If in triangle ABC, MN is parallel to BC, and M and N are points on AB and AC respectively. The area of quadrilateral MBCN = 130 cm2 . If AN : NC = 4 : 5., then the area of Triangle is MAN is:
Answer & explanation
Correct answer: option 3
Given,
the area of quadrilateral MBCN = 130
AN : NC = 4 : 5
Hence, AC = 4 + 5 = 9
In triangle ABC, if MN is parallel to BC, then
Area of \(\Delta \)AMN/Area of \(\Delta \)ABC = \( {(AN/AC) }^{2 } \)
Area of \(\Delta \)AMN/Area of \(\Delta \)ABC = \( {(4/9) }^{2 } \) = \(\frac{16}{81}\)
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 = 130
= 65 unit = 130
= 1 unit = 2 \( { cm}^{2 } \)
Therefore, Area of \(\Delta \)AMN = 16 x 2 = 32\( { cm}^{2 } \).