a region of land, 50m by 40m is to cross diagonally by a road 15 meters wide, wh

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** I calculate the area of the road to be 845.15 m² . Solution; To follow the method rough sketch a rectangle 50 wide by 40 high. Draw a line from the lower left corner to the top right corner. This is the centre line of the road crossing the plot diagonally. Sketch 7.5m lines on either side of this line, to give 15m width. . The plan is to calculate the distance from a corner that these lines cross the plot perimeter. Because the plot is not square they are different at the same corner, but are repeated at the opposite corner. Call the distance along the 50m side x, and the distance along the 40m side y. . The angle Ø the road centre makes with the lower 50m side is the angle having Tan = 40/50, => 38.6598 º Sketch a line from where the right road width line crosses 50m side, to form a perpendicular to the road centre line. In this small right-angle triangle , a side and an angle are known. the hypotenuse x must be calculated. Sin Ø = Opp/ Hyp Hyp = x = Opp / Sin Ø = 7.5 / Sine 38.6598 º = 12.0058m . Similarly calculating y, using (90 - 38.6598 º) for the angle, Hyp = y = 7.5 / Sin (90 - 38.6598 º) = 9.60486m . This is the sneeky bit. The area of the road = plot area - the area of the two triangles at either side of the road. Identical triangles to form a rectangle (50-x) by (40-y) Road Area = (50x40) - (50-x) by (40-y) = (50x40) - (50-12.0058) by (40-9.60468) = 845.15m² to 2 dps. .

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