# Rhombus

• May 29th 2007, 02:44 PM
jumpman23
Rhombus
The sides of a rhombus are 10cm long. If the lengths of the diagonals differ by 4, what is the area of the rhombus?
• May 29th 2007, 03:58 PM
Soroban
Hello, jumpman23!

Quote:

The sides of a rhombus are 10cm long.
If the lengths of the diagonals differ by 4, what is the area of the rhombus?

You're expected to know a few facts about a rhombus.

. . The diagonals of a rhombus are perpendicular.

. . If the diagonals are $d_1$ and $d_2$, the area is: . $A \:=\:\frac{1}{2}d_1d_2$
Code:

            A    10    B             * - - - - - *           / *      * /           /  *  *  /       10/    *O    / 10         /  *  *  /       / *      * /       * - - - - - *       D    10    C

We have: . $AB \,=\, BC \,=\, BD \,=\, DA \,=\, 10$

Let $AC\,= \,d$. .Then $BD\,=\,d + 4$

Then: . $AO \,= \,\frac{d}{2},\;BO \,= \,\frac{d+4}{2}$

In right triangle $AOB :\;\;AO^2 + BO^2\:=\:AB^2$

. . so we have: . $\left(\frac{d}{2}\right)^2 + \left(\frac{d+4}{2}\right)^2\;=\;10^2$

. . which simplifies to: . $d^2 + 4d - 192\:=\:0$

. . which factors: . $(d -12)(d + 16)\:=\:0$

. . and has roots: . $d \:=\:12,\,-16$

The diagonals are: . $AC \,=\, d \,= \,12,\;BD \,= \,d+4 \,= \,16$

Therefore, the area is: . $A \:=\:\frac{1}{2}(12)(16)\:=\:96$ cm².