Question

Triangle BCD, with vertices B(-9,-9), C(-4,-8), and D(-6,-3), is drawn inside a rectangle, as shown below.
What is the area of triangle BCD?

Answer 1

To find the area of triangle BCD, we can use the formula for the area of a triangle: A = (1/2) * base * height.

First, let's find the length of the base of triangle BCD. The base is the line segment BC, so we need to find the distance between points B and C.

Using the distance formula, which is given by d = sqrt((x2 - x1)^2 + (y2 - y1)^2), we have:

d = sqrt((-4 - (-9))^2 + (-8 - (-9))^2)
= sqrt(5^2 + 1^2)
= sqrt(25 + 1)
= sqrt(26)

So, the length of the base of triangle BCD is sqrt(26).

Next, we need to find the height of triangle BCD. The height is the perpendicular distance from point D to line BC.

Since D is not on line BC, we need to find the equation of the line passing through points B and C, and then find the distance between point D and this line.

The equation of a line passing through two points (x1, y1) and (x2, y2) is given by: y - y1 = (y2 - y1) / (x2 - x1) * (x - x1).

Using B(-9,-9) and C(-4,-8), we have:

(y - (-9)) = (-8 - (-9)) / (-4 - (-9)) * (x - (-9))
(y + 9) = (-1) / (5) * (x + 9)
(y + 9) = (-1/5) * (x + 9)
5 * (y + 9) = (-1) * (x + 9)
5y + 45 = -x - 9
x + 5y = -54

Now, let's find the perpendicular distance from point D(-6,-3) to this line.

The perpendicular distance from a point (x0, y0) to a line Ax + By + C = 0 is given by: d = |Ax0 + By0 + C| / sqrt(A^2 + B^2).

Using x + 5y = -54, we have:

d = |(1)(-6) + (5)(-3) + (-54)| / sqrt(1^2 + 5^2)
= |-6 - 15 - 54| / sqrt(1 + 25)
= |-75| / sqrt(26)
= 75 / sqrt(26)

So, the height of triangle BCD is 75 / sqrt(26).

Now, we can calculate the area of triangle BCD using the formula: A = (1/2) * base * height.

A = (1/2) * sqrt(26) * (75 / sqrt(26))
= (1/2) * 75
= 37.5

Therefore, the area of triangle BCD is 37.5 square units.