2022-01-05

Diagonal of a square and base of an equilateral triangle are on the same line

This question was uploaded on 05/01/22 on social media accounts.

Solution 1:

Blue area = A(ABC) - A(ADE)
\[A(ADE)=\frac12(AD)(ED)=\frac12(4\cdot\sin15)(4\cdot\cos15)\]
\[\Rightarrow A(ADE)=2\]
\[A(ABC)=\frac12(AB)(BC)=\frac12(2)(2\cdot\tan15)\]
\[\Rightarrow A(ABC)=4-2\sqrt3\]
\[\Rightarrow A_{Blue}=2-(4-2\sqrt3)=2\sqrt3-2\]
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