Practicing Success
In $\triangle ABC, \angle = 90^\circ, AB = 16 cm$ and $AC = 12 cm$. D is the midpoint of AC and $DE \perp CB$ at E. What is the area (in cm$^2$) of $\triangle$CDE? |
6.25 8.64 7.68 5.76 |
8.64 |
In \(\Delta \)ABC \( {BC }^{2 } \) = \( {AB }^{2 } \) + \( {AC }^{2 } \) \( {BC }^{2 } \) = \( {16 }^{2 } \) + \( {12 }^{2 } \) = 256 + 144 = 400 BC = 20 Area of \(\Delta \)ABC = \(\frac{1}{2}\) x 12 x 16 = 96 \( {cm }^{2 } \) As we know, \(\Delta \)ABC is similar to \(\Delta \)EDC \(\frac{Area\;of\;EDC}{Area\;of\;ABC}\) = \( {(\frac{DC}{BC}) }^{2 } \) \(\frac{Area\;of\;EDC}{96}\) = \( {(\frac{6}{20}) }^{2 } \) = \(\frac{36}{400}\) Area of \(\Delta \)EDC = \(\frac{36}{400}\) x 96 = 8.64 \( {cm }^{2 } \) |