Question

person runs along the boundary of an equilateral triangular field of each side 50 m. First side is covered in 50 seconds, second side is covered in 20 seconds and third side is covered in 80 seconds. Choose the corret statement. *
speed along second side is 2 m/s
speed along third is greater than 1 m/s
motion of the person is uniform
average speed of the person is 1 m/s



please answer the question​

Answer 1

Answer:

given the sides of the triangular park are 120m, 80m and 50m.

perimeter of the triangular park = 120m + 80m + 50m

= 250m

therefore it's semi-perimeter = 250/2

= 125m

\begin{gathered}\tt \small \: area \: of \: the \: triangular \: park \: by \\ \tt \small herons \: formula = \scriptsize {\sqrt{s(s - a)(s - b)(s - c)} } \\ \tt \footnotesize = \sqrt{125(125 - 120)(125 - 80)(125 - 50)} \\ \tt \footnotesize = \sqrt{125 \times 5 \times 45 \times 75} \\ \tt \footnotesize = \sqrt{2109375} \\ \tt \small = 1452.36 {m}^{2} \:\end{gathered}areaofthetriangularparkbyheronsformula=s(s−a)(s−b)(s−c)=125(125−120)(125−80)(125−50)=125×5×45×75=2109375=1452.36m2

hence, the gardener have to plant grass in 1452.36m²

now we have to find the cost of fencing the field with a barbed wire at the rate of rs 20 per m leaving a space of 3m wide for a gate.

therefore the gardener have to fence = 250 - 3

= 247m

so total cost of fencing at the rate of rs 20 per meter = 247 × 20

= rs 4940