If two triangles are similar with sides in ratio kkk, their areas are in ratio k2k^2k2. Formally, ar(△ABC)ar(△DEF)=(ABDE)2=(BCEF)2=(CAFD)2\dfrac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle DEF)} = \left(\dfrac{AB}{DE}\right)^2 = \left(\dfrac{BC}{EF}\right)^2 = \left(\dfrac{CA}{FD}\right)^2ar(△DEF)ar(△ABC)=(DEAB)2=(EFBC)2=(FDCA)2.