Area of a rhombus

A rhombus is a parallelogram, so base × height already works. But a rhombus has extra properties, and they give a second formula that is often much easier to use. It appears in the Śulba-Sūtras.

The extra properties: all four sides are equal, and the diagonals cut each other in half at right angles (they are perpendicular bisectors of each other).

Draw diagonal BD. It splits the rhombus into ∆ABD and ∆CBD, and because all the sides are equal, both are isosceles.

An isosceles triangle can be dissected into a rectangle of the same area: drop the perpendicular from the top, and fold the two halves outwards. Do that to each triangle and join the two rectangles into one rectangle WXYZ.

What are its sides? Following the pieces through: XW is the length of diagonal AC, and WZ is half the length of diagonal BD.

So Area of rhombus = Area of WXYZ = XW × WZ = AC × (BD/2) = ½ × AC × BD. In words: area of a rhombus = ½ × product of the diagonals.

There is a second way to the same answer, with no dissection. The diagonals are perpendicular, so AO is the height of ∆ADB on base BD, and CO is the height of ∆CDB on the same base.

Area(∆ADB) = ½ × AO × BD and Area(∆CDB) = ½ × CO × BD. Adding: ½ × BD × (AO + CO) = ½ × BD × AC, because AO + CO is the whole diagonal AC. The same formula, reached twice: a good reason to trust it.