Question 1188279: https://imgur.com/c7zQ0QS In the diagram, AD is the median of triangle ABC. Find the area of triangle ABC. Found 2 solutions by Edwin McCravy, greenestamps:Answer by Edwin McCravy(20060) (Show Source):
We drew in the altitude AE in green to the right. The green line is the
altitude of both triangles ABD and ADC.
Both triangles, ABD and ADC, have equal bases because AD is a median of triangle
ABC. So BD = DC. Let x = BD = DC, then BC = 2x
By the law of cosines,
x=0; x-22cos(B)=0
x = 22cos(B)
Ignore x=0, so x = 22cos(B)
Also by the law of cosines,
Divide through by 4
Substitute 22cos(B) for x
Angle B cannot be obtuse. It must be acute, so we ignore
the negative sign
Using the SAS formula for the area of triangle ABC
We find sin(B) from
We find BC = 2x = (2)( 22cos(B) ) = 44cos(B) = =
Edwin
I will pirate the drawing from the response from the other tutor and solve the problem by a very different method.
The given figure consists of a triangle with a segment ("cevian") from one vertex to the opposite side.
Stewart's Theorem is often useful in solving problems where that is the given information.
Strategy:
(1) find x using Stewart's Theorem
(2) find the altitude of triangle ABC knowing that triangle ABD is isosceles
(3) find the area of triangle ABC using base (2x) and the altitude
(1) Stewart's Theorem
Stewart's Theorem with the given figure gives us this equation:
or
Clearly reject x=0, since it would be nonsense; so
(2) Since triangle ABD is isosceles, triangle AEB is a right triangle with one leg and hypotenuse 11, making the altitude
(3) The area of triangle ABC is then one-half base times height:
