Question

The edges of a triangle board are 12cm,16cm and 20cm. it is to be painted on both sides the cost of painting it 25paise per cm^2 is.

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Answer 1

Appropriate Question :-

The edges of a triangle board are 12cm,16cm and 20cm. It is to be painted on both sides. The cost of painting it 25paise per cm^2 is _______

$\large\underline{\sf{Solution-}}$

Given that,

The edges of a triangle board are 12cm,16cm and 20cm.

Let assume that the required triangle be ABC such that

AB = c = 12 cm

BC = a = 16 cm

CA = b = 20 cm

Consider,

$\rm \: {a}^{2} + {c}^{2}$

$\rm \: =  \: {12}^{2} + {16}^{2}$

$\rm \: =  \: 144 + 256$

$\rm \: =  \: 400$

$\rm \: =  \: {20}^{2}$

$\rm \: =  \: {b}^{2}$

$\rm\implies \: {a}^{2} + {c}^{2} = {b}^{2}$

$\rm\implies \: \triangle \: ABC \: is \: right \: angle \: \triangle \: right \: angled \: at \: B$

So,

$\rm \: Area_{(\triangle\:ABC)} = \dfrac{1}{2} \times AB \times BC$

$\rm \: Area_{(\triangle\:ABC)} = \dfrac{1}{2} \times 12 \times 16$

$\rm \: Area_{(\triangle\:ABC)} = 96 \: {cm}^{2}$

Now, Total area to be painted is

$\rm \: Area_{(to \: be \: painted)} = 96 \times 2 = 192 \: {cm}^{2}$

Now, given that,

$\rm \: Cost \: of \: painting \: {1 \: cm}^{2} = 25 \: paisa \:$

So,

$\rm \: Cost \: of \: painting \: {192 \: cm}^{2} = 192 \times 25 = 4800\: paisa \:$

We know, I rupee = 100 paisa

So,

$\rm \: Cost \: of \: painting \: {192 \: cm}^{2} = \dfrac{4800}{100} = Rs \: 48 \:$

Hence,

The edges of a triangle board are 12cm,16cm and 20cm. It is to be painted on both sides. The cost of painting it 25paise per cm^2 is Rs 48.

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$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$