Consider the trapezium $ABCD$ such that the sides $AB$ and $DC$ are parallel, $AB=4$, $DC=10$, and the diagonals $AC$ and $BD$ are perpendicular to each other. Sides $DA$ and $CB$ are extended to meet each other at $E$. Angle $DEC$ is equal to $\frac{\pi}{4}$. Find the area of the trapezium.
Source: Pathfinder for Mathematics
I could solve using coordinate geometry but I wanted to know if there was an alternate solution using similarity of triangles since the formulas I used have not been taught till this exercise in the book. My solution:
Consider trapezium ABCD. Let $D(0,0), C(10,0), B(x+4,y)$ and $A(x,y)$. Let $m_1, m_2, m_3$ and $m_4$ be slopes of $BD, AC, AD$ and $BC$ respectively. Therefore,
$$m_1m_2 = -1$$ $$\tan\left(\frac{\pi}{4}\right)=\left|\frac{m_3-m_4}{1+m_3m_4}\right|$$
Solving will give
$y=\frac{20}{3}$ and we do not have to calculate $x$ since it will automatically cancel out when calculating area. Applying basic formula $$\Delta = [ADC]+[ABC]=\frac12\cdot (AB+DC) \cdot h= \frac12\cdot ((x+4-x)+(10)) \cdot y=\frac{140}{3}$$
Best Answer
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Define point $X$ as the intersection of diagonals $AC$ and $BD$.
Let $AX=x$ and $BX=y$.
$$\triangle AXB\sim DXC \implies CX=\frac{5y}{2} \;\;\text{and}\;\; DX=\frac{5x} {2}.$$ Using the Pythagorean Theorem ($AC\perp BD$), $$x^2+y^2=16\implies x^4+y^4=256-2x^2y^2\tag{1}$$ $$AD=\frac{\sqrt{4x^2+25y^2}}2 \;\;\text{and}\;\;BC=\frac{\sqrt{4y^2+25x^2}}2$$ Since $AB\parallel CD$, $$\triangle AEB\sim \triangle DEC \implies AE=\frac{\sqrt{4x^2+25y^2}}3 \;\;\text{and}\;\;BE=\frac{\sqrt{4y^2+25x^2}}3$$ Using the cosine rule in $\triangle AEB$, $$\begin{align*} 16&=\frac{4x^2+25y^2}{9}+\frac{4y^2+25x^2}{9}-2\cdot\frac{\sqrt{4x^2+25y^2}}3 \cdot\frac{\sqrt{4y^2+25x^2}}3\cdot \cos 45^{\circ} \\ 16&=\frac{29}{9}(x^2+y^2)-\frac{\sqrt{100(x^4+y^4)+641x^2y^2}}{9}\cdot \sqrt{2} \end{align*}$$ Simplifying the above equation using $(1)$, we have, $$x^2y^2=\frac{160}{21}$$ Area of trapezium is , $$[ABCD]=\frac{1}{2}\cdot \frac{7x}{2}\cdot \frac{7y}{2}=\frac{49}{8}x^2y^2=\frac{140}{3}.$$