Question

the area of trapezium is 120 sq cm and the perpendicular distance between is parallel sides is 8 cm find the length of the parallel sides if their difference is 6 CM

Answer 1

Let the parallel lines be a and b


Given that difference of them is 6 cm

=> a - b = 6


Now we know that area of a trapezium =
$\frac{1}{2} (a+ b)h$

So, here given that area = 120 and
height = 8

=>
$\frac{1}{2} (a + b) \times 8 = 120 \\ \\ = > (a + b) \times 4 = 120 \\ \\ = > (a + b) = \frac{120}{4} \\ \\ = > (a + b) = 30$


So a + b = 30


But we are given that a - b = 6

Adding them,


a + b + a - b = 30 + 6

=> 2a = 36

=> a = 36/2

=> a = 18 cm

Now a + b = 30

=> 18 + b = 30

=> b = 30 - 18

=> b = 12cm


So first parallel side = 18 cm

another = 12 cm


Hope it helps dear friend

Answer 2

area of trapezium=1/2 *sum of parallel sides *distance between then

so, let the greater side be x and the smaller parallel side be x-6 cm      

120sq.cm=1/2*(x+x-6)*8      

120=4*(2x-6)

30=2x-6

2x=36

x=18

so the parallel sides are 18cm and 12 cm        

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