Answer to Question #256880 in Geometry for tay

Question

a region of land, 50 m by 40 m is to cross diagonally by a road 15 meters wide.what is the roads are?

Answer 1

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 center 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 center 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 center line.

In this small right-angle triangle , a side and an angle are

known. the hypotenuse x must be calculated.

$Sin Ø = \frac{Opp}{ Hyp}$

$Hyp = x = \frac{Opp }{ Sin Ø} =\frac{ 7.5 }{ Sine \space 38.6598 º} = 12.0058m$

.

Similarly calculating y, using (90 - 38.6598 º) for the angle,

$Hyp = y = \frac{7.5 }{ Sin \space (90 - 38.6598 º)} = 9.60486m .$

This is the sneaky 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

$= (50\times40) - (50-x) \space by\space (40-y)\\ = (50\times 40) - (50-12.0058)\space by\space (40-9.60468)\\ = 845.15m² \space to\space 2 dps.$