Question

A thin metal iron•sheet is rhombus in shape, with each side 10 m. If one of its diaginals is 16 m, find the cost of painting its both sides at the rate of Rs 6 per $m^2$. Also, find the distance between the opposite sides of this rhombus.

Answer 1

Side of rhombus shaped iron sheet =10 m and one diagonals (AC)=16 m

Join BD diagonal which bisects AC at O

$\because$ The diagonals of a rhombus bisect eachother at right angle

$\therefore AO=OC=\frac{16}{2}=8m.$

Now in right $\Delta AOB$

$A{B}^2=A{O}^2+B{O}^2\Rightarrow (10{)}^2=(8{)}^2+B{O}^2$

$\Rightarrow 100=64+B{O}^2\Rightarrow B{O}^2=100-64=36=(6{)}^2$

$\therefore BO=6m$

$\therefore BD=2\times BO=2\times 6=12m$

Now, area of rhombus $=\frac{d_1\times d_2}{2}$

$=\frac{16\times 12}{2}=96m^2$

Rate of painting =Rs 6 per $m^2$

$\therefore$ Total cost of painting both sides,

$=2\times 96\times 6=Rs1152$

Distance between two opposite sides,

$=\frac{Area}{Base}=\frac{96}{10}=9.6m$