how can u prove the area of a trapezium is half the sum of the parallel sides X by the distance between them?

Question

i know to split the trapezium into a rectangle and 2 triangles but then i cant manipulate the algebra sufficiently to end up with the formula for calculating the area of a trapezium

Answer 1

Let the parallel sides be a (shorter) and b (longer) and the perpendicular distance between them be h.

If the trapezium is split into two triangles and a rectangle, and you join the two triangles together,

Area of the rectangle = a * h = ah
Area of the triangle = [(b-a)*h] / 2

Area of trapezium
= ah + (b-a)h/2
= ah + bh/2 - ah/2
= ah/2 + bh/2
= (h/2)(a+b)
= (1/2)(a+b)(h)

Answer 2

Divide the trapezium into a rectangle and 2 triangles, label the parallel sides a and b and the height h, work out the area of all 3 using algebra, ie rectangle = axh, etc, you should end up with a+b x h / 2.

Answer 3

Break up the trapezium into a rectangle, and two triangles.

Let the short side be a and the long side be a+b+c. Let the height be y.

Area of rectangle = xy
area of triangle1 = 1/2 *a*y
area of triange2 = 1/2*b*y
total area = y * (x + [a+b]/2)
= y * [2x + a + b] / 2
= y * [x + (x + a + b) ] / 2

Answer 4

Draw a trapezium ABCD such that AB < DC
AB and DC horizontal.
Draw perpendicular AE (E lies on DC)
Draw perpendicular BF (F lies on DC)
Let AE = BF = h (height)
Trapezium is now made up from rectangle ABFE and triangles ADE and BFC
LET AB = L, DC = B, DE = x, FC = y
Area triangle ADE = (1/2).x.h
Area triangle BFC = (1/2).y.h
Area rectangle ABFE = L.h
Total area = L.h + (1/2).(h).(x + y)
Now x + y = B - L
Total area = L.h +(1/2).h.((B - L)
Total area = (1/2).L.h + (1/2).B.h
Total area = (1/2).(L + B).h (as required)

Answer 5

If algebra is letting you down think of it this way.

2 copies of my trapezium fit together to make a parallelogram (turn one upside down)

Use the formula for the area of a parallelogram...

Check out the link below for more help

Answer 6

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