Triangle ABC has AB = BC = 5 and AC = 6. Let E be the foot of the altitude from B to \(\overline{AC}\) and let D be the foot of the altitude from BC to . Compute the area of triangle DEC.
+23245 Since triangle(ACB) is a right triangle with base AC, point E will be the midpoint of the base.
Look at right triangle(AEB). The hypotenuse is AB = 5; one side is AE = 3, so the other side, BE = 4 (Pythagorean Theorem).
Therefore, the area of triangle(ABC) = ½·AC·BE = ½·6·4 = 12 and the area of triangle(BEC) = 6.
Now I want to find the length of AD.
AD is the height of triangle(ABC) if BC is the base.
Area of triangle(ABC) = 12 = ½·BC·AD ---> 12 = ½·5·AD ---> 12 = 2.5·AD --->
AD = 4.8
I will use the length of AD to find the length of BD.
Triangle(ABD) is a right triangle with hypotenuse AB = 5 and side AD = 4.8.
Therefore, using the Pythagorean Theorem, BD = 1.4.
Since BC = 5 and BD = 4, CD = 3.6.
If you look at triangle(BCE) with base BC, triangle(DCE) with base DC, and tiangle(BDE) with base BD, they all have the same height. (It isn't drawn, but you can draw it from point E perpendicular to BC.
Area of triangle(BCE) = area of triangle(DCE) + area of triangle(BDE)
Since the length of DC = 3.6 and the length of BD = 1.4 and length of CD = 5, the length of CD = 3.6/5 the length of BC.
So, the area of triangle(DCE) = 3.6/5 · area of triangle(BCE)
---> area of triangle(DCE) = 3.6/5 · 6 = 4.32
