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?
To find the maximum area of triangle DEF, we need to determine the position of D on the line segment BC that will result in the maximum area.
Given that base BC has a length of 30 and a height of 24, the area of triangle ABC is (1/2) * base * height = (1/2) * 30 * 24 = 360.
Since DE is parallel to AB, the ratio of the length of DE to BC will be the same as the ratio of the height of triangle DEF to the height of triangle ABC. Let's denote the height of triangle DEF as h.
Therefore, the height of triangle DEF is h = (24 / 30) * DE = 4/5 * DE.
Similarly, since FD is parallel to AC, the ratio of the length of FD to AB will be the same as the ratio of the height of triangle DEF to the height of triangle ABC. Let's denote the length of FD as x.
Therefore, the length of FD is x = (30 / 24) * h = 5/4 * h.
The area of triangle DEF can be calculated as (1/2) * DE * FD = (1/2) * (4/5 * DE) * (5/4 * h) = (1/2) * DE * h.
To maximize the area of triangle DEF, we need to find the maximum value of DE * h.
Since DE and h are both positive, to maximize their product, we need to maximize both DE and h.
The maximum value of DE is the length of BC, which is 30.
The maximum value of h can be obtained by drawing a perpendicular from point B to AC. Let's denote the foot of the perpendicular as G. Since triangle ABC is obtuse at B, G will be the point on AC that is closest to B.
Using the Pythagorean theorem, we can calculate the length of BG as sqrt(BC^2 - CG^2) = sqrt(30^2 - 24^2) = sqrt(900 - 576) = sqrt(324) = 18.
Therefore, the maximum value of h is BG, which is 18.
Now, we can calculate the maximum area of triangle DEF using the formula:
Area of triangle DEF = (1/2) * DE * h = (1/2) * 30 * 18 = 270.
So, the maximum area of triangle DEF is 270 square units.
To find the maximum value of the area of triangle DEF as point D varies within the line segment BC, we first need to understand the relationships between the different elements of the triangle.
1. Triangle ABC: This is an obtuse triangle with an obtuse angle at B. We know that its base BC has a length of 30, and the height from B to the line containing AC is 24.
2. Points D and E: D is a point on the line segment BC, and E is a point on AC such that DE is parallel to AB. This means that the triangles BDE and ABC are similar.
3. Point F: F is a point on AB such that FD is parallel to AC. This means that the triangles AFD and ABC are similar.
To find the maximum value of the area of triangle DEF, we need to determine the point on line segment BC that maximizes the area of triangle BDE. Since triangles BDE and ABC are similar, as D moves along line segment BC, the area of triangle BDE will be proportional to the square of the ratio of the corresponding sides.
Let x be the distance from B to D. Then, the length of BD can be expressed as (30 - x), and the length of DE can be expressed as (24(x/30)) because the triangles BDE and ABC are similar.
The area of triangle BDE can be calculated using the formula:
Area = (1/2) * base * height
In this case, the base is DE and the height is BD.
Area of triangle BDE = (1/2) * (24(x/30)) * (30 - x)
To find the maximum value of this area, we need to find the maximum value of this expression.
We can start by simplifying the equation by factoring out common terms:
Area of triangle BDE = (1/2) * (24/30) * (x(30 - x))
Area of triangle BDE = 12/15 * (30x - x^2)
Area of triangle BDE = 4/5 * (30x - x^2)
Now, we have an expression that represents the area of triangle BDE. To find the maximum value, we can take the derivative of this expression with respect to x and set it equal to zero:
d(Area of triangle BDE)/dx = 4/5 * (30 - 2x)
Setting this equal to zero:
4/5 * (30 - 2x) = 0
30 - 2x = 0
2x = 30
x = 15
Therefore, the maximum value of the area of triangle DEF occurs when D is located halfway between B and C, at a distance of 15 units from point B on line segment BC.