please help me to solve this problem
Answers
AE=11-5=6
DE=UNDER ROOT 100-36=64=8
AREA OF TRAPEZIUM=1/2×(5+11)×8
=1/2×16×8=8×8=74m^2
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It is given that the length of AB is 11cm , length of DC( which is parallel to AB or EB ) is 5 cm, also the length of AD is 10 cm
As DC is parallel to AB or EB, DE should be perpendicular to DC and AB.
Hence, Length of DC is equal to the length of EB.
Now, Length of EB = 5 cm.
And, therefore length of AE will be the difference of length of AB and length of EB.
= > Length of AE = length of AB - length of EB
= > Length of AE = 11 cm - 5 cm
= > Length of AE = 6 cm
By Pythagoras Theorem :
= > AD^2 = AE^2 + ED^2
= > AD^2 - AE^2 = ED^2
= > ( 10 cm )^2 - ( 6 cm)^2 = ED^2
= > 100 cm^2 - 36 cm^2 = ED^2
= > 64 cm^2 = ED^2
= > 8 cm = ED
Now,
Sum of lengths of parallel sides = DC + AB = 5 cm + 11 cm = 16 cm
Length of perpendicular = ED = 8 cm
From the properties of quadrilateral wr we that the area of trapezium is half of the product of sum of lengths of parallel sides and the perpendicular.
Therefore,
= > Area of this trapezium = 1 / 2 x ( 16 cm ) x 8 cm
= > Area of this trapezium = 16 x 4 cm^2
= > Area of this trapezium = 64 cm^2
Hence the required area of the trapezium is 64 cm^2.