Area of triangle ABC: A=?
Formula to calculate the area of a triangle: A=(1/2)bh
In this case, we can take:
Base of the triangle: b=AC=5
Height of the triangle: h=CB=?
Triangles ABC and ACD are similars, because they have two congruent angles:
Angle DAC is a common angle and
Angle ADC = 90° = Angle ACB (the angle ADC is 90° because the angle BDC is 90°)
Then, the triangles ABC and ACD must have proportional sides:
CB/AC=CD/AB
AC=5; CD=?; AD=2
We can find CD in triangle ACD using the Pythagoras Theorem:
CD=sqrt (AC^2-AD^2)
CD=sqrt (5^2-2^2)
CD=sqrt (25-4)
CD=sqrt (21)
Now we can find CB:
CB/AC=CD/AD
Replacing the known values:
CB/5=sqrt(21)/2
Solving for CB. Multiplying both sides of the equation by 5:
5(CB/5)=5[sqrt(21/2)]
CB=5sqrt(21)/2
Now we can calculate the area of the triangle ABC:
A=(1/2)(AC*CB)
A=(1/2)(5)(5sqrt(21)/2)
A=[1*5*5sqrt(21)]/(2*2)
A=25srt(21)/4
A=25(4.582575695)/4
A=28.64109809
Rounded to the nearest hundredth
A=28.64
Answer: The area of triangle ABC is 28.64 square units