Question

The edges of a triangular board 12cm,16cm and 20cm. Find the cost of painting the board at the rate of 15 paise per cm square.

Answer 1

Here is your answer...

Answer 2

Given :

• 1st edge of the Triangle = 12 cm
• 2nd edge of the Triangle = 16 cm
• 3rd edge of the Triangle = 20 cm
• Rate of painting = 15 paise cm²

To Find :

• Cost of Painting = ?

Solution :

★Formula Used :

• ${\underline{\boxed{\pmb{\sf{ S = \dfrac{a + b + c}{2} }}}}}$

• ${\underline{\boxed{\pmb{\sf{ Area = \sqrt{s (s - a)(s - b)(s - c)} }}}}}$

Where :

• s = Semi - Perimeter
• a = Side 1
• b = Side 2
• c = Side 3

$\begin{gathered} \\ \qquad{\rule{150pt}{1pt}} \end{gathered}$

★Calculating the Semi - Perimeter :

$\begin{gathered} {\implies{\qquad{\sf{ S = \dfrac{a + b + c}{2} }}}} \\ \\ \\ \ {\implies{\qquad{\sf{ S = \dfrac{12+ 16 + 20}{2} }}}} \\ \\ \\ \ {\implies{\qquad{\sf{ S = \dfrac{12 + 36}{2} }}}} \\ \\ \\ \ {\implies{\qquad{\sf{ S = \dfrac{48}{2} }}}} \\ \\ \\ \ {\implies{\qquad{\sf{ S = \cancel\dfrac{48}{2} }}}} \\ \\ \\ \ {\qquad \; \; {\therefore \; {\underline{\boxed{\orange{\pmb{\frak{ Semi - Perimeter = 24\; cm }}}}}}}} \end{gathered} </p><p>$

$\begin{gathered} \\ \qquad{\rule{150pt}{1pt}} \end{gathered}$

★Calculating the Area :

$\begin{gathered} {\dashrightarrow{\qquad{\sf{ Area = \sqrt{s (s - a)(s - b)(s - c)} }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Area = \sqrt{24 (24 - 12)(24 - 16)(24 - 20)} }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Area = \sqrt{24 \times 12 \times 8 \times 4} }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Area = \sqrt{12 \times 2 \times 12 \times 2 \times 2 \times 2 \times 2 \times 2 } }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Area =12 \times 2\times 2 \times 2 }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ Area = 24 \times 4 }}}} \\ \\ \\ \ {\qquad \; \; {\therefore \; {\underline{\boxed{\purple{\pmb{\frak{ Area = 96 \; {cm}^{2} }}}}}}}} \end{gathered} </p><p>$

$\begin{gathered} \\ \qquad{\rule{150pt}{1pt}} \end{gathered}$

★Calculating The Cost :

$\begin{gathered} {\longmapsto{\qquad{\sf{ Cost = Area \times Rate }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ Cost = 96 \times 15 }}}} \; \; \; \; \; \bigg\lgroup {\pink{\sf{Rs. \; 1 = 100 \; Paise }}} \bigg\rgroup \\ \\ \\ \ {\longmapsto{\qquad{\sf{ Cost = 96 \times \dfrac{15}{100} }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ Cost = 96 \times \cancel\dfrac{15}{100} }}}} \\ \\ \\ \ {\longmapsto{\qquad{\sf{ Cost = 0.15 \times 96 }}}} \\ \\ \\ \ {\qquad \; \; {\therefore \; {\underline{\boxed{\red{\pmb{\frak{ Cost = ₹ \; 14.4}}}}}}}} \end{gathered} </p><p></p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered} \\ \qquad{\rule{150pt}{1pt}} \end{gathered}$

❛❛ Cost of Painting the Triangular board is ₹ 14.4 .❜❜

$\begin{gathered} \\ {\underline{\rule{300pt}{9pt}}} \end{gathered}$