Two Wrongs Make A Right?

Let $h_1$ be the height from $A$ . Nihar's equation gives

$\frac{\left(AB\right)h_1}{2}=14$

Let $h_2$ be the height from $C$ . Andrew's equation gives

$\frac{\left(BC\right)h_2}{2}=56$

Let $K$ be the area of $\triangle ABC$ .

Both equation contain valuable information but the information is not directly useful to us. If we knew $AB$ and $h_2$ or $BC$ and $h_1$ , we can find $K$ . Only way for us to get any information is by combining the two equations.

$\frac{\left(AB\right)h_1}{2} \times \frac{\left(BC\right)h_2}{2} = 14 \times 56$

On rearranging,

$\frac{\left(AB\right)h_2}{2} \times \frac{\left(BC\right)h_1}{2} =784\\ K^2 =784\\ K=28$

Let $a=AB$ and $b$ be the altitude from $A$ . Let $c=BC$ and $d$ be the altitude from $C$ . We know that $ab=14*2=28$ and $cd=56*2=112$ .

Also, $ad$ and $bc$ are both the area of a triangle since the area is any base times the altitude to that base. Thus, $ad=bc$ . If we multiply $ab$ and $cd$ , we get $abcd = 28*112=3136$ . We can rearrange the expression into $adbc=3136$ because multiplication is commutative. Since $ad=bc$ , $(ad)^2=3136$ so $ad=56$ . Since $ad$ is the base times the height, and you have to divide the product by $2$ to get the area, $ad*0.5=56*0.5=28$ which is the final answer.

1. $\dfrac{1}{2}cx=14$ $\implies$ $cx=28$ $\implies$ $c=\dfrac{28}{x}$

2. $\dfrac{1}{2}ay=56$ $\implies$ $ay=112$ $\implies$ $a=\dfrac{112}{y}$

The true area must be $A=\dfrac{1}{2}cy$ or $A=\dfrac{1}{2}ax$ .

Since the two areas are equal,

$A=A$

$\dfrac{1}{2}cy=\dfrac{1}{2}ax$

$cy=ax$

However, $c=\dfrac{28}{x}$ and $a=\dfrac{112}{y}$

We substitute

$\dfrac{28}{x}(y)=\dfrac{112}{y}(x)$

$28y^2=112x^2$

$y^2=4x^2$

$y=2x$

We substitute $y=2x$ in $A=\dfrac{1}{2}cy$ , we have

$A=\dfrac{1}{2}(c)(2x)=cx=$ $\color{#3D99F6}\large\boxed{28}$ $\boxed{\text{answer}}$