the edges of triangular board are 3 4 and 5 m. if the cost of painting is 10 rupees per m^2, then find the total cost of painting at only one face
Answers
Step-by-step explanation:
Let AB = a = 3m; AC = b = 4m and BC = c = 5m.
According to Heron's Formula;
Area = \sqrt{s (s - a) (s - b) (s - c)}s(s−a)(s−b)(s−c)
where s = semi-perimeter = \dfrac{a + b + c}{2}2a+b+c = \dfrac{3 + 4 + 5}{2} = \dfrac{12}{2} = 623+4+5=212=6
(s - a) = 6 - 3 = 3
(s - b) = 6 - 4 = 2
(s - c) = 6 - 5 = 1
Substituting in Heron's Formula;
Area = \sqrt{6\times 3 \times 2 \times 1 } = \sqrt{36} =6m^{2}6×3×2×1=36=6m2
Cost for painting 1m² Area of board = ₹10
Cost of painting 6m² Area of board = 6 × 10 = ₹60/-
Answer: Cost of painting the Triangular board on one face = ₹60/-
Additional Information:
For finding the Area of a Traingle when Base and Height is given, we use the Formula: \dfrac{1}{2} \times Base \times Height21×Base×Height
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