Question

find the area of given trapezium.
please solve it in a copy

Answer 1

ABCD is the trapezium where AB is parallel to DC
we draw CE //AD from point C
parallelogram ADCE formed in which AD //CE and DC//AE
AE= 20cm CE =10cm
BE=AB-AE
BE=32-20=12cm
Now we will find area of triangle BCE
area of triangle BCE say A1
A1=
$\sqrt{s(s - a)(s - b)(s - c)}$
where A=BE=12cm,B =EC=10cm,C=BC=16cm
S=
$\frac{a + b + c}{2}$
S=
$\frac{ 12 + 10 + 16}{2}$
S=
$\frac{38}{2}$
S=19
A1=
$\sqrt{19(19 - 12)(19 - 10)(19 - 16)}$
$\sqrt{ 19 \times 7 \times 9 \times 3}$
$\sqrt{ 3591}$
A1=59.92
here we need to find the height of parallelogram AECD which is CM to calculate the area of trapezium

area of triangle BCE=
$\frac{1}{2 } \times base \times height$
59.92=1/2*12*h
59.92=6h
h=
$\frac{ 59.92}{6}$
h= 9.98=CM
area of trapezium=
$\frac{1}{2} \times (ab + cd) \times cm$
$\frac{1}{2} \times (32 + 20) \times 9.98$
$\frac{1}{2} \times 52 \times 9.98$
$26 \times 9.98$
259.48