Problem:
What is the maximum area of an isosceles trapezoid that has legs of length 1 and one base twice as long as the other?
Answer Choices:
A. 45β
B. 78β
C. 452ββ
D. 23β
E. 433ββ
Solution:
Let the bases have lengths x and 2x. Extend the legs of the isosceles trapezoid to form two isosceles triangles with vertex angle ΞΈ, as shown. with vertex angle ΞΈ, as shown.
Because the ratio of the bases is 1:2, the two congruent sides of the smaller isosceles triangle have length 1 . The area of the trapezoid is 43β the area of the large isosceles triangle, so its area is maximized when the area of the large isosceles triangle is maximized. The area of the large isosceles triangle as a function of ΞΈ is 21ββ 22β sinΞΈ=2sinΞΈ. The maximum area of this isosceles triangle is 2 , occurring when ΞΈ=2Οβ. Therefore the maximum area of the trapezoid is 43ββ 2=(D)23ββ.
OR
Let h be the height of the trapezoid, let x and 2x be the base lengths, and let Ξ± be the base angle, as shown.
Then sinΞ±=h and 2cosΞ±=x. The area of the trapezoid is 2x+2xββ h, which can be written as a function of Ξ± :
3cosΞ±β sinΞ±=23βsin(2Ξ±).
Because sin(2Ξ±)β€1, the maximum area is (D)23ββ, which occurs when sin(2Ξ±)=1 and Ξ±=4Οβ.
OR
Let h be the height of the trapezoid, and let x and 2x be the base lengths (see the figure for the second solution). Then 12=h2+(2xβ)2, so
h=1β(2xβ)2β
The area A of the trapezoid is\
so
A=2x+2xβ1β(2xβ)2β
A2=49x2β(1β4x2β)=49x2ββ169x4β.
Let u=x2. Then A2=β169βu2+49βu, which has a maximum when
u=2β (β169β)β49ββ=2
Substituting u=2 gives A2=49β and A=(D)23ββ.
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