Question

area of trapezium is 120 CM square and the distance between the parallel side is 8 cm . find the length of the parallel side is their difference is 6 CM

Answer 1

$\: hey \: mate \: here \: is \: your \: answer \:$

$\textbf{Given that}$ Area Of A Trapezium Is 120 Cm Sqr And The Distance Between The Parallel Side

Is 8cm . Now We Need To Find length of the parallel side if their differences is 6cm.

$\textbf{Solution}$ ......

Now in Question it's given that ..

$height \: (h) \: = \: 8cm \: \: ( \: height \: is \: distance \: between \: parallel \: lines \: )$

Now Parallel Lines Electric Greater Line be = X

So According To Question It's Said That There Difference Is = 6cm

So The length of smaller line would be = X - 6

Now we know that area of a Trapezium is

$\frac{1}{2} \times (a + b) \: \times h \: (where \: a \: and \: b \: are \: \: length \: of \: parallel \: lines \: )$

$\: and \: h \: is \: height \:$

Therefore Now Putting Values Of a ,b And b As X , X - 6 , & 8cm Respectively...

That is ➡️

$\frac{1}{2} \times( (x) + (x - 6)) \times 8$

That is ➡️

$\frac{1}{2} ( \: 2x - 6) \times 8$

Now taking. 2 common from 2x-6 we have

$\frac{1}{2} 2(x - 3) \times 8$

Now cancelling 1/2 and 2 we have ....

$(x - 3) \times 8$

But In Question It's Given That Area Of Trapezium = 120 Cm Sqr

So we have ..

$(x - 3) \times 8 = 120 {cm}^{2}$

That is ...

$(x - 3) = \frac{120}{8} \: ( \: taking \: 8 \: to \: rhs)$

This is equal to ....

$x - 3 = 15cm$

So we have ....

$x = 18cm$

Hence the length of greater parallel line = X = 18cm

and

length of smaller parallel line = X-6 = 18-6 = 12cm

$\textbf{Therefore required answer is ...}$

18cm and 16 cm .

$hope \: its \: help \: full$

Thought of the day...

"Success Is Not A Thing Which Comes In Your Life It's The Thing Which You Need To Follow In Your Life . "

Thought made by Me ....

$be \: brainly$

Answer 2

area of trapezium=1/2(sum of parallel side)×h
=1/2(6+8)×h=120
1/2×14×h=120
7h=120
h=120/7
h=17-1/7