Question

W.
the other,
The area of a trapezium is 300 m². The perpendicular distance between the two parallel sides is 15 m. If
the difference of the parallel sides is 16 m, find the length of the parallel sides.
give the answer pls​

Answer 1

Given :

⬤ Area of Trapezium is 300 m².

⬤ Perpendicular Distance between two Parallel Sides is 15 m .

⬤ Difference of the Parallel Side is 16 m .

To Find :

⬤ Length of Two Parallel Sides .

Formula Used :

$\bf[\:{Area \: of \: Trapezium = \dfrac{1}{2}\:(Sum \: of \: Parallel \: Sides) \times{h}}\:]$

• h = height

Solution :

Let :

• One Parallel Side be x .
• Other Parallel Side be x - 16 .

Now :

300 = 1/2 × (x + x - 16) × 15

300 = 1/2 × (2x - 16) × 15

(300 × 2) = 30x - 240

600 = 30x - 240

600 + 240 = 30x

840 = 30x

840/30 = x

28 = x

Therefore , The Value of x is 28 .

Hence ,

One Parallel Side = x

= 28

Other Parallel Side = x - 16

= 28 - 16

= 12

Therefore , The Two Parallel Sides are 28 m and 12 m .

Answer 2

Step-by-step explanation:

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$\text{\Large\underline{\red{Question:-}}}$

$\Longrightarrow$ ● The area of a trapezium is 300 m². The perpendicular distance between the two parallel sides is 15 m. If the difference of the parallel sides is 16 m, find the length of the parallel sides.

$\text{\Large\underline{\orange{To\ find:-}}}$

● In this question we have to find the length of the parallel sides.

$\text{\Large\underline{\green{Given:-}}}$

$\sf We\ have = \begin{cases} \sf\pink{Area\ of\ trapezium\ is\ 300m^{2.}} \\ \\ \sf\pink{Perpendicular\ distance\ between\ two\ parallel\ sides\ is\ 15m.} \\ \\ \sf\pink{Difference\ of\ the\ parallel\ side\ is\ 16m.} \end{cases}$

$\text{\large\underline{\blue{Formula\ to\ be\ used:-}}}$

$\sf\orange{(Area\ of\ trapezium\ =\ \dfrac{1}{2}\ (Sum\ of\ parallel\ sides)\ \times\ h)}$

$\text{\Large\underline{\purple{Solution:-}}}$

● Let the one parallel side be x.

● And the other parallel side be x - 16.

$\sf{\underline{\ddag\qquad According\ to\ the\ question \qquad\ddag}}$

$\sf \: \: \: \: : \implies {300\ =\ \dfrac{1}{2}\ \times\ (x\ +\ x\ -\ 16)\ \times\ 15}$

$\sf \: \: \: \: : \implies {300\ =\ \dfrac{1}{2}\ \times\ (2x\ -\ 16)\ \times\ 15}$

$\sf \: \: \: \: : \implies {(300\ \times\ 2)\ =\ 30x\ -\ 240}$

$\sf \: \: \: \: : \implies {600\ =\ 30x\ -\ 240}$

$\sf \: \: \: \: : \implies {600\ +\ 240\ =\ 30x}$

$\sf \: \: \: \: : \implies {840\ =\ 30x}$

$\sf \: \: \: \: : \implies {\cancel \dfrac{840}{30}\ =\ x}$

$\sf \: \: \: \: : \implies {28\ =\ x}$

$\sf \: \: \: \: : \implies {\purple{\underline{\fbox{x\ =\ 28.}}}}\bigstar$

The value of x is 28.

$\sf{\underline{\ddag\qquad Therefore \qquad\ddag}}$

$\sf \Longrightarrow {One\ parallel\ side\ =\ x\ =\ 28m.}$

$\sf \Longrightarrow {Other\ parallel\ side\ =\ x\ -\ 16\ =\ 28\ -\ 16\ =\ 12m.}$

Hence:-

● The two parallel sides are 28m and 12m.

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