A Residence Association is planning to develop a park in the colony. The rectangular park of dimensions 75m × 25m consists a path of uniform width along its boundary in the out side to develop walking habit among the residents of the colony. To beautify the park, 5 square flower beds are made in the park by spending 10% of the total cost of 5 lakh rupees for Rs.2500 per sq m.
1. The expression for finding the area of the path
2. The cost of constructing a 1m wide path at Rs. 1250 per sq m
3.The distance travelled by a person if he completes 10 rounds along the inner boundary of the path
4. The side of each square flower bed
Answers
Answer:
the layer of the earth's atmosphere which contains a high concentration of ions and free electrons and is able to reflect radio waves. It lies above the mesosphere and extends from about 80 to 1,000 km above the earth's surface.
Expression for finding the area of the path is 4+200x.
GIVEN: Dimension of park = 75*25m;
TO FIND: The expression for finding the area of the path;
SOLUTION:
Park of dimensions 75m × 25m
Path outside of width x m, all side
Hence, outer boundaries 75 + 2x , 25 + 2x
Area of path = ( 75 + 2x) (25 + 2x) - 75 * 25
= 150x + 50x + 4x ² + 75 * 25 - 75 * 25
= 4x² + 200x
The expression for finding the area of the path = 4x² + 200x
1m wide path Hence area of path = 4(1)² + 200(1)
= 204 m²
Rate = 1250 per sq m
Hence cost = 204 * 1250 = 2,55,000
Distance travelled by a person if he completes 10 rounds along the inner boundary of the path = 10 * 2 ( 75 + 25) = 2000 m = 2 km
5 square flower beds are made in the park by spending 10% of the total cost of 5 lakh rupees for Rs.2500 per sq m.
Spend = (10/100) 500000 = 50000
5 square flower beds
=> cost on one flower bed =50000/5 = 10000
Rs.2500 per sq m.
Hence area of one flower bed = 10000/2500 = 4 m²
The side of each square flower bed = √4 = 2 m
PROJECT CODE #SPJ2
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