Question

two parallel sides of an isosceles trapezium are 6 cm and 14 cm respectively if the length of each non parallel side is 5 cm find the area of the trapezium

Answer 1

Let the trapezium be ABCD where AB and CD are the parallel sides.

Now,

AB=6 cm

CD=14 cm

AD=BC=5 cm

Construct a line parallel to BC from A to a point E on DC.

NOw, ar(ADE)= 12 sq. cm (By Heron's Formula)

ar(ADE)= 1/2*b*h

Therefore the height is 6 cm.

ar(ABCE)= b*h

=6cm*6cm

=36 sq.cm

Hence, area of ABCD is=12+36 i.e. 48 sq. cm.

Answer 2

HOLA MATES !!!

HOPE THIS HELPS YOU...

Answer:-
Let us consider a trapezium ABCD with 2 non parallel sides AB =CD = 5cm and 2 parallel sides BC = 6cm and AD = 14cm . Then when 2 non parallel sides are of equal length then parallel sides will be of formation such that when a bisector is drawn through one side it even bisects the other side .So , when we draw a perpendicular through B to AD or C to AD then we can get the perpendicular height 'h' = AL.

Now since AB and CD are equal AD will be equally distributed on both sides of BC so (14 - 6)/2 = 4cm = BL on both sides of BC .

Then by using Pythagoras theorerm,
=> AB² = AL² + BL²
=> 25 = h² + 16
=> h = 9cm
Then height of trapezium, h = 9cm


Step-by-step explanation:-
Now let us apply the formula for area of trapezium,
=> A = h(a+b)/2

=> A = 9(6+14)/2

=> A = 9*10

=> A = 90square cm

Hence , the area of the trapezium ABCD is 90square cm.









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