Problem 10 · Two triangles overlapping along the hypotenuse
Union of areas: $2 \cdot 14$ minus the overlap triangle.
Integer answer, at most 4 digitsThe legs of two equal right triangles measure 4 cm and 7 cm. The figure shows the two triangles overlapping so that their hypotenuses coincide, with $AD = BC = 4$ cm and $AC = BD = 7$ cm. Find the total area of the figure. (If the answer is a fraction, write the numerator and denominator of the irreducible fraction consecutively. For example, if the answer is $2/3$, write 23.)
Reasoned solution
Key idea: area of the union $=$ sum of the two areas minus the overlap, which is triangle $ABX$ with $X$ the crossing point of $AC$ and $BD$.
Each triangle has area $\dfrac{4 \cdot 7}{2} = 14$, and the common hypotenuse is $AB = \sqrt{16+49} = \sqrt{65}$.
Put $A = (0,0)$ and $B = (\sqrt{65}, 0)$. By the symmetry of the figure (the two triangles are mirror images about the perpendicular bisector of $AB$), the point $X$ lies on $x = \tfrac{\sqrt{65}}{2}$.
Vertex $C$ has coordinates $\left(\tfrac{49}{\sqrt{65}},\ \tfrac{28}{\sqrt{65}}\right)$ (projection $AC^{2}/AB$ and height $\tfrac{4\cdot 7}{\sqrt{65}}$), so line $AC$ has slope $\dfrac{28}{49} = \dfrac{4}{7}$. At $x = \tfrac{\sqrt{65}}{2}$: