Asked by Anonymous
a trapezoid abcd has parallel sides ab and dc of lengths 8 and 22. If both diagonals AC and BD are of length 17, what is the area of the trapezoid
Answers
Answered by
Steve
the excess length of dc over ab is 22-8=14
SO, there are two triangles of base 7 and hypotenuse 17 at the ends of the trapezoid.
The height h of the trapezoid is thus
h^2 = 17^2 - 7^2 = 240
h = √240
The area of the trapezoid
A = (8+22)/2 * √240
= 15*4√15 = 60√15
SO, there are two triangles of base 7 and hypotenuse 17 at the ends of the trapezoid.
The height h of the trapezoid is thus
h^2 = 17^2 - 7^2 = 240
h = √240
The area of the trapezoid
A = (8+22)/2 * √240
= 15*4√15 = 60√15
Answered by
Steve
Oops. I was considering the 17 as the length of the edges, not the diagonals.
The height h is given by
15^2 + h^2 = 17^2
h=8
A = (8+22)/2 * 8 = 120
The height h is given by
15^2 + h^2 = 17^2
h=8
A = (8+22)/2 * 8 = 120
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