There may be simpler ways of doing, but this is what came to mind for me.
In fact, the problem boils down to finding the height h, and the offset of the two parallel sides, k.
Let ABCD be the vertices of the quadrilateral where AB and DC are parallel sides.
AB=4 centimetres, and
DC=10 centimetres.
Drop a perpendicular from A to the long side DC to intersect at E.
Assume for the moment E is between D and C, so that DEC is a straight line.
Let
k=distance DE, and
h=distance between the two parallel sides = length AE
Consider each of the oblique sides as the hypothenuse of a right triangle,
for ADE,
AE²+DE²=AD²
h²+k²=8² ..... (1)
For the side BC,
(10-k-4)²+h²=BC²
36-12k+k²+h²=144....(2)
Subtract (1) from (2)
36-12k-8²=144
k=-44/12=-11/3
Meaning that the point E is situated on the extension of the line CD.
Substitute k into (1) gives
h=sqrt(8²-(11/3)²)
=sqrt(455)/3
=7.110243
Can you take it from here?
I have no idea how to do this problem I thought you would have to do something with the diagonals but not sure...
The parallel sides of a trapezoid are 4 and 10 centimeters long and the oblique sides are 8 and 12 centimeters long. Fi9nd the angles and the area of the trapezoid
1 answer
1 answer