Question

ABCD is a trapezium in which AB//DC,AB-78cm,CD-52cm,AD-28cm and BC-30cm. find the area of the trapezium

Answer 1

to do this problem we draw perpendiculars from C and D to B and A. we get a rectangle and two triangles.join the two triangles .Now find the area of the new triangle using herons formula.first we find s.
s=(a+b+c)/2=42
then area=
$\sqrt{s(s - a)(s - b)(s - c)}$
area=336 cm square
next to find height we use the formula area=1/2*base*height
336=1/2*26*height
height=336/13
now we find area of rectangle=52*336/13=1344
hence the total area of trapezium=1344+336=1680cm sq

Answer 2

From the figure.
In triangle CBM - by Herons formula

$s = a + b + c \div 2$
$28 + 30 + 26 \div 2$
$84 \div 2 = 42$
Now area of triangle by Herons formula
$\sqrt{42(42 - 28)(42 - 30)(42 - 26)}$
$\sqrt{42 \times 1412 \times 16}$
$\sqrt{2 \times 3 \times 7 \times 2 \times 7 \times 4 \times 3 \times 4 \times 4}$

$4 \times 2 \times 2 \times 7 \times 3$
$8 \div 42 = 336cm {}^{2}$
Area of parallelogram = Base multiplied by height
$parallelogram = b \times h$
We don't have height
To find height
$\: area \: \: = 1 \div 2 \times base \: \times height \:$
$336 =1 \div 2 \times 13 \times h \:$
$336 = 13 \times h$
$336 \div 13 = h$
$25.84 = h$
Now we can find the area of parallelogram
$parallelogram = base \times height \:$
Parallelogram
$parallelogram \: = 52 \times 25.84$
$= 1343.68cm {}^{2}$

Add bath the areas
$1343.68 + 336 = 1679.68$
Area of trapezium =half multiplied by addinsion of two parallel sides multiplied by height
$area \: of \: trapezium \: = 1 \div 2 \times sum \: of \: two \: parallel \: sides \: \times height \:$
Trapezium
$trapezium \: = 1 \div 2 \times (78 + 52) \times 25.84$
$= 1679.68$