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?
Answer & explanation
Correct answer: option 2
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 \( {cm }^{2 } \)