Problema Solution
An isoceles trapezoid has sides the length of 9, 10, 21,10. Find the distance that separates the parallel sides, then find the length of the diagonals. Finally find the angles of the trapezoid to the nearest tenth of a degree.
Answer provided by our tutors
make a good drawing and use Pythagorean Theorem
h = the distance that separates the parallel sides, h>0
h^2 = 10^2 - ((1/2)(21 - 9))^2
by solving we find:
h = 8
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d = the length of each diagonal, d>0
d^2 = 8^2 + ((21 - (1/2)(21 - 9))^2
d = 17
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let A and B represent the angles
sin A = h/10
sin A = 8/10
A = 53.1 degrees
sin B = (1/2)(21 - 9)/10
sin B = 0.6
B = 36.9 +90
B = 126.9 degrees
another way to find B is to use that the sum of the angles of the isosceles trapezoid is 360:
2(A + B) = 360
2(53.1 + B) = 360
by solving we find:
B = 126.9 degrees
click here to see the step by step solution of the equation: