Recall that in a rhombus, the diagonals are not only perpendicular to each other, but also bisect one another.

Thus, we can find the lengths of half of each diagonal and use that in the Pythagorean Theorem to find the length of the side of the rhombus.

First, find the lengths of half of each diagonal.

\(\displaystyle \text{Half Diagonal 1}=\frac{40}{2}=20\)

\(\displaystyle \text{Half Diagonal 2}=\frac{46}{2}=23\)

Now, use these half diagonals as the legs of a right triangle that has the side of the rhombus as its hypotenuse.

\(\displaystyle \text{Side of Rhombus}=\sqrt{(\text{Half Diagonal 1})^2+(\text{Half Diagonal 2})^2}\)

Plug in the lengths of the half diagonals to find the length of the rhombus.

\(\displaystyle \text{Side of Rhombus}=\sqrt{20^2+23^2}=\sqrt{929}=30.48\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

Recall that in a rhombus, the diagonals are not only perpendicular to each other, but also bisect one another.

Thus, we can find the lengths of half of each diagonal and use that in the Pythagorean Theorem to find the length of the side of the rhombus.

First, find the lengths of half of each diagonal.

\(\displaystyle \text{Half Diagonal 1}=\frac{42}{2}=21\)

\(\displaystyle \text{Half Diagonal 2}=\frac{50}{2}=25\)

Now, use these half diagonals as the legs of a right triangle that has the side of the rhombus as its hypotenuse.

\(\displaystyle \text{Side of Rhombus}=\sqrt{(\text{Half Diagonal 1})^2+(\text{Half Diagonal 2})^2}\)

Plug in the lengths of the half diagonals to find the length of the rhombus.

\(\displaystyle \text{Side of Rhombus}=\sqrt{21^2+25^2}=\sqrt{1066}=32.65\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

Recall that in a rhombus, the diagonals are not only perpendicular to each other, but also bisect one another.

Thus, we can find the lengths of half of each diagonal and use that in the Pythagorean Theorem to find the length of the side of the rhombus.

First, find the lengths of half of each diagonal.

\(\displaystyle \text{Half Diagonal 1}=\frac{90}{2}=45\)

\(\displaystyle \text{Half Diagonal 2}=\frac{100}{2}=50\)

Now, use these half diagonals as the legs of a right triangle that has the side of the rhombus as its hypotenuse.

\(\displaystyle \text{Side of Rhombus}=\sqrt{(\text{Half Diagonal 1})^2+(\text{Half Diagonal 2})^2}\)

Plug in the lengths of the half diagonals to find the length of the rhombus.

\(\displaystyle \text{Side of Rhombus}=\sqrt{45^2+50^2}=\sqrt{4525}=67.27\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

Recall that in a rhombus, the diagonals are not only perpendicular to each other, but also bisect one another.

Thus, we can find the lengths of half of each diagonal and use that in the Pythagorean Theorem to find the length of the side of the rhombus.

First, find the lengths of half of each diagonal.

\(\displaystyle \text{Half Diagonal 1}=\frac{52}{2}=26\)

\(\displaystyle \text{Half Diagonal 2}=\frac{58}{2}=29\)

Now, use these half diagonals as the legs of a right triangle that has the side of the rhombus as its hypotenuse.

\(\displaystyle \text{Side of Rhombus}=\sqrt{(\text{Half Diagonal 1})^2+(\text{Half Diagonal 2})^2}\)

Plug in the lengths of the half diagonals to find the length of the rhombus.

\(\displaystyle \text{Side of Rhombus}=\sqrt{26^2+29^2}=\sqrt{1517}=38.95\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

Recall that in a rhombus, the diagonals are not only perpendicular to each other, but also bisect one another.

Thus, we can find the lengths of half of each diagonal and use that in the Pythagorean Theorem to find the length of the side of the rhombus.

First, find the lengths of half of each diagonal.

