### 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)

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### Solution in Detail

Observe the triangles AMN and ABC

[1] Given that MN ∥ BC, so $\displaystyle \angle M = \angle B$

[2] Given that MN ∥ BC, so $\displaystyle \angle N = \angle C$

$\displaystyle \therefore $ by AA rule, $\displaystyle \Delta ABC \sim \Delta AMN$

Ratio of similar sides $\displaystyle \frac{4k + 5k}{4k} = \frac 94$

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$\displaystyle \therefore \frac{\text{ar(ABC)}}{\text{ar(AMN)}} = \bigg(\frac 94\bigg)^2$

$\displaystyle \therefore \text{ar(ABC)} = \frac{81}{16} \times \text{ar(AMN)}$

But triangle ABC is sum of triangle AMN and the quadrilateral MNCB whose area given as 130

For simplicity let, $\displaystyle x = \text{ar(AMN)}$

$\displaystyle \therefore 130 + x = \frac{81}{16} \times x$

Solving, $\displaystyle x = \text{ar(AMN)} = 32\:\underline{Ans}$

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