Question

In the trapezium ABCD, P and Q are the mid-points of AC and BD respectively AB || CD with AB > CD. Then PQ is equal to ______.

विकल्प

• `1/2 (AB - CD)`

• `1/2 (AB + CD)`

• `1/3 (AB - CD)`

• `1/3 (AB + CD)`

Answer 1

In the trapezium ABCD, P and Q are the mid-points of AC and BD respectively AB || CD with AB > CD. Then PQ is equal to `underlinebb(1/2 (AB - CD))`.

Explanation:

Place AB and CD horizontally.

Let A = (0, 0)

B = (AB, 0)

D = (x, h) 

And C = (x + CD, h) 

Mid-point of AC is `P = ((x + CD)/2, h/2)`

And mid-point of BD is `Q = ((x + AB)/2, h/2)`

Their y-coordinates are equal and the x-distance is

`((x + AB) - (x + CD))/2`

= `(AB - CD)/2`

So, `PQ = 1/2 (AB - CD)`.