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Class 10 > Learn Mathematics > A triangular park has sides 120 m, 80 m and 50 m. A...
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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.

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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))...
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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)) => 375 sqrt(15) sq.m read less
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Perimeter = 120+80+50 = 250 m Leaving 3 m for the gate, the length to be fenced = 247 m Cost of fencing = 247 x 20 = Rs 4940 Area of the park = 1/2 x base x Altitude = 0.5 x 80 x 50 = 2000 sq m
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Area of a triangle =s (s-a)(s-b)(s-c); where s=(120+80+50)÷2=125 Area to plant grass= 125×5×45×75=21,09,375sq.m. Length of barbed wire=perimeter of the field-space for door =120+80+50-3=247m Cost of fencing=247×20= Rs.4940
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using Henon's(s-a)(s-b)(s-c)=?125×5×45×75=(25×15)?15=375?15sq.m Cost of fence without gate=(250-3)×20=247×20=RS.4940
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Length of garden (perimeter)=120+80+50=250 Now s=(a+b+c)/2=125 Area=?s(s-a)(s-b)(s-c)=?125×5×45×75=(25×15)?15=375?15sq.mt Cost of fence without gate=(250-3)×20=247×20=RS.4940/-
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