\(\displaystyle \text{Half Diagonal 1}=\frac{58}{2}=29\)

\(\displaystyle \text{Half Diagonal 2}=\frac{60}{2}=30\)

Now, use these half diagonals as the legs of a right triangle that has the side of the rhombus as its hypotenuse.

\(\displaystyle \text{Side of Rhombus}=\sqrt{(\text{Half Diagonal 1})^2+(\text{Half Diagonal 2})^2}\)

Plug in the lengths of the half diagonals to find the length of the rhombus.

\(\displaystyle \text{Side of Rhombus}=\sqrt{29^2+30^2}=\sqrt{1741}=41.73\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

Recall that in a rhombus, the diagonals are not only perpendicular to each other, but also bisect one another.

Thus, we can find the lengths of half of each diagonal and use that in the Pythagorean Theorem to find the length of the side of the rhombus.

First, find the lengths of half of each diagonal.

\(\displaystyle \text{Half Diagonal 1}=\frac{100}{2}=50\)

\(\displaystyle \text{Half Diagonal 2}=\frac{108}{2}=54\)

Now, use these half diagonals as the legs of a right triangle that has the side of the rhombus as its hypotenuse.

\(\displaystyle \text{Side of Rhombus}=\sqrt{(\text{Half Diagonal 1})^2+(\text{Half Diagonal 2})^2}\)

Plug in the lengths of the half diagonals to find the length of the rhombus.

\(\displaystyle \text{Side of Rhombus}=\sqrt{50^2+54^2}=\sqrt{5416}=73.59\)

Make sure to round to \(\displaystyle 2\) places after the decimal.

A rhombus is defined to be a quadrilateral with four congruent sides. 

Parallelogram \(\displaystyle ABCD\) gives the lengths of two of its opposite sides to be congruent, but this is characteristic of all parallelograms. No information is given about the other two sides, so the figure need not be a rhombus.

Opposite sides of a parallelogram are congruent, so 

\(\displaystyle CD = AB = 12\)

and 

\(\displaystyle AD = BC = 12\)

All four sides are congruent to one another. It follows by definition that Parallelogram \(\displaystyle ABCD\) is a rhombus.

Below are two quadrilaterals marked \(\displaystyle ABCD\) with \(\displaystyle \overline{AC}\) drawn.

The quadrilateral on the left has four congruent sides and is by definition a rhombus. The quadrilateral on the right is not a rhombus, since not all four sides are congruent.

In both cases, \(\displaystyle \overline{AB} \cong \overline{AD}\), \(\displaystyle \overline{CB} \cong \overline{CD}\), and, by the reflexive property, \(\displaystyle \overline{AC} \cong \overline{AC}\). By the Side-Side-Side Congruence Postulate, it can be proved that \(\displaystyle \bigtriangleup ABC \cong \bigtriangleup ADC\) in both diagrams.

Therefore, Quadrilateral \(\displaystyle ABCD\) need not be a rhombus.

As we are establishing whether or not \(\displaystyle \bigtriangleup ABC \cong \bigtriangleup DEF\), then \(\displaystyle A\), \(\displaystyle B\), and \(\displaystyle C\) correspond respectively to \(\displaystyle D\), \(\displaystyle E\), and \(\displaystyle F\).

If we examine the sides and the angle of \(\displaystyle \bigtriangleup ABC\) in the congruence statements - \(\displaystyle \overline{AB}\), \(\displaystyle \overline{AC}\), and \(\displaystyle \angle C\) - we see that these are two sides and a nonincluded angle. But if we examine the sides and angles of \(\displaystyle \bigtriangleup DEF\) -\(\displaystyle \overline{DE}\), \(\displaystyle \overline{DF}\), and \(\displaystyle \angle E\) - we see that these are two sides and an included angle.

The only congruence postulate dealing with two sides and an angle is the SAS Congruence Postulate, which requires congruence between two sides and the included angle of both triangles. This theorem does not apply, so we cannot prove the triangles congruent.