Question

the parallel sides of a trapezium are 24 CM and 52 CM and other sides are 26 CM and 30 CM find the area 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 = 28cmCE=AD=26cm and BC=30cmnow in ∆CEB, we using formulaAREA 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=12cmput it in the formula we get √{42×14×16×12} = 336cm^2we get the area of ∆CEB=336cm^2also, using formula AREA of ∆CEB=1/2×EB×CF336cm^2=(1/2×26×CF)cm^2336cm^2=(13×CF)cm^2there foreCF= 336/13cmit is the height of the trapezium plzzzz mark me as brainlist i did so much hard work

Answer 2

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