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


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