Dragan

To solve this problem, we can use a Triangle similarity theorem or trigonometry. 

4m/12m = 2m/x      x = 6

Nick's shadow is 6m long

Lamppost's shadow is 18m long   

area of ΔPEF + area of ΔPGH = area of ΔPFG

Area of trapezoid EFGH = (8 + 21) * 2 = 58 u²  

The sum of pentagon's interior angles is 540º

540º / 5 = 108º

∠ x = ( 180º - 108º ) / 2  

Why have you answered the question if the answer is known?

btw, your answer is correct!!!

In how many ways can 6 girls and 2 boys sit in a row if the 2 boys insist on sitting next to each other?

1/     GGGGGGBB

2/     GGGGGBBG

3/     GGGGBBGG

4/     GGGBBGGG

5/     GGBBGGGG

6/     GBBGGGGG

7/     BBGGGGGG

A = 6 * [sqrt( 15² - 12² )] + 2 * 12  

( 7x - 9 ) + ( 5x - 11 ) = ( 9x + 1 )         solve for x

7w + 3 = 104 / 2  

Quadrilateral PQRS is a trapezoid with bases PQ and RS. The median MN meets the diagonals PQ and RS at X and Y, respectively. If SR=20 and XY=7, find PQ.

Let's draw a vertical line that intersects point W on SR and point Z on PQ. Line WZ passes through point X.

Let the line segment WZ be 14 units.   (btw, that can be any positive value)

Find an angle XRW                ∠XRW = arctan(XW / XR) = 27.40757544º

PQ = 2[XZ / tan(∠XRW)] + XY = 34 units

After number 6,  #9 and  #2 leave.