Question

find the area of the trapezium in which parallel sides are 25 cm and 10 cm and non parallel sides are 14 cm and 13 cm

Answer 1

Hello,
let's look at the figure.
AB=10 cm;DF=25cm;AD=14cm; BF=13cm
we calculate the CF:
CF=DF-CD=25-10=15 cm

For ΔBCF,
we calculate the perimeter of triangle
P=BC+BF+CF=14+13+15=42 cm
we calculate the  semiperimeter of the triangle:
p=P/2=42/2=21 cm
we use the formula of Heron:
At=√p×(p-BC)×(p-BF)×(p-CF);
=√21(21-14)(21-13)(21-15)
=√21×7×8×6=√7056= 84 cm²
we calculate the height of triangle:
BE=2At:CF=(2×84):15=168:15=11.2 cm

we calculate the area of Trapezium
A =[(AB+DF)×BE):2=[(10+25)×11.2]:2=(35×11.2):2=392:2=196 cm²

bye :-)

Answer 2

Answer:

in trapezium ABCE

ABCD is a parallelogram and bdc is a triangle an isosceles triangle whose side BD and BC is equal to 10 cm and DC is 12 centimetre height is 48 CM

area of trapezium is half into sums of parallel sides into height

half into 13 + 25 into 48

how into 48 into 24 is equal to 812