In ΔABC, MN || BC, the area of quadrilateral MBCN=130 sq cm. If AN : NC = 4 : 5, then the area of ΔMAN is:
Answer & explanation
Correct answer: option 3
Given,
The area of quadrilateral MBCN = 130 \( { cm}^{2 } \).
AN : NC = 4 : 5
Hence, AC = 4 + 5 = 9
In \(\Delta \)ABC, If MN is parallel BC, then
\(\frac{area\;of\;AMN}{area\;of\;ABC}\) = \( {(\frac{AN}{AC}) }^{2 } \)
\(\frac{area\;of\;AMN}{area\;of\;ABC}\) = \( {(\frac{4}{9} )}^{2 } \) = \(\frac{16}{81}\)
Area of \(\Delta \)AMN = 16 unit and area of \(\Delta \)ABC = 81 unit
Now, Area of quadrilateral MBCN = area of \(\Delta \)ABC - area of \(\Delta \)AMN
= (81 - 16) unit = 130 \( { cm}^{2 } \)
= 65 unit = 130 \( { cm}^{2 } \)
= 1 unit = 2 \( { cm}^{2 } \)
Therefore, area of \(\Delta \)AMN = 16 x 2 = 32 \( { cm}^{2 } \).