Hi Alaska D
A Primitive Pythagorean Triple is present in the given information and the answer choices. The diagonal along with one of the choices are two pieces of the triple and you can use them to find the missing piece.
h is 35
r is 6 (Diameter of the cylinder is 12)
122 +352 = 372
Checking
Total Surface Area of a Cylinder = 2∏rh + 2∏r2
I’ve factored out 2∏r below
2∏rh(h + r)
2*∏*6(35+6) = 12∏(41)=492∏
FIrst it is always good to make a drawing or at least see some images of the Diagonal of a Right cylinder. The next thing that should come out from drawing it is a Right Triangle. From there an application of the Pythagorean Theorem can be made. Mark M’s first equation actually illustrates that. Notice
372 - 4r2 = h2
4r2 = (2r)2
2r is the diameter of a circle in your problem the base
372 - (diameter)2 = h2
At this point you can actually plug in the given selections to see which one works
While you don’t know the diameter or the height, you should know some Primitive Pythagorean Triples.
Since this is an SAT Math problem, I have another idea about what the information given may be trying to trigger. Typically with these tests you get a set of formulas but largely they expect a lot of cumulative knowledge based a year of Geometry and all that came before it.
The Diagonal of a Right cylinder is 37 from the top corner to bottom opposite corner.
This Diagonal is also the hypotenuse of a Right Triangle and it is a positive integer and a piece of a Primitive Pythagorean Triple. The Right Triangle formed consists of the Diameter(2r) of the Cylinder, the height and the Hypotenuse of 37
Primitive Pythagorean Triples of integers a, b, and c share no common factors they are not multiples of another known triple such as 3,4,5 that’s why they are “Primitive”
An SAT may expect one to know maybe the first ten or twelve Pythagorean Triples or a couple of the special patterns (and there are some) that yield them.
One of these patterns for a Primitive says if a is the smallest number, and is an even positive integer then b and c are both odd and c = b + 2
If c = b + 2 then c - 2 = b
a2 +(c -2)2 = c2
a2 + (37 - 2)2 = 372
a2 + 352 = 372,
a2 = 372 - 352 = 1369 -1225 =144
a = √144 = 12
I noticed that your problem gave the diagonal as an integer and the answer choices were also integers so it was actually giving two parts of a Pythagorean Triple while strongly encouraging you to draw a Right Triangle. The Special Pattern above is not the only one that exists there are more and of course this not the only way to approach the problem. However, recognizing numbers from a Pythagorean Triple quickly on a test could save some time.
Sorry for the long note.
Kristen S.
05/21/23