To calculate the surface area of the given prism, we need to find the areas of all its faces and sum them up.

Prism details:

• Base shape: A right trapezoid with bases of 6 cm and 8 cm and height 14 cm.
• Height of the prism: 14 cm.

Step-by-step calculation:

1. Area of one trapezoidal base: The formula for the area of a trapezoid is: $A = \frac{1}{2} \times (b_1 + b_2) \times h$ Substituting $b_1 = 6$, $b_2 = 8$, and $h = 14$: $A = \frac{1}{2} \times (6 + 8) \times 14 = \frac{1}{2} \times 14 \times 14 = 98 \, \text{cm}^2$ There are two such bases, so the total area for the bases is: $2 \times 98 = 196 \, \text{cm}^2$

2. Area of the lateral faces: There are 4 lateral faces, all rectangles. Their areas are calculated as:

   • Rectangle 1: $6 \, \text{cm} \times 14 \, \text{cm} = 84 \, \text{cm}^2$
   • Rectangle 2: $8 \, \text{cm} \times 14 \, \text{cm} = 112 \, \text{cm}^2$
   • Two side faces (14 cm × the slant height). The slant height isn't provided explicitly, but from the trapezoid’s geometry, we can calculate it as the hypotenuse of the triangle formed by half the difference in base lengths (1 cm) and the height (14 cm): $\text{Slant height} = \sqrt{(14)^2 + (1)^2} = \sqrt{196 + 1} = \sqrt{197} \, \text{cm}$ So, the area of one slant side is: $14 \times \sqrt{197} \, \text{cm}^2$ There are two such faces, so their combined area is: $2 \times 14 \times \sqrt{197} = 28 \sqrt{197} \, \text{cm}^2$

3. Total surface area: Combine all the areas: $\text{Total surface area} = 196 + 84 + 112 + 28\sqrt{197} \, \text{cm}^2$ Simplifying: $\text{Total surface area} \approx 392 + 28 \sqrt{197} \, \text{cm}^2$

If you'd like, I can refine this further or clarify any step. Do you have any questions or need detailed expansion?

Related questions:

1. How is the slant height of a triangle calculated?
2. How do you calculate the area of irregular shapes like a trapezoid?
3. What is the significance of lateral area in 3D geometry?
4. How can surface area be applied in real-life contexts?
5. How do approximation and square root calculations affect precision?