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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.
Round your answers to the nearest hundredth.
You must show all of your work to receive credit.

Sagot :

kripalimalviya99

Answer:

Let X represent the shorter base, and drop the heights from shorter to longer base.

 

Two congruent right triangles are formed, because the trapezoid is isosceles and

the heights are the same.

 

The base angles are then 70 degree each, as they total 140.

 

So the height is 7* sin 70 = 6.5778483455....

 

The base of each right triangle is 11 - x/2 

 

Now the algebra gets ugly.

 

By pythagorean theorem:

 

(11 - x/2)^2 + (7sin70)^2 = 7^2

 

(11-x/2)^2 + 49sin70^2 = 49

(11 - x/2)^2 = 49 - 49sin70^2

(11 - x/2)^2 = 49(1 - sin70^2)

(11 - x/2)^2 = 49 cos70^2  <-- trig identity : sin^2 + cos^2 = 1 ---> cos^2 = 1 - sin^2 

121 - 11x + x^2/4 = 49 cos70^2

484 - 44x + x^2 = 196 cos 70^2

x^2 - 44x + 484 - 196cos70^2 = 0

x^2 - 44x + C = 0 where C = 484 - 196cos70^2 = 461.07235

 

By quadratic formula:

X = 44 +OR- square-root( (-44)^2 - 4*C) / 2

= 44 +OR- square-root( 91.71)/2

 

The two roots are:

26.78282 AND 17.2117331

The first one is disqualified and we must reject it

because if used will make the base of the

right triangles negative. Remember this is

supposed to be the smaller base. So it 

cannot be greater than 22.

 

SO the smaller base is 17.2117331

 

So 11 - x/2 = overhang = 2.39413345

 

Drawing the diagonal, it's length can

be found by pythagorean theorem.

square-root(17.2117331)^2 + (7sin70)^2)=18.4258472....<-- diagonal

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