# An isosceles trapezoid has an area of 40 m^2 and an altitude of 2m. Its two bases have a ratio of 2 is to 3. What are the lengths of the bases and one diagonal of the trapezoid?

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by kaehamor

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by kaehamor

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Where a and b are parallel bases with ratio of 2:3, and h is the height or altitude.

Area: 40 m²

a = 2x

b = 3x

h = 2 m

Equation:

40 m² = ¹/₂ (2x + 3x) (2m)

40 m² = ¹/₂ (5x) (2 m)

40 m² = ¹/₂ (10 m)(x)

40 m² = 5m (x)

40 m² ÷ 5m = 5m(x) ÷ 5m

8 m = x

Substitute 8 m to x in parallel bases a and b:

Base a = 2x ⇒ 2(8 m) = 16 m

Base b = 3x ⇒ 3(8 m) = 24 m

Diagonal of Isosceles Trapezoid, using Pythagorean Theorem for solving the diagonal (hypotenuse).

Diagonal =

=

=

=

=

= 2 meters

≈ 20.099 meters

Please click image below for solution with illustration.