SOLUTION: Find valu of 'x' in a trapezium where parallel sides are 50m and 26m respectively. Total area is 1254 sqm . Area 1 of upper portion is 618sqm and area2 of lower portion is 636sqm.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Find valu of 'x' in a trapezium where parallel sides are 50m and 26m respectively. Total area is 1254 sqm . Area 1 of upper portion is 618sqm and area2 of lower portion is 636sqm.       Log On


   

Question 1202022: Find valu of 'x' in a trapezium where parallel sides are 50m and 26m respectively. Total area is 1254 sqm . Area 1 of upper portion is 618sqm and area2 of lower portion is 636sqm. Height of whole trapezium is 33m. x is height of lower portion of trapezium. Trapezium is divided by a segment in two unequal portions.
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Here is a crude sketch based on the given information:

                   26
        ************************
       *                         *
      *                            *
     *   A = 618                     *
    *                                  *
   *************************************** -----------
  *                                        *
 *      A = 636                              * height x
************************************************-----
                    50

The area of the trapezoid is 1254, which is (altitude) times (average of bases):

1254 = h((50+26)/2)
1254 = 38h
h = 1254/38 = 33

Since the height of the lower portion is x, the height of the upper portion is 33-x.

Since the formula for the area of a trapezoid is (altitude) times (average of bases), for convenience let 2y be the length of the segment parallel to the two bases. We then have this sketch:

                   26
        ************************----------------------
       *                         *
      *                            *
     *   A = 618                     *    height 33-x
    *                 2y               *
   *************************************** -----------
  *                                        *
 *      A = 636                              * height x
************************************************-----
                    50

Then for the areas of the two trapezoids we have

%2833-x%29%28%282y%2B26%29%2F2%29=618 [1]
x%28%282y%2B50%29%2F2%29=636 [2]

Simplify [1]

%2833-x%29%28y%2B13%29=618
33y%2B429-xy-13x=618
-xy-13x%2B33y=189 [3]

Simplify [2]

x%28y%2B25%29=636
xy%2B25x=636 [4]

Add [3] and [4]

12x%2B33y=825
4x%2B11y=275 [5]

Solve [5] for x and substitute in [4]

I leave the rest to you. The numbers work out to give a very ugly irrational value for x, which leads me to believe that the given numbers are not right....