What is the area of the blue triangle?

Relevant wiki: Length and Area - Composite Figures - Intermediate

The two triangles $ABF$ and $ADE$ each have a height and a base side in common with the parallelogram, so that their areas are both equal to half the parallogram area. If we divide the triangles into the areas $A_1$ , $A_2$ , $A_3$ , $A_4$ and $A_5$ as shown above, the result is $\begin{aligned} A_{ADE} &= A_1 + A_2 + A_3 = \frac{1}{2} A_\text{ABCD} \\ A_{ABF} &= A_3 + A_4 + A_5 = \frac{1}{2} A_\text{ABCD} \\ \Rightarrow \quad A_5 &=A_1 + A_2 - A_4 \\ &= 6 \end{aligned}$

W.K.T , If a triangle and a parallelogram are on the same base and between same parallels, then the area of the triangle is half of the area of the paralellogram. So, from the above statement - $ar(ABF) = \frac{1}{2} × ar(ABCD)\cdots (1)$ Similarly, $ar(AED) = \frac{1}{2} × ar(ABCD)\cdots (2)$

$\therefore$ , From equations (1) and (2) $ar(ABF) = ar(AED)\cdots (3)$ Now, Let the area of the region enclosed between the orange shaded regions be $'y'$ So, from equation (3) we get, $63 + y + x = 48 + y + 21$ $63+ x = 69$ $x = 69 - 63$ $\boxed{x = 6}$

Relevant wiki: Length and Area - Composite Figures

We must realize that $\Delta ADE$ and $\Delta ABF$ have the same area (it is half of the area of the parallelogram).

This is because the product of their bases and heights are the same due to their relation to the parallelogram.

Note that the quadrilateral is common among both figures.

Hence,

$48 + 21 = 63 + x$ $69 = 63 + x$ $x = \boxed{6}$

Let us assume that the white quadrilateral between the marked areas as y.(refer diagram if you have problem). Now let us shift point E to point C.(the point to observed is that the area of triangle AED doesn't change in doing so,as base and hieght are both constant). Similaly let us shift point F to point C. What we get to see is that a digonal is formed from A to the opposite point which you may call A,C or F. What is to be observed is that the area of the triangles formed are equal. (Diagonals bisect parallelogram).hence we proove that area of AED and AFB are equal. Now, let us come back to the original diagram. Remember, we assumed the white space as y. We get an equation. 48+y+21=x+y+63. Y cancels from both sides, and x=6, and that's our answer!!!!

I feel like a freak, I visualised it (I do a lot like this that way). I simply looked at X as being close to a third of 21. Then mentally rotated X into the corners of 21 and concluded it wasn't quite a third, but definitely more than a quarter. that doesn't leave many whole integers in that bracket that make sense. It had to be 6.

The triangles $ABF$ and $DEA$ have the same area, which is half of the parallelogram $ABCD$ 's area. Both triangles contain that entire empty space in the middle, implying their leftover area when said empty space is removed is the same. Therefore, 48+21=63+x, implying x=6.

The triangles are all somewhat similar. 63-48=15, so 21-15=6. Further more each triangle is made of the same components so their area are equal.

Well I just added 48 and 21 and then subtracted the result from 63

x+W+63 = 48+W+21 ==> x=48+21-63 ==> x=6

As an alternate way to show $\triangle ABF$ and $\triangle ADE$ are congruent is that by moving one point on a triangle parallel to its base it’s area doesn’t change. By moving point $E$ up to point $C$ we can clearly see that it’s area is one half the parallelogram, and similarly with $F$ and $C$ . Once we know the areas are congruent we know that,

$48 + 21 + A_{quad} = 63 + A_{quad} + x, x = \boxed{6}$

The answer has to come from the numbers 21, 48 and 63, so the only answer that made sense was 48+21-63! Bad maths, good logic