Question
A rhombus shaped sheet with perimeter $40\ cm$ and one diagonal $12\ cm,$ is painted on both sides at the rate of $Rs. 5$ per $m^2.$ Find the cost of painting.
✓
Answer
Let $\text{ABCD}$ be a rhombus having each side equal to $x \ cm.$
$i.e., AB = BC = CD = DA = x \ cm$ Given, perimeter of a rhombus $= 40$
$\therefore\text{ AB}+\text{BC}+\text{CD}+\text{DA}=40$
$\Rightarrow\ \text{x}+\text{x}+\text{x}+\text{x}=40$
$\Rightarrow4\text{x}=40$
$\Rightarrow\text{x}=\frac{40}{4}$
$\therefore\ \text{x}=10\text{cm}$ In $\triangle\text{ABC},$
let $\text{a}=\text{AB}=10\text{ cm},\text{ b}=\text{BC}$
$=10\text{ cm}$ and $\text{c}$
$=\text{AC}=12\text{ cm}$ Now,
semi$-$perimeter of a $\triangle\text{ABC},\text{ s}=\frac{\text{a}+\text{b}+\text{c}}{2}$
$=\frac{10+10+12}{2}$
$=\frac{32}{2}$
$=16\text{ cm}$
$\therefore$ Area of $\triangle\text{ABC}=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$ [by Heron's formula$]$
$=\sqrt{16(16-10)(16-10)(16-12)}$
$=\sqrt{16\times6\times6\times4}$
$=4\times6\times2$
$=48\text{ cm}^2$
$\because$ Cost of painting of the sheet of $1\ cm^2 = Rs. 5\ cm$
$\therefore$ Cost of painting of the sheet of $96\ cm^2 = 96 \times 5 = Rs. 480$
Hence, the cost of the painting of the sheet for both sides $= 2 \times 480 = Rs. 960$
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