KVPY SA Exam 19-Nov-2017 Question Paper

Question : 4 of 80

Marks: +1, -0

Let ABC be a triangle and M be a point on side AC closer to vertex C than A. Let N be a point on side AB such that MN is parallel to BC and let P be a point on side BC such that MP is parallel to AB. If the area of the quadrilateral BNMP is equal to

5/18

th of the area of triangle ABC, then the ratio AM/MC equals.

5
6
$18/5$
15/2

Validate

Solution:

To find

{AM}/{MC}=α/{a-α}

.....(i)
∴ ΔBAC ~ ΔPMC

\PM/\CM=\BA/\AC

k/{a-α}=\BA/a

\BA=\ak/{a-α}

∴ Area of ΔABC

=1/2\AC.\AB

=1/2a.\ak/{\a-α}

...(ii)
Area of quad. BNMP = Ar trap. AMPB – ar ΔAMN ...(iii)
Also ΔNAM ~ ΔBAC

\AN/\AM=\AB/\AC

\AN/α=\ak/{(a-α)a}

\AN={\kα}/{a-α}

Put in (iii)
Area quad. BNMP

=1/2(\AB+\PM)×\AM

-1/2\AM×\AN

=1/2{|{ak}/{a-α}+k|}×

α-1/2α×{kα}/{a-α}

=1/2[{2akα-kα^2-kα^2}/{a-α}]

={akα-kα^2}/{a-α}=kα

Given

αk =5/18 . 1/2 . {a^2k}/{a-α}

36/5=a^2/{α(a-α)}

36aα - 36α^2 = 5a^2

5a^2 -36aα + 36α^2 = 0

5a^2 -30aα- 6aα + 36α^2 = 0

5a (a - 6α) - 6α(a-6α) = 0

5a = 6α

a = 6α

Not true as M is near to C than A

∴α/{a-α}=1/{a/α-1}

=1/{6/5-1}=5

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