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Area Of Rhombus

Theorem: Area of a rhombus is equal to one half of the product of diagonals.

Prerequisites:
Properties of rhombus (proof)
Area of right triangle (proof)

Proof:

Let ABCD be a rhombus having diagonals AC and BD of lengths d_1 and d_2 respectively, intersecting at a point E.
Since in a rhombus, diagonals bisect each other at right angles. Hence,

\qquad\quad\; AE = CE = \dfrac{d_1}{2}\\ \qquad\quad BE = DE = \dfrac{d_2}{2}

Now consider \triangle ABE.  Since, \angle AEB is a right angle, hence using the formula for area of a right triangle, area of \triangle ABE is equal to:

\qquad\quad\begin{align} ar (\triangle ABE) &= \dfrac{1}{2}\dfrac{d_1}{2}\dfrac{d_2}{2}\\[4pt] &= \dfrac{d_1 d_2}{8}\end{align}

Here ar stands for area. Similarly, area of other three triangles, i.e., \triangle BCE, \triangle CDE and \triangle DAE = \dfrac{d_1 d_2}{8}.

\begin{align}\qquad\quad\text{Area of ABCD} &= ar (\triangle ABE) + ar (\triangle BCE) + ar (\triangle CDE) + ar (\triangle DAE)\\ &= 4\times\dfrac{d_1 d_2}{8}\\ &= \dfrac{d_1 d_2}{2}\end{align}

Q.E.D.


Recommended:
Area of triangle
Area of rectangle
Properties of rhombus

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