maths problem including an isosceles trapezium?

Question

a diagram shows a field which a farmer wants to fence.
THIS IS THE KIND OF ARRANGEMENT THE CORNERS ARE IN:
.....D.......C

A..................B

SO IF U JOINED UP THE LETTERS, THEY ARE THE SHAPE ITS IN.

The field is in the shape of an isosceles trapezium in which AB is parallel to DC and AD = BC

AB=80m and CD =60m
The distance between the parallel sides is 30m.

Calculate the length of fencing the farmer will need. (basically the perimeter of the trapezium)

Answer 1

Subtract AB-CD = 20m and half of that would be 10m. (This is the area from AD1 + C1B (as shown below).

.......D..........C

A...D1.........C1....B

Now solve using the laws of right triangles.

AD1=10M
DD1=30M

a^2 + b^2 = c^2

100+900 = 1000
square root of 1000 = 31.6m

31.6m x 2 (for CB and AD) + 60 (CD) + 80 (AB) =203.25M of fencing.

Answer 2

You would use the pythagorean theorom to figure DA and CB.

a^2+b^2 = c^2
30^2+((80-60)/2)^2 = c^2 The bases of the right triange are 30 m (the distance between parallel sides) and the other is (AB-CD)/2
900+100 = C^2
1000 = C^2
C = 31.62

31.62 is the length of the sides of the trapezoid. Double that to get the total. 63.24 then add the other two sides. The answer is 203.24....

Answer 3

Drop perpendiculars from D and C to AB at E and F respectively.

Then DC = EF = 60 so AE = FB must both = 10m
Since DE = 30m we get AD =sqrt(30^2+10^2) = 31.62m
So total perimeter = 2* 31.62 + 80+60 = 203.25m