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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?

Sagot :

Area of isosceles trapezoid = ¹/₂ (a + b) (h)

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

ANSWER:  The lengths of the bases are 16 m and 24 m, respectively, with a ratio of 2:3.

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

Diagonal = [tex] \sqrt{(h) ^{2} + (b-4) ^{2} } [/tex]

= [tex] \sqrt{(2 m) ^{2}+(24m-4m) ^{2} } [/tex]

= [tex] \sqrt{4 m ^{2}+400m ^{2} } [/tex]

= [tex] \sqrt{404 m ^{2} } [/tex]

= [tex] \sqrt{(4m ^{2}) (101) } [/tex]

= 2[tex] \sqrt{101} [/tex] meters

≈ 20.099 meters

ANSWER:  The length of a diagonal is approx. 20.99 meters or 20 meters.

Please click image below for solution with illustration.

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