Practicing Success
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: |
40 cm2 65 cm2 32 cm2 45 cm2 |
32 cm2 |
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\( { cm}^{2 } \). |