Question

find the height of a trapezium in which parallel sides are 25cm 77cm and non - parallel sides and 26cm and 60cm given the area of the trapezium as 1644cm

Answer 1

Hello,
let's look at the figure.
AB=60 cm;DF=77cm;AD=25cm; BF=26cmwe calculate the CF:
CF=DF-CD=77-60=17 cm

For ΔBCF,
we calculate the perimeter of triangle
P=BC+BF+CF=25+26+17=68 cm
we calculate the  semiperimeter of the triangle:
p=P/2=68/2=34 cm
we use the formula of Heron:
At=√p×(p-BC)×(p-BF)×(p-CF);
=√34(34-25)(34-26)(34-17)
=√34×9×8×17=√41616= 204 cm²
we calculate the height of triangle:
BE=2At:CF=(2×204):17=408:17=24 cm

we calculate the area of Trapezium
A =[(AB+DF)×BE):2=[(60+77)×24]=(137×24):2=3288:2=1644 cm²

bye :-)

Answer 2

Hey friends here is ur answer

Draw a line from vertex C and parallel

to AD, which meet AB at E.

∵ AE II DC and AD II EC

∴ AECD is a parallelogram

EC = AD = 25 cm

AE = DC = 60 cm

EB = AB - AE

= 77 - 60

= 17 cm

Area △BEC

s = 25 + 26 + 17 / 2

= 34 cm

Area △BEC = √(34 * 9 * 8 * 17)

= 17 * 3 * 2 * 2

= 204 cm²

Let h cm is the length of perpendicular to EB.

Area △BEC = 1/2 * 17 * h

= 204 cm²

h = 204 * 2 / 17

= 24 cm Ans.

Area trap. ABCD = 1/2 * (60 + 77) * 24

= 137 * 12

= 1644 cm²

Hence Proved.

PLEASE MARK IT AS BRAINLIEST.