frac{5}{2}$

$5\sqrt2$

$2\sqrt5$

Explanation

Let the shorter diagonal be , and the longer diagonal be . The longer dimension is twice as long as the other diagonal. Write an expression for this.

$d2=2\cdot d1$

Write the area of the rhombus.

$A=\frac{d1 \cdot d2}{2}$

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

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

$\sqrt5=d1$

Rhombus_1

is a rhombus with side length . Diagonal $\overline{BD}$ has a length of . Find the length of diagonal $\overline{AC}$ .

$10\sqrt{2}:cm$

$2\sqrt{61}:cm$

$2\sqrt{34}:cm$

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

$\overline{AD} = 10:cm$

$\overline{ED} = 6:cm$

Using the Pythagorean Theorem,

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

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

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

$\overline{AE} = 8:cm$

$\overline{AC} = 16:cm$

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

$\small 42$

$\small 13$

$\small 21$

$\small 26$

$\small 1775$

Explanation

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

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

$\small q=\frac{2 \cdot 273}{13}$

$\small q=42$

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

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

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 to find the length of diagonal $\overline{AC}$ . From the given information, each of the sides of the rhombus measures meters and $\overline{BD} = 10$ .