Problem Two:
In this problem, we are asked to find the area of an irregular trapezoid. The
trapezoid has a vertex at the origin, "x-equals-zero
y-equals-zero."
From the origin, one edge extends to the point, "x-equals-seven
y-equals-fifteen."
From this point, another edge extends horizontally to the right for six units.
This edge terminates at the point, "x-equals-thirteen
y-equals-fifteen.
From the point, "thirteen
fifteen," a third edge extends vertically
downward for fifteen units, terminating at the point, "x-equals-thirteen
y-equals-zero."
The final edge extends from this point back to the origin along the x-axis.
The figure may be divided into two parts. The first part is a right triangle with
one acute vertex at the origin, a second acute vertex at the point, "seven
fifteen,"
and the right angle at the point, "zero
seven." The second part
of the figure is a rectangle with a height of seven units, and a base extending
along the x-axis from the point, "zero
seven," to the point, "zero
thirteen."
The right triangle has a base equal to seven units and a height equal to fifteen
units. The area of a triangle is one-half of the product of its base and its height.
One half of seven units times fifteen units is fifty-two-and-one-half square units.
The rectangle has a base equal to six units and a height equal to fifteen units.
The area of a rectangle is the product of its base and its height. Six units times
fifteen units equals ninety square units. The total area is the sum of the parts.
"Fifty-two-and-one-half square units," plus, "ninety square units,"
equals, "one-hundred-and forty-two-and-one-half-square-units."
Expressing all units in feet, the area is one-hundred-and forty-two-and-one-half-square-feet.
Expressed as an integral, the area is again the sum of two parts. The first part
is the area between the x-axis and the line from the origin to the point, "seven
fifteen."
The equation of this line is, "y equals fifteen-over-seven x." The corresponding
integral is, "The integral from x-equals-zero to x-equals-seven of fifteen-over-seven-x-dee-x."
The value of this integral, again using feet for our unit of measure, is fifty-two-and-one-half
square feet. The second part is the area between the horizontal line y-equals-fifteen
and the x-axis between x equals seven and x equals thirteen. The corresponding
integral is, "The integral from x-equals-seven to x-equals-thirteen of fifteen-dee-x."
The value of this integral is ninety square feet. Again, the area is the sum of
the two integrals, or one-hundred-and-forty-two-and-one-half square feet.