Math worksheet for calculating missing diagonals in rhombuses using area and side length formulas.

To solve the problems involving the missing diagonals of a rhombus, we use the formula for the area of a rhombus:

\[
\text{Area} = \frac{1}{2} \times d_1 \times d_2
\]

where \( d_1 \) and \( d_2 \) are the lengths of the diagonals.

Let's solve each problem step by step.

---

1) If \( WY = 70 \, \text{yd} \), find \( XZ \).

Given:
- \( WY = 70 \, \text{yd} \)
- Area = \( 13.25 \, \text{yd}^2 \)

Let:
- \( d_1 = WY = 70 \, \text{yd} \)
- \( d_2 = XZ \)

Using the area formula:
\[
\text{Area} = \frac{1}{2} \times d_1 \times d_2
\]
\[
13.25 = \frac{1}{2} \times 70 \times d_2
\]

Solve for \( d_2 \):
\[
13.25 = 35 \times d_2
\]
\[
d_2 = \frac{13.25}{35}
\]
\[
d_2 = 0.38 \, \text{yd}
\]

Thus, \( XZ = 0.38 \, \text{yd} \).

---

2) If \( FH = 6 \, \text{ft} \), find \( EG \).

Given:
- \( FH = 6 \, \text{ft} \)
- Area = \( 15 \, \text{ft}^2 \)

Let:
- \( d_1 = FH = 6 \, \text{ft} \)
- \( d_2 = EG \)

Using the area formula:
\[
\text{Area} = \frac{1}{2} \times d_1 \times d_2
\]
\[
15 = \frac{1}{2} \times 6 \times d_2
\]

Solve for \( d_2 \):
\[
15 = 3 \times d_2
\]
\[
d_2 = \frac{15}{3}
\]
\[
d_2 = 5 \, \text{ft}
\]

Thus, \( EG = 5 \, \text{ft} \).

---

3) If \( KM = 35 \, \text{in} \), find \( LN \).

Given:
- \( KM = 35 \, \text{in} \)
- Area = \( 700 \, \text{m}^2 \)

Let:
- \( d_1 = KM = 35 \, \text{in} \)
- \( d_2 = LN \)

Using the area formula:
\[
\text{Area} = \frac{1}{2} \times d_1 \times d_2
\]
\[
700 = \frac{1}{2} \times 35 \times d_2
\]

Solve for \( d_2 \):
\[
700 = 17.5 \times d_2
\]
\[
d_2 = \frac{700}{17.5}
\]
\[
d_2 = 40 \, \text{in}
\]

Thus, \( LN = 40 \, \text{in} \).

---

4) If \( VT = 7 \, \text{ft} \), find \( SU \).

Given:
- \( VT = 7 \, \text{ft} \)
- Area = \( 80.5 \, \text{ft}^2 \)

Let:
- \( d_1 = VT = 7 \, \text{ft} \)
- \( d_2 = SU \)

Using the area formula:
\[
\text{Area} = \frac{1}{2} \times d_1 \times d_2
\]
\[
80.5 = \frac{1}{2} \times 7 \times d_2
\]

Solve for \( d_2 \):
\[
80.5 = 3.5 \times d_2
\]
\[
d_2 = \frac{80.5}{3.5}
\]
\[
d_2 = 23 \, \text{ft}
\]

Thus, \( SU = 23 \, \text{ft} \).

---

5) If \( BD = 16 \, \text{in} \), find \( AC \).

Given:
- \( BD = 16 \, \text{in} \)
- Area = \( 192 \, \text{in}^2 \)

Let:
- \( d_1 = BD = 16 \, \text{in} \)
- \( d_2 = AC \)

Using the area formula:
\[
\text{Area} = \frac{1}{2} \times d_1 \times d_2
\]
\[
192 = \frac{1}{2} \times 16 \times d_2
\]

Solve for \( d_2 \):
\[
192 = 8 \times d_2
\]
\[
d_2 = \frac{192}{8}
\]
\[
d_2 = 24 \, \text{in}
\]

Thus, \( AC = 24 \, \text{in} \).

---

6) If \( VX = 49 \, \text{yd} \), find \( UW \).

Given:
- \( VX = 49 \, \text{yd} \)
- Area = \( 90.5 \, \text{yd}^2 \)

Let:
- \( d_1 = VX = 49 \, \text{yd} \)
- \( d_2 = UW \)

Using the area formula:
\[
\text{Area} = \frac{1}{2} \times d_1 \times d_2
\]
\[
90.5 = \frac{1}{2} \times 49 \times d_2
\]

Solve for \( d_2 \):
\[
90.5 = 24.5 \times d_2
\]
\[
d_2 = \frac{90.5}{24.5}
\]
\[
d_2 = 3.7 \, \text{yd}
\]

Thus, \( UW = 3.7 \, \text{yd} \).

---

7) The length of one of the diagonals of a rhombus is 38 inches. Find the length of the other diagonal if the area is 646 square inches.

Given:
- One diagonal \( d_1 = 38 \, \text{in} \)
- Area = \( 646 \, \text{in}^2 \)

Let:
- \( d_2 \) be the other diagonal

Using the area formula:
\[
\text{Area} = \frac{1}{2} \times d_1 \times d_2
\]
\[
646 = \frac{1}{2} \times 38 \times d_2
\]

Solve for \( d_2 \):
\[
646 = 19 \times d_2
\]
\[
d_2 = \frac{646}{19}
\]
\[
d_2 = 34 \, \text{in}
\]

Thus, the other diagonal is \( 34 \, \text{in} \).

---

8) The area of a rhombus is 125 square yards. If one of the diagonals measures 10 yards, find the length of the other diagonal.

Given:
- One diagonal \( d_1 = 10 \, \text{yd} \)
- Area = \( 125 \, \text{yd}^2 \)

Let:
- \( d_2 \) be the other diagonal

Using the area formula:
\[
\text{Area} = \frac{1}{2} \times d_1 \times d_2
\]
\[
125 = \frac{1}{2} \times 10 \times d_2
\]

Solve for \( d_2 \):
\[
125 = 5 \times d_2
\]
\[
d_2 = \frac{125}{5}
\]
\[
d_2 = 25 \, \text{yd}
\]

Thus, the other diagonal is \( 25 \, \text{yd} \).

---

Final Answers:

\[
\boxed{
\begin{array}{ll}
1) & XZ = 0.38 \, \text{yd} \\
2) & EG = 5 \, \text{ft} \\
3) & LN = 40 \, \text{in} \\
4) & SU = 23 \, \text{ft} \\
5) & AC = 24 \, \text{in} \\
6) & UW = 3.7 \, \text{yd} \\
7) & \text{Other diagonal} = 34 \, \text{in} \\
8) & \text{Other diagonal} = 25 \, \text{yd} \\
\end{array}
}
\]