Area of triangle ABC: A=?
A=(1/2)(AC*BC)
AC=5.0
BC=?
In triangle ACD: Suppose angle CAD=x and angle ACD=y→x+y=90°
As angle ACB is 90° and angle ACD is y, then the angle DCB must be equal to x.
Then triangles ACD and DCB are similars, because they have two congruent angles: Angle DAC (x) with angle DCB (x); and angle ADC (90°) with angle CDB (90°). Then theirs sides must be proportionals:
In triangle DCB: BC (hypotenuse) / CD (adjacent to angle DCB, x)
In triangle ACD: AC (hypotenuse) / AD (adjacent to angle CAD, x)
BC/CD=AC/AD
BC=?
CD=?
AC=5.0
AD=2.0
We can find CD in right 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)
BC/CD=AC/AD
BC/sqrt(21)=5/2
Solving for BC
BC=5*sqrt(21)/2
Area of triangle ABC: A=(1/2)(AC*BC)
A=(1/2){(5)*[5*sqrt(21)/2]}
A=(1/2)[25*sqrt(21)/2]
A=25*sqrt(21)/4
A=25*(4.582575695)/4
A=28.64109809
Rounded to the nearest hundreth:
A=28.64 square units
Answer: The area of triangle ABC is 28.64 square units