Geometry problem #60059

MN:NB=1:2 => area triangle MNC=21

No need to draw lines. ∆MNC ≈ ∆NAB, and the areas of similar triangles go as the square of the corresponding lengths.

• Length AB = 2MC, so the area of ∆NAB = 4 times the area of ∆MNC. • ∆ABC is ½ of the parallelogram and ∆BMC = ¼ of the parallelogram.

Let “x” be the unknown area:

∆ABC = 4x + 42 = 2∆BMC = 2(x + 42) = 2x + 84 x = 21. QED

$[AB]||[DC]$ hence $MNC$ and $ANB$ is similar with ratio of $\frac{1}{2}$

So $2|MN|=|NB|$ .

Since heights of $MN$ and $NB$ are equal (from point $B$ ): $A(MNC)=2A(BNC)=21$

? = x, x + 42 = 1/2(4x + 42), x + 42 = 2x + 21, x = 21 ,