The longer base of an isosceles trapezoid measures 18 ft. The nonparallel sides measure 8 ​ft, and the base angles measure 75 degrees.​a) Find the length of a diagonal.​b) Find the area.

Answer:a) The length of the diagonal is 17.71 feetb)  The area of the trapezoid is 123.14 feet²Step-by-step explanation:* Lets explain how to solve the problem- Look to the attached figure- ABCD is an isosceles trapezoid∵ DC is the longer base with length 18 feet∵ AD and BC are the two non-parallel sides with length 8 feet∵ ∠ ADC and ∠ BCD are the bases angles with measure 75°- AE and BF are ⊥ DC# In Δ BFC∵ m∠BFC = 90° ⇒ BF ⊥ CD∵ m∠C = 75°∵ BC = 8∵ sin∠C = BF/BC∴ sin(75) = BF/8 ⇒ multiply both sides by 8∴ BF = 8 × sin(75) = 7.73∵ cos∠C = CF/BC∴ cos(75) = CF/8 ⇒ multiply both sides by 8∴ CF = 8 × cos(75) = 2.07# In Δ BFD∵ m∠BFD = 90°∵ DF = CD - CF ∴ DF = 18 - 2.07 = 15.93∵ BD = √[(DF)² + (BF)²] ⇒ Pythagoras Theorem∴ BD = √[(15.93)² + (7.73)²] = 17.71a)∵ BD is the diagonal of the trapezoid* The length of the diagonal is 17.71 feetb)- The area of any trapezoid is A = 1/2 (b1 + b2) × h, where b1 and b2   are the barallel bases and h is the height between the two bases∵ b1 is CD∴ b1 = 18∵ b2 is AB∵ AB = CD - (CF + DE)∵ ABCD is an isosceles trapezoid∴ CF = DE∴ AB = 18 - (2.07 + 2.07) = 13.86- BF is the perpendicular between AB and CD∴ BF = h∴ h = 7.73∵ A = 1/2 (18 + 13.86) × 7.73 = 123.14 * The area of the trapezoid is 123.14 feet²