The figure below shows a barn in the shape of two congruent pentagonal prisms that intersect at right angles and have a common center. The ends of the prisms are made of a 12 foot by 7 foot rectangle surmounted by an isosceles triangle with sides 10 feet, 10 feet, and 12 feet. Each prism is 30 feet long. Find the volume of the barn in cubic feet.

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1. **Volume of One Pentagonal Prism:**
   - The end of each prism is a combination of a rectangle and an isosceles triangle.
   - The rectangle has dimensions \(12 \text{ ft} \times 7 \text{ ft}\).
   - The isosceles triangle has sides \(10 \text{ ft}, 10 \text{ ft}, 12 \text{ ft}\).

2. **Area of the Isosceles Triangle:**
   - Using Heron's formula to find the area of the triangle:
     \[
     s = \frac{10 + 10 + 12}{2} = 16 \text{ ft}
     \]
     \[
     \text{Area} = \sqrt{s(s-10)(s-10)(s-12)} = \sqrt{16 \times 6 \times 6 \times 4} = \sqrt{2304} = 48 \text{ ft}^2
     \]

3. **Volume of the Top Part of the Prism:**
   - The top part is a triangular prism with base area \(48 \text{ ft}^2\) and length \(30 \text{ ft}\):
     \[
     \text{Volume}_{\text{top}} = 48 \times 30 = 1440 \text{ ft}^3
     \]

4. **Volume of the Bottom Part of the Prism:**
   - The bottom part is a rectangular prism with dimensions \(12 \text{ ft} \times 7 \text{ ft} \times 30 \text{ ft}\):
     \[
     \text{Volume}_{\text{bottom}} = 12 \times 7 \times 30 = 2520 \text{ ft}^3
     \]

5. **Total Volume of One Pentagonal Prism:**
   \[
   \text{Volume}_{\text{prism}} = 1440 + 2520 = 3960 \text{ ft}^3
   \]

6. **Volume of Two Pentagonal Prisms:**
   \[
   \text{Volume}_{\text{two prisms}} = 2 \times 3960 = 7920 \text{ ft}^3
   \]

7. **Volume of the Common Center:**
   - The common center is a combination of a pyramid and a rectangular box.
   - The base of the pyramid is a square with side \(12 \text{ ft}\) and height \(8 \text{ ft}\):
     \[
     \text{Volume}_{\text{pyramid}} = \frac{1}{3} \times 12 \times 12 \times 8 = \frac{1152}{3} = 384 \text{ ft}^3
     \]
   - The rectangular box has dimensions \(12 \text{ ft} \times 12 \text{ ft} \times 7 \text{ ft}\):
     \[
     \text{Volume}_{\text{box}} = 12 \times 12 \times 7 = 1008 \text{ ft}^3
     \]

8. **Total Volume of the Common Center:**
   \[
   \text{Volume}_{\text{common center}} = 384 + 1008 = 1392 \text{ ft}^3
   \]

9. **Final Volume of the Barn:**
   \[
   \text{Volume}_{\text{barn}} = 7920 - 1392 = 6528 \text{ ft}^3
   \]

The final answer is \(\boxed{6528 \text{ ft}^3}\).