Question 4

$$P$$ is a point inside $$ABCD$$ such that $$PA=2$$, $$PB=4$$, $$PC=5$$ and $$PD=6$$. The maximum area of quadrilateral $$ABCD$$ is

Solution

Split the quadrilateral into triangles $$PAB,PBC,PCD$$ and $$PDA$$. Their total area is at most $$\frac{1}{2}(2\mathbin{\times}4+4\mathbin{\times}5+5\mathbin{\times}6+6\mathbin{\times}2)=35$$ because each included sine is at most $$1$$. This value is attained when the four angles at $$P$$ are right angles, so the maximum area is $$35$$.

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$$P$$ is a point inside $$ABCD$$ such that $$PA=2$$, $$PB=4$$, $$PC=5$$ and $$PD=6$$. The maximum area of quadrilateral $$ABCD$$ is

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