Question

The parallel side of a trapezium are 25 cm and 13 cm its non parallel sides are equal each being 10 cm find the area of the trapezium

Answer 1

area of trapezium =1/2h(a+b)

h=10cm

a=25cm

b=13cm

area=1/2×10(25+13)cm

=5(38)cm

=190cmsq

Answer 2

Answer:

${\huge{\underline{\underline{\sf{\blue{Solution :-}}}}}}$

${\huge{\underline{\underline{\sf{\blue{Given :-}}}}}}$

DC = 13 cm ,

AB = 25 cm ,

DA = CB = 10 cm

Through C , Draw CM parallel to DA meeting AB at M .

Now,

AM = DC = 13 cm,

MB = 25 cm - 13 cm = 12 cm

Draw CN _|_ MB

Since,

$\triangle \: CMB \: is \: an \: isosceles \: triangle \: , so \: CN \: bisects \: MB$

∴ MN = NB = 6 cm,

Now,

$from \: right \: \triangle \: CMN,$

$CN = \sqrt{CM ^{2} - MN ^{2} } = \sqrt{10 ^{2} - 6 ^{2} } = \sqrt{100 - 36 = \sqrt{64} } = 8 \: cm$

∴ Area of trapezium ABCD

$= \frac{1}{2} \times height \: \times (sum \: of \: the \: parallel \: sides \: )$

$= \frac{1}{2} \times 8 \times (13 + 25) \: cm ^{2}$

$= 4 \times 38 \: cm ^{2} = 152 \: cm \: ^{2}$