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 trapezium

Answer 1

Answer:

30 cm²

Step-by-step explanation:

Let the trapezium be ABCD as shown in the figure.

Given the length of the parallel sides are 6cm and 14cm. The length of the parallel sides is 5cm.

So AB = 6cm ; CD = 14cm; BD = AC = 5cm

To find the area we need the height of the trapezium. Drawing perpendiculars AP and BQ from AB to CD will give us a rectangle ABPQ. Hence AB= PQ=6cm. Also AP=BQ

Now consider right angle triangles ΔACP and ΔBQD

Using pythagoras theorem we know that

AC²=CP²+AP² and BD²=BQ²+QD²

CP²=AC²-AP² and QD²=BD²-BQ²

As, AC = BD and AP= BQ ; So CP = QD = (14-6)/2=4

Now we can get the height of the trapezuim.

AC²=AP²+CP²

AP²=5²-4² = 25-16=9

AP = √9=3

So the area of the trapezium will be:

1/2(6+14)*3 = 30cm²

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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