Question

the parallel sides of trapezium are 24cm and 52cm and the other sides are 26cm and 30cm. find the height of the trapezium

Answer 1

let ABCD be trapezium
where CE||AD and CF (perpendicular on) AB
now EB=(AB-AE)=(AB-DC)
=(52-24)cm = 28cm
CE=AD=26cm and BC=30cm
now in ∆CEB, we using formula

AREA OF TRIANGLE=√{S(S-a)(S-b)(S-c)}
where S=1/2(sum of all sides of triangle)

S=1/2(28+26+30)cm= 42cm
(S-a)=(42-28)cm=14cm
(S-b)=(42-26)cm=16cm
(S-c)=(42-30)cm=12cm

put it in the formula we get
√{42×14×16×12} = 336cm^2
we get the area of ∆CEB=336cm^2

also, using formula

AREA of ∆CEB=1/2×EB×CF
336cm^2=(1/2×26×CF)cm^2
336cm^2=(13×CF)cm^2
there fore
CF= 336/13cm
it is the height of the trapezium