A triangular park ABC has sides 120m,80m and 50m (see the fig) .A gardener Dhania has to put a fence all around it and also plant grass inside.
What is the length of the longest side?
a) 300 m b) 120m c) 80m d) 50 m
Find the semi perimeter of the park ABC.
a) 125 m b) 350 m c) 900 m d) 375 m
What is the area of the park?
a) 375√ m2 b) 3375√ m2 c) 33750√ m d) 475√ m2
What is the cost of fencing the park with barbed wire at the rate of
Rs.20 per metre leaving aspace 3m wide for a gate on one side ?
a) Rs.4940 b) Rs.9000 c) Rs.4950 d) Rs.4500
Answers
Answer:
\bf \large \it Hey \: User!!!HeyUser!!!
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}
areaofthetriangularparkby
heronsformula=
s(s−a)(s−b)(s−c)
=
125(125−120)(125−80)(125−50)
=
125×5×45×75
=
2109375
=1452.36m
2
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