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
