In △ UQV |UV|\(^2\) = 6\(^2\) + 4\(^2\) (Pythagora's theorem) |UV| = \(\sqrt{52}\) = 7.21cm Also, in △ WVR |WV|\(^2\) = 2\(^2\) + 2\(^2\) = 4 + 4 = 8 (same reason as above) |WV| = \(\sqrt{8}\) = 2.83 Therefore, the area of △ UVW = \(\frac{1}{2}\) x 2.83 x 7.21 = 10.21 ≈ 10cm\(^2\)