ABCD is a square of side 4 cm. If E is a point in the interior of the square such that ΔCEDis equilateral, then area of Δ ACE is

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We have the following diagram. Since ΔCED is equilateral, Therefore, EC=CD=DE=4 cm Now, $\angle$ ECD=$60^\circ$ Since AC is diagonal of sqr.ABCD Therefore, $\angle$ACD=$45^\circ$ Therefore we get, $\angle$ECA=$\angle$ECD-$\angle$ACD =$60^\circ$-$45^\circ$ =$15^\circ$ Now, in ΔACE, draw a perpendicular EM to the base AC. So in Δ EMC sin $15^\circ$=(EM)/(EC) =(EM)/4 Therefore, EM=sqrt2(sqrt3-1) Now in ΔAEC ar (ΔAEC=$\frac{1}{2}$) (AC)(EM) =4(sqrt3-1)cm^2 [Solution diagram: 64102_solution_1.png]

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