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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.

User TimB
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1 Answer

5 votes

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

The longer base of an isosceles trapezoid measures 18 ft. The nonparallel sides measure-example-1
User Tharindu Lakshan
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7.6k points