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A rhombus sheet whose perimeter is $32\ m$ and whose one diagonal is $10\ m$ long, is painted on both sides at the rate of $Rs.\ 5\ per\ m^2$. Find the cost of painting.
Given:
A rhombus sheet whose perimeter is $32\ m$ and whose one diagonal is $10\ m$ long, is painted on both sides at the rate of $Rs.\ 5\ per\ m^2$.
To do:
We have to find the cost of the painting.
Solution:
Length of each side of the rhombus $=\frac{32}{4}$
$= 8\ m$
Length of one of the diagonals $AC = 10\ m$
In $\triangle ABC$,
$a=8\ m, b= 8\ m, c= 10\ m$
$s=\frac{a+b+c}{2}$
$=\frac{8+8+10}{2}$
$=\frac{26}{2}$
$=13$
Area of the triangle $ABC=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{13(13-8)(13-8)(13-10)}$
$=\sqrt{13 \times 5 \times 5 \times 3}$
$=5 \sqrt{39} \mathrm{~m}^{2}$
Area of one sides of sheet $=2 \times 5 \sqrt{39}$
$=10 \sqrt{39} \mathrm{~m}^{2}$
Area of both sides of the sheet $=2 \times 10 \sqrt{39}$
$=20 \sqrt{39}$
$=20 \times 6.25$
$=125.0 \mathrm{~m}^{2}$
Rate of polishing both the sides of the sheet $= Rs. \5$ per $\mathrm{m}^{2}$
Total cost of polishing $=Rs. 125 \times 5$
$= Rs.\ 625$.
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