ABCD is a trapezoid with line BC perpendicular to line AB and line BC perpendicular to line CD. AB=13 BC=12 CD=8.A line segment is drawn from A to E, which is the midpoint of line CD. What is the area of triangle AED?

Question

Aurora Ayala · Accepted Answer

#S_(triangle AED)=24#

The trapezoid described is somewhat as the figure below:

The formula of the area of the triangle is
#S_(triangle AED)=(base*height)/2#

Since AF is perpendicular to the triangle's side DE, and AF=BC=12, we have:
#S_(triangle AED)=(DE*AF)/2=(4*12)/2=24#

Parker Allen · Answer

undefined To find the area of triangle AED, we first need to calculate the lengths of AE and ED, then we can use the formula for the area of a triangle.

Given that E is the midpoint of CD, AE is equal to half of AB, and ED is equal to half of CD.

$$ AE = \frac{AB}{2} = \frac{13}{2} = 6.5 $$

$$ ED = \frac{CD}{2} = \frac{8}{2} = 4 $$

Now, we can use the formula for the area of a triangle, which is given by:

$$ Area = \frac{1}{2} \times base \times height $$

In this case, the base is AE and the height is ED. Substituting the values:

$$ Area = \frac{1}{2} \times 6.5 \times 4 = 13 $$

Therefore, the area of triangle AED is 13 square units.