# Thread: Area of rhombus

1. ## Area of rhombus

I am having problems doing the following (please have a look at the attached figure):

Find the area of a rhombus whose side is 15 cm long while one of the diagonals is 24 cm long. Here AD=15 cm & AC=24 cm.
(Hint: Find OD using Pythagoras theorem)

Thanks,

Ron

2. Hello rn5a
Originally Posted by rn5a
I am having problems doing the following (please have a look at the attached figure):

Find the area of a rhombus whose side is 15 cm long while one of the diagonals is 24 cm long. Here AD=15 cm & AC=24 cm.
(Hint: Find OD using Pythagoras theorem)

Thanks,

Ron
Following my answer to your previous post, the other way of finding the area of a rhombus is to use the formula:
Area of rhombus = $\displaystyle \tfrac12$ product of the lengths of the diagonals
The reason this formula works is because the diagonals of a rhombus are perpendicular, thus dividing the rhombus into four right-angled triangles. The area of each triangle is $\displaystyle \tfrac12$ base x height; when you multiply this by 4, you'll find that you get the formula above.

So, use the hint you're given to find OD (which is half the length of one of the diagonals), and then multiply it by the length of the other diagonal.

3. Area of rhombus = product of the lengths of the diagonals
The reason this formula works is because the diagonals of a rhombus are perpendicular, thus dividing the rhombus into four right-angled triangles. The area of each triangle is base x height; when you multiply this by 4, you'll find that you get the formula above.
Sorry but I couldn't exactly understand the above which I have quoted.

base x height x 4 = 2 base x height

So how does the Area of rhombus = product of the lengths of the diagonals?

Thanks,

Ron

The reason this formula works is because the diagonals of a rhombus are perpendicular, thus dividing the rhombus into four right-angled triangles. The area of each triangle is base x height; when you multiply this by 4, you'll find that you get the formula above.

5. Hello rn5a
Originally Posted by rn5a
Sorry but I couldn't exactly understand the above which I have quoted.

base x height x 4 = 2 base x height

So how does the Area of rhombus = product of the lengths of the diagonals?

Thanks,

Ron
In your diagram, if the diagonals meet at $\displaystyle O$, then the four triangles: $\displaystyle AOD, DOC, COB, BOA$ are congruent, and each is right-angled at $\displaystyle O$. The area of one of them, for example $\displaystyle AOD$, is $\displaystyle \tfrac12AO\times OD$. So the area of the rhombus is $\displaystyle 4\times\tfrac12AO\times OD=2\times AO \times OD$.

But $\displaystyle AO = \tfrac12 AC$ and $\displaystyle OD = \tfrac12 DB$. So the area of the rhombus $\displaystyle = 2\times\tfrac12 AC \times\tfrac12 DB=\tfrac12AC\times DB=\tfrac12$ the product of the diagonals.

,
,

,

,

,

,

,

,

,

# area of rhombus each side15cm and diagonal 24cm

Click on a term to search for related topics.