Question

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 $\frac{5}{18}$ths of the area of triangle ABC, then the ratio AM/MC equals.

• A. $5$
• B. $6$
• C. $\frac{18}{5}$
• D. $\frac{15}{2}$

Answer 1

Answer: (A)

$5$

Let AM be x and MC be y.

MN is parallel to BC.

$\angle ANM=\angle ABC$

$\angle AMN=\angle ACB$

MP is parallel to NB.

$\angle ANM=\angle MPC$

Therefore,

$\Delta ANM\sim \Delta MPC\sim \Delta ABC$

By theorem, ratio of areas of two similar triangle is equal to the ratio of the squares of their corresponding sides.

$\frac{area\Delta ANM}{area\Delta ABC}=\frac{\left({AM}^2\right)}{\left({AC}^2\right)}=\frac{x^2}{{(x+y)}^2}---\left(1\right)$

$\frac{area\Delta MPC}{area\Delta ABC}=\frac{\left({MC}^2\right)}{\left({AC}^2\right)}=\frac{y^2}{{(x+y)}^2}---\left(2\right)$

$area\Delta ANC+area\Delta MPC=area\Delta ABC-area\square NMCB=area\Delta ABC-\frac{5}{18}area\Delta ABC=\frac{13}{18}area\Delta ABC$

Adding equation (1) and (2):

$\frac{13}{18}=\frac{x^2+y^2}{{(x+y)}^2}$

$5x^2-26xy+5y^2=0$

$5x^2-25xy-xy+5y^2=0$

$(5x-1)(x-5y)=0$

$\frac{x}{y}=5OR\frac{x}{y}=\frac{1}{5}$

But x>y

Therefore answer is 5.