The two non-parallel sides of an isosceles trapezoid are each 7 feet long. the longer of the two bases measures 22 feet long. the sum of the base angles is 140°. a. find the length of the diagonal. b. find the length of the shorter base.

smaller base is 17.21 feet

Diagonal = 18.42 feet

Explanation:

suppose, x be the shorter base.

Now, drop the heights from shorter to longer base.

 thus,

Two congruent right triangles are formed

and since the trapezoid is isosceles, the heights are the same from both ends of the shorter base.

Now, it is given that the sum of base angles is 140° and the trapezoid is isosceles. thus, the base angles at both the ends will be 70°

Therefore,

we get the height as 7 × sin 70° = 6.57

 now,

The base of each right triangle is (22 - x)/2 = 11 - x/2

Now, applying the concept of Pythagoras theorem, we have

(11 - x/2)²+ (7×sin70°)² = 7²

 or

(11-x/2)² + 49×(sin70°)² = 49

or

(11 - x/2)² = 49 - 49×(sin70°)²

(11 - x/2)² = 5.73

or

(11 - x/2) = 2.39   and    (11 - x/2) = -2.39

or

x = 17.21       and    x = 26.78

since the shorter base cannot be more than 22 feet

Thus, the smaller base is 17.21 feet

 also,

The base of each right triangle is 11 - x/2  = 11 - (17.21/2) = 2.39 feet

 Now,the length of the diagonal can be found out by

diagonal² = (17.21)² + (7×sin70°)²

or

diagonal² =339.45

or

diagonal = 18.42 feet

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