Question 11
SSC-CGL 2020 Mar 3 Shift 1
In a triangle ABC, M and N are points on AB and AC respectively, such that MN is parallel to BC and AN : NC is 4 : 5. If the area of the quadrilateral MNCB is 130 sq. units, then what is the area of triangle MAN?
Solution in Brief
Triangles ABC and AMN are similar with ratio of sides as 9/4. So area ABC = 81/16 x area AMN. But area ABC = area AMN + 130. So area AMN + 130 = 81/16 x area AMN. Solving, area AMN = 32 sq. units ans!
Video Explanation (slow explanation)
Please watch this slow and easy explanation:
Solution in Detail
Observe the triangles AMN and ABC
[1] Given that MN ∥ BC, so
[2] Given that MN ∥ BC, so
by AA rule,
Ratio of similar sides
REMEMBER: The ratio of the areas of two similar triangles is equal to the square of the ratio of any pair of their corresponding sides
But triangle ABC is sum of triangle AMN and the quadrilateral MNCB whose area given as 130
For simplicity let,
Solving,
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