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Home » Area of Geometric Shapes » Area of Rhombus

Area of Rhombus

Last updated: Jan 4, 2020 by Ido Sarig · This website generates income via ads and uses cookies · Terms of use · Privacy policy

There are several ways to find the area of a rhombus. A rhombus is a special kind of parallelogram, in which all the sides are equal.

Because it is a parallelogram, we can find its area using the general formula for the area of a parallelogram - the length of a side times the hight to that side.

Geometry drawing: area of a parallelogram

But a rhombus is also a kind of a kite - a kite in which all the sides are equal. So we can also use the formula for an area of a kite to get the area of a rhombus.

We've seen that if we know the lengths of a kite's diagonals, the area of the kite is the product of the lengths of these diagonals, divided by two.

Let's derive this formula for a rhombus, directly.

Problem

ABCD is a rhombus. Find a formula for its area using the lengths of its diagonals |AC| and |BD|.

rhombus with diagonals

Strategy

We'll start out the same way as we did when we derived the formula for the area of a kite, by using the technique of partitioning the kite into simpler shapes. The diagonals have already partitioned the rhombus into four triangles, ΔADO, ΔABO, ΔCBO and ΔCDO.

We know the diagonals of a rhombus are perpendicular to each other, and they bisect each other. And we've also seen that the diagonals of a parallelogram divide it into 4 equal areas.

So all we have to do is find the area of one of these triangles (ΔADO, ΔABO, ΔCBO or ΔCDO), and multiply by 4 to get the area of the rhombus.

Let's take triangle ΔADO. The diagonals of the rhombus are perpendicular, so it is a right triangle, whose area is (leg x leg) / 2. And because the diagonals bisect each other, its two legs' lengths are equal to half of each diagonal. So its area is |AC|/2 times |BD|/2, divided by two.

The area of the rhombus is 4 times this, so it is 4 • (|AC|/2 • |BD|/2)/2., or after simplifying, |AC|• |BD|/2

Solution

(1) |AO|=|AC|/2 //diagonals of a parallelogram bisect each other
(2) |DO|=|DB|/2 //diagonals of a parallelogram bisect each other
(3) AreaAOD=|AO|•|DO|/2 // Area of a triangle
(4) AreaAOD=(|AC|/2 • |DB|/2)/2 //(1), (2), (3)
(5) AreaABCD = 4 • AreaAOD //Diagonals of a parallelogram divide it into 4 equal areas.
(6) AreaABCD = 4 • (|AC|/2 • |DB|/2)/2 //(4), (5)
(7) AreaABCD=|AC|• |DB|/2

« Converse of the Pythagorean Theorem
Heron's Formula »

About the Author

Ido Sarig is a high-tech executive with a BSc degree in Computer Engineering. His goal is to help you develop a better way to approach and solve geometry problems. You can contact him at [email protected]

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About

Welcome to Geometry Help! I'm Ido Sarig, a high-tech executive with a BSc degree in Computer Engineering and an MBA degree in Management of Technology. I'm here to tell you that geometry doesn't have to be so hard! My goal with this website is to help you develop a better way to approach and solve geometry problems, even if spatial awareness is not your strongest quality. Read More…

Geometry Topics

  • Area of Geometric Shapes
  • Circles
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    • Equilateral Triangles
    • Isosceles Triangles
    • Pythagorean Theorem
    • Right Triangles
    • Similar Triangles
    • Triangle Inequalities

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