Question

The parallel sides of a Trapezium are 20 cm and 10 cm and its non parallel sides are equal each being 13 cm find the area of the trapezium

Answer 1

Area of trapezium = 180cm^2

Answer 2

Given ....

ABCD is a Trapizium.
where....
AB = 20cm , DC =AE = 10cm , DA =CB =13cm.

CONSTRUCTION...= DRAW CB ll AD, meeting

AB at E.

AND...

DRAW CF perpendicular on AB.

NOW...

EB = (AB - AE) = (AB -DC)

EB = (20 - 10cm)
= 10cm.

Hence F is the mid point therefore ...

EF = 1/2 x EB
= 1/2 X 10cm
=5 cm.

NOW in Triangle CEF

CE = 13cm
EF = 5cm

THEREFORE applying pythagorous theorm ..

${H}^{2} = {P}^{2} + {B}^{2} \\ \\ {CE}^{2} = {CF}^{2} + {EF}^{2} \\ \\ {13}^{2} = {CF}^{2} + {5}^{2} \\ \\ 169 = {CF}^{2} + 25 \\ \\ 144 = {CF}^{2} \\ \\ \sqrt{144} = CF \\ \\ CF = 12cm.$

Hence the Height of the Trapizium is 12cm

Finally...

AREA OF TRAPIZIUM ...

1/2 X ( sum of parallel sides ) x Height.

$\frac{1}{2} \times (20 + 10)cm \times 12cm \\ \\ = \frac{1}{2} \times 30cm \times 12cm \\ \\ = 15cm \times 12cm \\ \\ = 180 {cm}^{2} \\ \\ \\ Therefore \: area \: of \: the \: Trapizium \: is \: 180 {cm}^{2} \\ \\ \\ \\ HIT \: A \: THNX \: IF \: YOU \: GAIN \: HELP$