Question

Open area of a house is in the shape of a trapezium with parallel sides in the ratio 3:5. Its area is
50 m2 and the perpendicular distance between the parallel sides is 5 metres. Find the length of two
parallel sides of that open area.​

Answer 1

Answer:

ANSWER

Let the parallel sides be 3x and 5x

Area of trap=21×Sum of parallel sides×h

So, 50=21×(3x+5x)×5

⇒50=4x×5

⇒x=2050=2.5

So, the first side=3x=3×2.5=7.5m

and the second side is =5x=5×2.5=12.5m

Answer 2

Question :

To find length of parallel sides of an open area in shape of a trapezium.

Answer :

Given :

• Ratio of parallel sides = 3:5

• Area of trapezium = 50 m²

• Perpendicular distance = 5 m

Let's do ::

${\underline{\boxed{\bf{area \: of \: trapezium \: = \frac{1}{2} \times (a + b) \times h }}}}$

• a and b are parallel sides
• h is the perpendicular distance.

Let, parallel sides be x, 3x and 5x

$\longrightarrow\sf \: \frac{1}{2}(a + b) \times h = 50 \\ \\ \longrightarrow\sf \: \frac{1}{2} (3x + 5x) \times 5 = 50 \\ \\ \longrightarrow \sf \frac{1}{2} \times 8x \times 5 = 50 \\ \\ \sf\longrightarrow \: 4x \: \times 5 = 50 \\ \\ \longrightarrow\sf \: 4x = \frac{50}{5} \\ \\ \sf\longrightarrow \: 4x = 10 \\ \\ \sf\longrightarrow \: x = \frac{10}{4} \\ \\ \longrightarrow\bf \: x \: = 2.5$

So, X is 2.5 m.

Now , let's find the parallel sides :

$\sf➝ \: 3x \\ \\ ➝\sf \: 3 \times 2.5 \\ \\ ➝\bf \: 7.5$

$\sf➝ \: 5x \\ \\ \sf➝ \: 5 \times 2.5 \\ \\ \bf \: ➝ \: 12.5$

The sides are 7.5 and 12.5.

Let's verify ::

$\sf \: = > \frac{1}{2} (a \: + b) \times h \: = 50$

$= > \sf \: \frac{1}{2} \times (7.5 + 12.5) \times 5 = 50$

$= > \sf \: \frac{1}{2} \times 20 \times 5 = 50$

$= > \sf \: \frac{1}{2} \times 100 = 50$

$\sf = > \: 50 = 50$

$\bf \: LHS \: = \: RHS \:$

Thus Solved !!