Question

The parallel sides of a trapezium are in the ratio 4:5. If the distance between the parallel sides
is 6 cm and the area is 81 cm sq, find the lengths of the parallel sides of the trapezium​

Answer 1

Given:

• The parallel sides of a trapezium are in the ratio 4:5
• The distance between the parallel sides (i.e. altitude of the trapezium) is 6 cm
• The are of the trapezium is 81 cm sq.

To find:

The length of the parallel sides of the trapezium

So,

Let the parallel sides of the trapezium be 4x cm and 5x cm

• The formula used to find the area of the trapezium is

1/2 * (sum of parallel sides) * altitude units sq.

⇒ 1/2 * (4x + 5x) * 6 = 81

[As The value of altitude = 6 cm, and area = 81 cm sq. is given]

⇒ 1/2 * (9x) * 6 = 81

⇒ 9x * 3 = 81

⇒ 27x = 81

⇒ x = 81 ÷ 27

⇒ x = 3

Thus,

The value of x is 3

Hence,

The parallel side of trapezium are

• 4x = 4*3 = 12 cm

• 5x = 5*3 = 15 cm

Answer 2

Given : The parallel sides of a trapezium are in the ratio 4:5. If the distance between the parallel sides is 6 cm and the area is 81 cm sq.

____________________________

Step-by-step explanation:

Let the Lengths of parallel sides of the trapazium be 4 cm and 5 cm respectively.

• As it is given in the question that distance between the parallel sides is 6 cm and we know that distance between the parallel side is nothing but height of the trapezium :]

Therefore, Height of the trapezium = 6 cm

_____________________

☯ $\underline{\boldsymbol{According\: to \:the\: Question\:now :}}$

$\bigstar\:\:\sf Area \: of \: trapezium \: = \: \dfrac{1}{2} \times (sum \: of \: parallel \: sides) \times Height\:\:\bigstar$

$:\implies \sf 81 = \dfrac{1}{\cancel{2}} \times (4x + 5x) \times \cancel{6}$

$:\implies \sf 81 = 4x + 5x \times 3$

$:\implies \sf 81 = 9x \times 3$

$:\implies \sf 9x = \dfrac{81}{3}$

$:\implies \sf 9x = 27$

$:\implies \sf x = \dfrac{27}{9}$

$:\implies \underline{ \boxed{\sf x = 3 }}$

_________________

$\dag \: \underline{\tt Therefore,length \: of \: parallel \: sides \: are: }$

$\bullet\:\:\textsf{4 * 3= \textbf{12 cm}}$

$\bullet\:\:\textsf{5 * 3= \textbf{15 cm}}$