Question

Find the area of a trapezium whose parallel sides are 25cm,13cm and other sides are 15cm each

Answer 1

as we have trapezium with parallel side AB and DC in which AB = 13 cm and BC=25 CM
So firstly draw a line to BC which is parallel to AD
as we get ABPD as a llgm and a BPC as a isoscales triangle
with side BC=BP=15cm and PC=12cm
as draw bisector of angle B which will perpendicular to PC
then by the Pythagoras theorem
.
${h}^{2} = {b \: }^{2} \: + {p}^{2} \\ {15}^{2} = {6}^{2} + {p}^{2} \\ 225 = 36 + {p}^{2} \\ p \: = \sqrt{225 - 36} \\ p = \sqrt{189} {cm}^{2}$


so we now know the height of the triangle an as we know that the height of triangle is equal to height of the trapezium so,

area of trapezium equal to half sum of parallel side multiplied by height
then ,
$= 1 \div 2(13 + 25) \times \sqrt{189} \\ = 1 \div 2 \times 38 \times \sqrt{189} \\ = 19 \sqrt{189} {cm}^{2}$
so here is required answer........