Question

Th
parallel sides.
The area of a trapezium is 248 sq. m and its height is 8 m. If one of the parallel side
is smaller than the other by 4 m, find two parallel sides.
whose area is 1.6 m², altitude is 0 ap10 dm ​

Answer 1

Answer:

The lengths of the parallel sides of the trapezium are 29 m & 33 m.

Step-by-step-explanation:

We have given that,

Area of a trapezium is 248 sq. m.

The height of the trapezium is 8 m.

We have to find the lengths of the two parallel sides.

Let the two parallel sides be x m and y m.

From the given condition,

One parallel side is smaller than the other by 4 m.

∴ x = y + 4 - - - ( 1 )

Now, we know that,

Area of trapezium = ( Sum of parallel sides ) * Height / 2

⇒ 248 = ( x + y ) * 8 / 2

⇒ 248 = ( x + y ) * 8 ÷ 2

⇒ 248 = ( x + y ) * 4

⇒ 4 ( x + y ) = 248

⇒ x + y = 62 - - - [ Dividing both sides by 4 ]

⇒ y + 4 + y = 62 - - - [ From ( 1 ) ]

⇒ y + y + 4 = 62

⇒ 2y = 62 - 4

⇒ 2y = 58

⇒ y = 58 ÷ 2

⇒ y = 29 m

Now,

x = y + 4 - - - ( 1 )

⇒ x = 29 + 4

⇒ x = 33 m

∴ The lengths of the parallel sides of the trapezium are 29 m & 33 m.

Answer 2

Given :-

Area of trapezium = 248 sq. m

Height = 8 m

One parallel side is smaller than other by 4

To Find :-

Parallel sides

Solution :-

Let the 1st parallel side be x and other parallel side x - 4

We know that

$\large \sf Area = \dfrac{1}{2} \times (sum \; of \; parallel \; side) \times h$

$\implies$ 248 = 1/2 × (x + x - 4) × 8

$\implies$ 248 = 4 × (x + x - 4)

$\implies$ 248/4 = x + x - 4

$\implies$ 62 + 4 = 2x

$\implies$ 66 = 2x

$\implies$ x = 66/2

$\implies$x = 33

Parallel sides are 33 m and 33 - 4 = 29 m