# Exercise ID343

Geometry → Two-dimensional geometric shapes → Quadrilaterals
[Level: ] [Number of helps: 0] [Number of pictures: 0] [Number of steps: 21] [Number of characters: 0]

Calculate the area of the parallelogram if its altitudes are 3 and $2\sqrt{3}$ cm, and the angle $between tham is$ 60°!

" display="block"/>
$\mathrm{sin}\alpha =\frac{{h}_{b}}{a}=\frac{{h}_{a}}{b}\phantom{\rule{0ex}{0ex}}$ $\mathrm{sin}60°=\frac{2\sqrt{3}}{a}=\frac{3}{b}=\frac{\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}$ $\frac{2\sqrt{3}}{a}=\frac{\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}$ $\frac{2}{a}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}$ $\frac{a}{2}=\frac{2}{1}\phantom{\rule{0ex}{0ex}}$ $a=4\phantom{\rule{0ex}{0ex}}$ $T=a·{h}_{a}=3·4\phantom{\rule{0ex}{0ex}}$ $T=12\phantom{\rule{0ex}{0ex}}$ x

$A=a·{h}_{a}=b·{h}_{b}$