Kuwait University Mathematics Aptitude Test study materials
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Triangular Courtyard Grass, Fence, and Fountain Clearance

Kuwait University Mathematics Aptitude Test · Put your knowledge to work in context.

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Triangular Courtyard Grass, Fence, and Fountain Clearance

A school in Kuwait City is redesigning a triangular courtyard. Two sides measure 18 m and 24 m, and the angle between them is 65 degrees. The third side is an existing wall that cannot be moved. The school wants to lay grass over the whole courtyard except for a circular fountain of radius 3 m, and to put a low fence along the two measured sides only. Grass costs 3.5 KD per square metre, and fencing costs 8 KD per metre. The maintenance team claims that the fountain will occupy less than 10% of the courtyard. The project budget for grass and fencing together is 3,000 KD. The design must keep the fountain at least 1 m away from every side of the courtyard. Determine whether the plan is feasible and whether the maintenance team's claim is correct.

1. Calculate the courtyard area, the length of the third side, the grass area after removing the fountain, and the total cost of grass and fencing. Is the budget met? Is the fountain under 10% of the courtyard?

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The area is

Area = (1/2)(18)(24) sin 65 = 216(0.9063) = 195.76 m^2.

By the cosine rule, the third side c satisfies

c^2 = 18^2 + 24^2 - 2(18)(24) cos 65 = 324 + 576 - 365.14 = 534.86,

so c = 23.13 m.

The fountain area is pi(3^2) = 28.27 m^2, so the grass area is 195.76 - 28.27 = 167.49 m^2. Grass costs 167.49(3.5) = 586.22 KD. The fence length is 18 + 24 = 42 m, costing 42(8) = 336 KD. Total cost is 586.22 + 336 = 922.22 KD, well under the 3,000 KD budget.

The fountain share is 28.27/195.76 = 0.1444, or 14.4%. The maintenance team's claim of less than 10% is false.

2. Can the 3 m radius fountain be placed so that it is at least 1 m from every side? Use the inradius or another distance argument.

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The largest circle that can fit inside a triangle is the incircle. Its radius is r = Area/s, where s is the semiperimeter. Here

s = (18 + 24 + 23.13)/2 = 32.565,

so r = 195.76/32.565 = 6.01 m.

The incenter is 6.01 m from each side. A 3 m radius fountain centered there leaves 6.01 - 3 = 3.01 m clearance to every side, which is more than the required 1 m. Therefore the fountain can be placed with at least 1 m clearance. A separate structural check would still be needed for fence posts or the wall, but geometrically the clearance condition is satisfied.