Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Calculating Area Orange.
yB = 6
Area OABC = 161/5 square units.
Let xB be a.
Calculating g, gradient of length AB.
(y2-y1)/(x2-x1) = g
g = (6-4)/(a-0)
g = 2/a
Therefore, h, gradient at the perpendicular is;
h = -1/g
h = -1/(2/a)
h = -a/2
Let xC be b.
It implies;
(6-0)/(a-b) = h
Calculating b.
6/(a-b) = -a/2
12 = -a²+ab
b = (a²+12)/a units.
Again, b is xC = OC.
c = xC-xB
c = ((a²+12)/a)-a
c = 12/a units.
Calculating a, observing trapezoid area formula and triangle area formula.
½(4+6)*a + ½(6*(12/a)) = 161/5
5a+(36/a) = 161/5
(5a²+36)/a = 161/5
25a²+180 = 161a
25a²-161a+180 = 0
It implies;
a ≠ 1.44 units.
a = 5 units.
a is xB.
Recall.
b = (a²+12)/a
And a = 5 units.
b = (5²+12)/5
b = 37/5 units.
b = 7.4 units.
b is OC = xC.
d = (37/5)-5
d = 7.4-5
d = 2.4 units.
e² = 2.4²+6²
e = √(41.76)
e = √(1044/25)
e = ⅕(6√(29)) units.
e is BC.
f = ½(e)
f = ⅗√(29) units.
f is half BC.
j² = 5²+2²
j² = 25+4
j = √(29) units.
j is AB.
Therefore, Area Orange is;
½*f*j
= ½*√(29)*⅗√(29)
= (3*29)/10
= 87/10 square units.
= 8.7 square units.
