XW = the full length of the diagonal AC
WZ = half the length of the other diagonal, ½ BD
Area of rhombus ABCD = Area of rectangle WXYZ
= XW × WZ
= AC × (BD/2)
= ½ × AC × BD
So the area of a rhombus is half the product of its diagonals.
Why it happens: the two isosceles triangles ∆ABD and ∆CBD become rectangles of heights AO and OC, and AO + OC = AC because O lies on AC. Both rectangles have the same width, half of BD, which is why they slot together into one.
Q2.
Area of rhombus ABCD can also be determined by finding the areas of ∆ADB and ∆CDB. What formula does this give us?
Answer
The diagonals are perpendicular, so AO and CO are heights on the common base BD.
Area (∆ADB) = ½ × AO × BD
Area (∆CDB) = ½ × CO × BD
Area of rhombus ABCD = Area (∆ADB) + Area (∆CDB)
= ½ × AO × BD + ½ × CO × BD
= ½ × BD × (AO + CO)
= ½ × BD × AC
The very same formula as before — reached without any cutting at all.
Why it happens: because AC ⊥ BD, the segment AO really is the perpendicular distance from A to the line BD, so it may be used directly as a height. Two different routes — dissection and splitting into triangles — must give the same area, and they do.