In the triangular right-prism on the right: The sides of the equilateral triangle are 12 m. The height of the prism is 20 m. Determine the: a) volume of the prism b) surface area of the prism.

We are given that $KO = OM = 8$ units, $JO = 6$ units, $LO = 15$ units, and $JL$ bisects $KM$ perpendicularly. This means that the diagonals $JL$ and $KM$ intersect at point $O$, $O$ is the midpoint of $KM$, and $JL \perp KM$.

A quadrilateral with perpendicular diagonals where one diagonal bisects the other is a kite. We can confirm this by checking the side lengths using the Pythagorean theorem:
In $\triangle JOK$: $JK^2 = JO^2 + KO^2 = 6^2 + 8^2 = 36 + 64 = 100 \implies JK = 10$ units.
In $\triangle JOM$: $JM^2 = JO^2 + OM^2 = 6^2 + 8^2 = 36 + 64 = 100 \implies JM = 10$ units.
In $\triangle LOK$: $KL^2 = LO^2 + KO^2 = 15^2 + 8^2 = 225 + 64 = 289 \implies KL = 17$ units.
In $\triangle LOM$: $LM^2 = LO^2 + OM^2 = 15^2 + 8^2 = 225 + 64 = 289 \implies LM = 17$ units.
Since $JK = JM$ and $KL = LM$, JKLM is a kite.
JKLM is a kite.

The diagonals of the kite are $KM$ and $JL$.

The area of a kite is given by the formula $\frac{1}{2} d_1 d_2$, where $d_1$ and $d_2$ are the lengths of the diagonals.

The base of the prism is JKLM, and its area is $168 \text{ units}^2$ (from part b i)).
The height of the prism is given as $LR = 25$ units.

The volume of a right-prism is given by the formula $\text{Area of base} \times \text{Height}$.

In the right-angled triangle $\triangle JOK$, $JO = 6$ units and $KO = 8$ units.

In the right-angled triangle $\triangle LOK$, $LO = 15$ units and $KO = 8$ units.

The side lengths of the kite are $JK = 10$ units, $JM = 10$ units, $KL = 17$ units, and $LM = 17$ units.

Area of base JKLM = $168 \text{ units}^2$ (from part b i)).
Height of prism $LR = 25$ units.

The formula for the total surface area of a right-prism is $2 \times \text{Area of base} + \text{Perimeter of base} \times \text{Height}$.