A triangular park has sides 120 m, 80 m and 50 m. A gardener has to put a fence all around it and also plant grass inside. How much area does he need to plant? Find the cost of fencing it with barbed wire at the rate of Rs. 20 per metre, leaving a space of 3 m wide for a gate on one side.
Answers
Perimeter of the garden=120+80+50 =250m As space of 3 m wide for a gate is left on one side, so the length of wire required=(250-3)=247m. Cost of fencing=247X20=Rs 4940. semiperimeter(s)=perimeter/2 =250/2=>125m Area of triangle is given by A= sqrt(s(s-a)(s-b)(s-c))=sqrt(125x5x45x75)=(25x15sqrt(15))..
Step-by-step explanation:
Given :-
A triangular park has sides 120 m, 80 m and 50 m. A gardener has to put a fence all around it and also plant grass inside. How much area does he need to plant
To find:-
Find the cost of fencing it with barbed wire at the rate of Rs. 20 per metre, leaving a space of 3 m wide for a gate on one side.
Solution:-
Given sides of a triangular park are 120 m , 80 m ,
50 m
Let a = 120 m
b = 80m
c=50m
Total area of the park to plant =
Area of the triangular park
=>area of the triangle
=>√[s(s-a)(s-b)(s-c)] sq.units
Where ,s =(a+b+c)/2
= >S = (120+80+50)/2
=>s=250/2
=>S = 125 m
Now area = √[125(125-120)(125-80)(125-50)]
=>√[125×5×45×75]
=>√(625×9×5×15×5)
=>25×3×5×√15
=>375√15 sq.m
Total area = 375√15 sq.m
Perimeter of the park = Sum of all sides
P= a+b+c
=>P= 120+80+50
=>P = 250 m
Perimeter of the given park = 250 m
and width of the gate = 3m
If fencing is left for gate so
Total Perimeter of the park leaving space for gate
=>250-3
=>247m
Cost of fencing the boarder of the triangular park per metre = Rs. 20
Total cost of fencing of 247 m
= 247×20
=4940
Total cost = Rs. 4940
Answer:-
1)the cost of fencing it with barbed wire at the rate of Rs. 20 per metre, leaving a space of 3 m wide for a gate on one side is Rs. 4940
2)Required area of the park to be planted plants is 375√15 sq.m