For the given pentagon abcde the diagonal

ec

∥

ab

. i, g, f, h are midpoints of

bc

,

cd

,

de

,

ea

respectively. the length of

fg

is 50% more than the length of ab. find the area of the quadrilateral hfgi, if a△adb = 16sq. in.

  28 in²

Step-by-step explanation:

Without constraining the problem unduly, we can make the assumption that AB = 2 inches. Then the altitude from AB to D is h in ...

  Area ABD = (1/2)(AB)h

  16 in² = (1/2)(2 in)(h)

  16 in = h . . . . . . . . . . . divide by 1 in

__

The altitude D to AB is the sum of the heights from D to EC (h1) and from AB to EC (h2). That is ...

  16 = h1 + h2

We also know that the height from FG to EC is 1/2 the height from D to EC, hence (1/2)h1. Likewise, the height to midsegment HI from either EC or AB is half the height from EC to AB, hence (1/2)h2. This means the total height of the quadrilateral HFGI is (1/2)h1 + (1/2)h2 = (1/2)(h1 +h2) = 8.

__

We are given that FG is 50% longer than AB, so its length will be ...

  FG = AB×(1 + .5) = (2 in)(1.5) = 3 in

Since FG is the mid-segment of triangle CDE, base EC is twice its length, or ...

  EC = 2×FG = 2(3 in) = 6 in

__

Mid-segment HI is the average of the base lengths of trapezoid ABCE, so is ...

  HI = (EC +AB)/2 = (6 + 2)/2 = 4

__

Now, we know the height and base lengths of trapezoid HFGI, so we can find its area as ...

  A = (1/2)(b1 +b2)h = (1/2)(3 in + 4 in)(8 in) = 28 in²

The area of quadrilateral HFGI is 28 square inches.

You can make any assumption you like about the dimension of AB, and the rest of the dimensions scale accordingly. The result is still the same.

sorry can't

step-by-step explanation:

i googled though and something came up but it was in science langueage not common english sooo