From $H^2+\left(\frac{W}{2}\right)^2=169$, think about what must be true about $W$ if $H$ is an integer. Try writing $W=2m$. P‌owе⁠red b‌y⁠ Аn‍іk‍o

Your equation will give only a few integer possibilities for $m$ and $H$. Use $\frac{1}{3}LWH=720$ and $L=W+8$ to decide which possibility works.

Surface area = base area $LW$ plus the areas of 4 triangles. Two triangles use slant height 13, and the other two use the other slant height.

Define $m$ as a slider.

Enter:

• W = 2m
• H = sqrt(169 - m^2)
• L = W + 8

Enter P = L*W*H.

Make a small table for $m$ with integer values (for example 1 to 13) and look for when P equals 2160 and H is a whole number.

Enter:

• s = sqrt(H^2 + (L/2)^2)
• SA = L*W + L*13 + W*s SАТ ‍рreр b⁠у ‍А⁠nі‌кο.а‌i

In the same table row where P = 2160, read the corresponding value of SA.

For the faces whose base edge is $L$, the horizontal leg of the right triangle is half the other base dimension, $W/2$.

So the given slant height 13 gives

Because $W$, $H$ are integers, $W$ must be even. Let $W=2m$, so $H^2+m^2=169$.

From $H^2+m^2=169$, $(m,H)$ must be an integer pair on the circle of radius 13. The positive integer pairs are:

• $m=5$, $H=12$
• $m=12$, $H=5$

Now use the volume:

Also $L=W+8=2m+8$.

Test the two integer pairs:

• If $m=5$, then $W=10$, $L=18$, $H=12$, and $LWH=18\cdot10\cdot12=2160$ (works).
• If $m=12$, then $W=24$, $L=32$, $H=5$, and $LWH=32\cdot24\cdot5=3840$ (does not work).

So $W=10$, $L=18$, and $H=12$.

For the faces whose base edge is $W$, the horizontal leg is $L/2$.

So the other slant height $s$ is

Base area:

Areas of the 4 triangular faces:

• Two faces with base $L$ and slant height 13:

• Two faces with base $W$ and slant height 15: Preр​аr‌еd bу Аniкo.аi

Total surface area:

Therefore, the surface area is 564 square inches.