What is the area of $\triangle ABC$ if area of $ABEF$ and $BCFD$ is 15 and 23 in the following diagram?

Question

$ABEF$ and $BCFD$ parallelograms have areas respectively $15$ and $23$. Find the area of $\triangle ABC$.

Source: Bangladesh Math Olympiad 2014 Junior Category

Is only areas of two parallelograms enough to find the area of the triangle? I can solve the math. Can anyone help me with a hint, please?

Answer 1

Hint:

Let's say $DB=FC=a$, $BE=AF=b$ and $h$ being the height of both parallelograms. Then we can find the area of the parallelograms and the triangle in terms of $a,b,h$:

$$\begin{cases}ah = 23 \\ bh = 15 \\ A=\dfrac{1}{2}(a+b)h\end{cases}$$

where $A$ is the area of the triangle.

Answer 2

Connect AD and CE. Area of BNFM = Area of △AMD + Area of △ENC. so the area of triangle would simply be $\frac{15+23}{2} = 19$

Answer 3

Observe $AC=AF+FC$ and both parallelograms and the triangle have the same height $h$. Therefore, $$ \textrm{Area of triangle }=\frac{1}{2}(AF+FC)h=\frac{1}{2}(\textrm{Area first parall + Area second parall})=19. $$

Answer 4

Observe that BDM and CFN are congruent, likewise AFM and BEN are congruent. We know that $A(BCFD)=A(ABEF)+8$, therefore $A(CFN)=A(AFM)+4$. Let $x=A(AFM)$, and from the original statement $A(BMFN)=15-2x$. Go from there.

Answer 5

$$S_{\Delta ABC}=\frac{AC\cdot h_b}{2}=\frac{(AF+FC)h_b}{2}=\frac{AF\cdot h_b+FC\cdot h_b}{2}=\frac{15+23}{2}=19.$$

Answer 6

The indicated triangles are congruent (also stated by Rhys Hughes):

$\hspace{1cm}$

Hence: $$\begin{cases} 2x+z=15\\ 2y+z=23 \end{cases} \Rightarrow 2(x+y+z)=38 \Rightarrow x+y+z=19.$$