Triangle ABC has side lengths AB = 6, BC = 10, CA = 8.
If D is the midpoint of BC and angle ADE = 90 degrees, what is the area of triangle ADE?
Since DB = CD then DB = 5
If we let A = (0,0) , then using similar triangles we have that the coordinates of D = (3 , 4)
Then AD = sqrt(3^2 + 4^2) = sqrt (25) = 5
And the area of triangle ADE = (1/2) AD ( ED) = (1/2) (5) ED = (5/2)ED ( 1)
Also the altitude of ADE = 3
So its area also = (1/2) 3(EA) = (3/2) EA (2)
Equating (1) and (2) we have that
5ED = 3EA
ED = (3/5) EA
Using the Pythagorean Theorem
EA^2 - ED^2 = AD^2
EA^2 - [( 3/5) EA ] ^2 = 5^2
EA^2 - (9/25)EA^2 = 25
(16/25) EA^2 =25
EA^2 = (25* 25) /16 = 625/16
EA = sqrt (625/16) = 25 / 4
So....the area of ADE = (1/2)(EA)(altitude) = (1/2) (25/4)(3) = 75/8 units ^2 = 9.375 units^2