In the adjoining figure, $$(\triangle ABC)$$ is an isosceles right-angled triangle. $$(BE)$$ is perpendicular to $$(AD)$$. If $$(BE) = 1\text{ cm}$$, then the area in $$\text{cm}^2$$ of quadrilateral $$(ABCD)$$ is
Solution
Place $$A = (0,0)$$, let $$AD$$ be horizontal and write $$B = (u,1)$$ because $$BE = 1$$. Since $$\triangle ABC$$ is right isosceles at $$B$$, a rotation of $$\overrightarrow{BA}$$ through $$90^\circ$$ gives $$C = (u+1,1-u)$$, and hence $$D = (u+1,0)$$. The shoelace formula gives the area as $$\frac{1}{2}\left|u(1-u)-(u+1)-(1-u)(u+1)\right| = 1$$.
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