Mathematics Question and Solution
Let AB = a.
Let BE = b.
It implies;
2a+b = 392
b = 392-2a
BC = ½(b-70)
BC = ½(392-2a-70)
BC = ½(322-2a)
Therefore;
BC = DE = 161-a
c = ½(70)+161-a
c = 196-a
Where c is half BE.
d² = a²-(196-a)²
d² = a²-38416+392a-a²
d² = 392a-38416
d = √(392a-38416)
Where d is the height of the triangle ABE.
e² = √(392a-38416)²+35²
e² = 392a-38416+1225
e² = 392a-37191
e = √(392a-37191)
Where e is AC = AD.
It implies;
Calculating a (AB = AE)
AB+BC = AD+CD
a+161-a = √(392a-37191)+70
161-70 = √(392a-37191)
91 = √(392a-37191)
91²+37191 = 392a
392a = 45472
a = 116 units.
Therefore;
AB = AE = 116 units.
BC = 161-a, and a = AB = 116 units.
Therefore;
BC = 161-116
BC = DE = 45 units.
BE = BC+CD+DE
BE = 45+70+45
BE = 160 units.
BE is the base of the triangle ABE.
d = √(392a-38416)
Where d is the height of the triangle ABE.
And a = AB = 116 units.
Therefore;
d = √(392(116)-38416)
d = 84 units.
It implies;
Area Triangle ABE is;
½(160*84)
= 6720 square units.
