1. Jun 12, 2010

### thereddevils

Is it true that the area of quadrilateral in general is the product of its diagonal divided by 2 ? Does this include rhombus or parallellogram ?

2. Jun 12, 2010

### Staff: Mentor

Did you mean "product of its diagonals divided by 2"? If so, this isn't true.

As a counterexample, consider a rectangle whose width is w and length l. The length of the diagonal is sqrt(w^2 + l^2). The product of the diagonals is w^2 + l^2, and half that is (1/2)(w^2 + l^2) != lw.

If that's not what you meant, what did you mean?

3. Jun 12, 2010

### thereddevils

thanks Mark, yes thats what i meant.

I came across this question asking to find the area of quadrilateral and i have the values of its diagonals. I got the answer coincidentally by multiplying its diagonals and halved it.

What's the correct formula ?

4. Jun 12, 2010

### ocohen

no this is not true.

5. Jun 13, 2010

### Axiom17

Refer to this quick diagram:

http://yfrog.com/afpf4j

OK so you know the values of $A$ and $B$ right, and you need to find the area of the shape.

Obviously the area is given by $X \times Y$.

Also the length of diagonal $A$, which is equal to the diagonal length $B$, is obviously given by $A=B=\sqrt{X^{2}+Y^{2}}$

$$\frac{A\times B}{2}=\frac{(\sqrt{X^{2}+Y^{2}})\times (\sqrt{X^{2}+Y^{2}})}{2}=\frac{X^{2}+Y^{2}}{2}$$

Hence can see:

$$\frac{X^{2}+Y^{2}}{2}\neq XY$$

Although like you said you can 'accidently' get the correct answer, if for example $X=Y=2$ but it's not true in general.

Hope that helps

6. Jun 14, 2010

### thereddevils

thanks ! So that method only works for quadrilaterals with all equal sides ie squares and rhombus.

7. Jun 14, 2010

### Axiom17

I Believe that to be the case.

If the sides are equal:

$$\frac{A\times A}{2}=\frac{(\sqrt{X^{2}+X^{2}})\times (\sqrt{X^{2}+X^{2}})}{2}=\frac{2X^{2}}{2}=X^{2}$$

Hence clearly:

$$X^{2}=X\times X$$

So the method would work. But surely it's much easier to just use $Area=X\times Y$