Question

find the area of a Trapezium whose parallel sides are 25 cm and 10 cm respectively and non parallel sides are 14 cm and 13 cm respectively

Answer 1

Let ABCD be the trapezoid
with AB || CD.
AB= 10. BC= 14 CD = 25 DA = 13
Draw the altitude from A which intersects line DC at X , so the altitude of the trapezoid is h.
Draw the line parallel to BC through A which intersects at Y.
In DAY , DA= 13 AY= 14 YD = 15 (the latter is the difference between the bases becauseis a parallelogram).
Since AY^2 < DA^2 + YD^2, the angle at D is acute, therefore lies between D and Y.
We then define DX = a, and with X between D and Y we then have DY =15-a. Apply the Pythagorean Theorem to each of the right triangles ADX.....
AD^2 = AX^2+DX^2
13^2 = a^2 + h^2
169 = a^2 + h^2 -------------------> 1 equation
now in triangle AYX....
AY^2 = AX^2 + XY^2
14^2 = h^2 + (15-a)^2
196 = h^2 + 225 - 30a + a^2
-29 = h^2 -30a + a^2 ----------------------> 2 equation

by solving both equation we get
a = 6.6
and h= 11.2

then....we can easily find area of trapezium
= {(AB + DC)/2} × h
= (10+25)/2 × 11.2
= 17.5 × 11.2
= 196 cm^2

HOPE THIS HELP YOU :)