Mathematics Question and Solution
Calculating area x.
Let 3a be the height of the ascribed triangle.
Therefore, considering the 6 square units inscribed triangle area.
½*(3a)*b = 6
3ab = 12
b = (4/a) units.
b is the base of the 6 square units triangle area.
c = y+b
c = (y+(4/a)) units.
Calculating y, the base of the area x, its height is ⅓(3a) units which is a.
3a ~ a
c ~ y
It implies;
3 ~ 1
(y+(4/a)) ~ y
Cross Multiply.
3y = y+(4/a)
2y = 4/a
y = (2/a) units.
Again, y is the base of area x with a it's height.
d = b+y
d = (4/a)+(2/a)
d = (6/a) units.
d is the base of the ascribed triangle.
Therefore, x the required area is;
½(3a)(6/a) = 2+6+x
½(18) = 8+x
9 = 8+x
x = 1 square unit.
Or
½(a)(2/a) = x
½(2) = x
x = 1 square unit.
