Question

Let ABCD is a trapezium whose areais ( a² - b²). AB =a. DC = b and AB || CD then the distancw between the parallel sides is ?

1 ) a - b
2) 2(a - b)
3) a + b
4) 2(a + b )

explanation pls ​

Answer 1

Given

Area of trapezium ABCD

AB = a, DC = b and AB| |CD

To find

Distance between parallel sides

Solution

Area of trapezium

Substituting the value

Area of trapezium ABCD

So, option 2) 2(a - b) is correct answer.

Answer 2

Answer:

Given

• Area of trapezium ABCD
• $= (a {}^{2} - b {}^{2} )$
• AB = a, DC = b and AB| |CD

To find

• Distance between parallel sides

Solution

Area of trapezium

$= \frac{(sum \: of \: parallel \: sides)height \: between \: parallel \: sides}{2}$

Substituting the value

Area of trapezium ABCD

$(a {}^{2} - b {}^{2} ) = \frac{(a + b)height}{2} \\ (a + b)(a - b) = \frac{(a + b)height}{2} \\ \frac{2(a + b)(a - b)}{(a + b)} = height \\ height = 2(a - b)$

So, option 2) 2(a - b) is correct answer.