Triangle ABC has an obtuse angle at
B, base
BC has
length equal to 30 and height equal to
24.
D is a point on
the line segment BC and
E is a point on AC such that DE∥AB. F is a point on AB such that FD∥AC. As D
varies within line segment BC, what is the maximum value of area of triangle DEF?
90
To find the maximum value of the area of triangle DEF as point D varies along line segment BC, we need to identify the location on line segment BC that will result in the largest possible area for triangle DEF.
Given that BC has a length of 30 and a height of 24, we can start by calculating the area of triangle ABC. The formula for the area of a triangle is:
Area = (1/2) * base * height
Plugging in the values, we can calculate the area of triangle ABC:
Area(ABC) = (1/2) * 30 * 24
= 360
Since triangle ABC is obtuse and has a base and height, it can be divided into two triangles, DEF and FEC, which are similar to triangle ABC.
Now, let's try to find the relationship between triangle ABC and triangle DEF.
Given that DE is parallel to AB and DF is parallel to AC, we can conclude that triangle DEF is similar to triangle ABC. This means that triangle DEF has the same shape, but possibly different dimensions.
Since the two triangles are similar, the ratio of their areas is equal to the square of the ratio of their corresponding side lengths.
Area(DEF) / Area(ABC) = (DE^2 / AB^2) = (DF^2 / AC^2)
Let's denote the length of line segment BD as x, where x ranges from 0 to 30, as D varies within line segment BC.
Since DE is parallel to AB, triangle DBE is similar to triangle ABC. This means that the ratio of the lengths of their corresponding sides is the same. Thus, we can write:
DE / AB = DE / (x + 30) = 24 / 30
Simplifying the equation, we find:
DE = 24 * (x + 30) / 30
= 24x/30 + 24
Similarly, since FD is parallel to AC, triangle FDC is similar to triangle ABC. This gives us:
DF / AC = DF / 24 = (30 - x) / 30
Rearranging the equation, we get:
DF = 24 * (30 - x) / 30
= 24 - 24x/30
Now that we have expressions for the lengths of DE and DF, we can calculate the area of triangle DEF using the formula:
Area(DEF) = (1/2) * DE * DF
Substituting the values, we get:
Area(DEF) = (1/2) * (24x/30 + 24) * (24 - 24x/30)
Simplifying further, we obtain:
Area(DEF) = (12x + 720 - 16x)/30 * (24 - 16x)/30
= (720 - 4x^2)/225
To find the maximum value of Area(DEF), we need to find the vertex of the parabola defined by the equation (720 - 4x^2)/225.
The vertex of a parabola given by the equation y = ax^2 + bx + c is located at x = -b/2a. In our case, a = -4, b = 0, and c = 720/225.
Substituting these values, we find:
x = -0 / (2 * -4)
= 0
Therefore, the maximum value of Area(DEF) occurs when x = 0.
Substituting this into the equation for Area(DEF), we have:
Area(DEF) = (720 - 4 * 0^2) / 225
= 720 / 225
= 32/5
= 6.4
Hence, the maximum value of the area of triangle DEF is 6.4.