Hello, jumpman23!
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 $\displaystyle d_1$ and $\displaystyle d_2$, the area is: .$\displaystyle A \:=\:\frac{1}{2}d_1d_2$ Code:
A 10 B
* - - - - - *
/ * * /
/ * * /
10/ *O / 10
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D 10 C
We have: .$\displaystyle AB \,=\, BC \,=\, BD \,=\, DA \,=\, 10$
Let $\displaystyle AC\,= \,d$. .Then $\displaystyle BD\,=\,d + 4$
Then: .$\displaystyle AO \,= \,\frac{d}{2},\;BO \,= \,\frac{d+4}{2}$
In right triangle $\displaystyle AOB :\;\;AO^2 + BO^2\:=\:AB^2$
. . so we have: .$\displaystyle \left(\frac{d}{2}\right)^2 + \left(\frac{d+4}{2}\right)^2\;=\;10^2$
. . which simplifies to: .$\displaystyle d^2 + 4d - 192\:=\:0$
. . which factors: .$\displaystyle (d -12)(d + 16)\:=\:0$
. . and has roots: .$\displaystyle d \:=\:12,\,-16$
The diagonals are: .$\displaystyle AC \,=\, d \,= \,12,\;BD \,= \,d+4 \,= \,16$
Therefore, the area is: .$\displaystyle A \:=\:\frac{1}{2}(12)(16)\:=\:96$ cm².