A) 96 sq cm
B) 84 sq cm
C) 72 sq cm
D) 60 sq cm
Correct Answer: C
Solution :
Produce the median AM to D such that \[GM=MD\]and join to B and C. |
Now GBDC is a parallelogram. Note that the sides of the |
\[\Delta \,GDC\]are 6, 8, 10 \[\Rightarrow \]\[\angle GDC=90{}^\circ \] |
\[\left. \begin{matrix} \begin{matrix} Area\,\,of\,\,\Delta \,ADC=\frac{12.8}{2}=48 \\ Area\,\,of\,\,\Delta \,MDC=\frac{3.8}{2}=12 \\ \end{matrix} \\ \end{matrix} \right]\] |
\[\Rightarrow \] Area of \[\Delta \,AMC=36\] |
\[\Rightarrow \] Area of \[\Delta \,ABC=72\,\,c{{m}^{2}}\] |
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