Question : 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 MAN is:
Option 1: 40 cm2
Option 2: 65 cm2
Option 3: 32 cm2
Option 4: 45 cm2
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Correct Answer: 32 cm 2
Solution :
Given that MN is parallel to BC, triangle ABC is similar to triangle MAN.
The ratio of their areas is the square of the ratio of their corresponding sides.
⇒ AN : NC = 4 : 5
⇒ AC = AN + NC = 4 + 5 = 9
Therefore, the ratio of the areas of triangle MAN to triangle ABC,
$(\frac{AN}{AC})^2 = \frac{4}{9}= \frac{16}{81}$
The area of quadrilateral MBCN is the area of triangle ABC minus the area of triangle MAN.
The area of triangle MAN is 16 units and the area of triangle ABC is 81 units, then the area of quadrilateral MBCN = 81 – 16 = 65 units.
Given that the area of quadrilateral MBCN is 130 cm
2
,
⇒ 65 units = 130 cm
2
⇒ 1 unit = 2 cm
2
Therefore, the area of triangle MAN = 16 units × 2 cm
2
/unit = 32 cm
2
Hence, the correct answer is 32 cm
2
.
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