Hard SAT Geometry Problem with Solution Today I would like to solve the SAT geometry problem I posted yesterday. You can see the original post here: Hard SAT Geometry Problem Level 5 Geometry ABCD shown above is a square, m∠DFE = 60°, EF = 3, and CF = 4. What is the area of square ABCD ? Solution: DF is the hypotenuse of a 30, 60, 90 triangle (ΔDEF). Since EF = 3, we have DF = 6. We can now use the Pythagorean Theorem to get DC2 = DF2 – CF2 = 62 – 42 = 36 – 16 = 20. Notes: (1) The area of a square is A = s2, where s is the length of a side of the square. (2) There is no need to find DC here because the area of the square is equal to DC2, and that’s what we found. (3) The 30, 60, 90 triangle and the Pythagorean Theorem are both given to you at the beginning of each SAT math section. Share this problem with your Facebook friends that are preparing for the SAT If you’re preparing for the SAT, you may want to check out the Get 800 collection of SAT math books. Comments comments