Answer:
a) The length of the diagonal is 17.71 feet
b) The area of the trapezoid is 123.14 feet²
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] = 17.71
a)
∵ BD is the diagonal of the trapezoid
* The length of the diagonal is 17.71 feet
b)
- 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