A rectangle is a four sided figure with opposite side equal and parallel and all sides perpendicular with adjacent sides.
Divide the rectangle into unit squares and if necessary partial squares.
The area of the rectangle can be determined by counting these unit squares.
By inspection (a useful phrase which means I am not going to give a rigorous mathematical proof ) this is equal to
g x h
i.e. length x width.
Consider triangles ABD, CDB.
AB=CD, AD=CB, ^DAB=^BCD=1 rt. ang. ... (Properties of rectangle)
So
∆ABD ≡ ∆CDB ... (2 sides & included angle equal)
Area of ∆ABD = ∆CDB
Area of ABCD = gh ... (see above)
So
Area of rt. ang. triangle = ½(gh)
ABCD is a parallogram, with AB = g
perpendicular distance between AB & CD = h
Extend BA to E such that ED is perpendicular to BE
and
Extend DC to F such that FD is perpendicular to DF
AB = DC ... (Opp. sides of parallelogram equal)
^ABC = ^EAD ... (Cor. angles of parallelogram)
= ^ADC ... (Alt. angles of parallelogram)
= ^BCF ... (Cor. angles of parallelogram)
^CBF = 1 rt. ang. - ^ABC
= 1 rt. ang. - ^ADC
= ^EDA
So
∆EAD ≡ ∆CFB ... (2 sides & included angle equal)
So
EA = w = CF
EBFD is a rectangle .. (Opp. sides equal and adj. sides perp.) So
Area of EBFD = h(w + g) ... (see above)
But
Area EAD + Area FCB = ½(wh) + ½(wh)
= wh
And
Area ABCD = Area EBFD - (Area EAD + Area FCB)
= wh + gh - wh = gh
The area of a parallelogram is equal to the base times the perpendicular height.
Note this is also the area of a rectangle.
So all parallelograms, including rectangles, on the same base between the same parallel lines have the same area.
This is an alternative proof to Parallelograms on same base between parallels are equal
ABCD is a parallelogram
whose base is g and perpendicular height h.
Draw DB
Consider triangles ABD, CDB
BD = DB ... (Common side)
^ADB = ^CBD ... (Alt. angles AD parallel CB)
^ABD = ^CDB ... (Alt. angles AD parallel CB)
∆ABD ≡ ∆CDB ... (2 angles & included side equal)
So
Area of ∆ABD = ∆CDB
But
Area of ABCD = gh ... (See above)
So
Area of ∆ABD = ∆CDB = ½(gh)
This is the same result proven above for right angle triangle. So we have shown that it is the same for all triangles.
So area of a triangle is half base times height.
This is an alternative proof to Triangles on same base between parallels are equal
ABCD is a trapezium with DC, length a, parallel to AB, length b, and height h.
Extend AB until it meets a line through C parallel to AD at E making parallelogram AECD.
AE = DC ... (opp. sides of parallelogram are equal)
So
BE = a-b
Area ABCD = Area AECD - Area CBE
= ah - ½(a-b)h
= ½(a+b)h
The are of a trapezium is half the sum of the lengths of the parallel sides times height.