Question

Abcd is a trapezium and p q are mid points on the diagonals ac and bd respectively then pq is equal to

Answer 1

Given : Abcd is a trapezium and p q are mid points on the diagonals ac and bd

To find :  PQ

Solution:

ABCD is a trapezium

Let say AB ║ CD

Draw a line PX   ( X being mid point of BC)

CP/AP = CX/BX = 1

Hence PX ║ AB

=> ΔCPX ≈ ΔCAD

=> PC/AC =  QX/ AB

=> 1/2 = QX/AB

=> QX = AB/2

Now join QX

=> DP/BP = CX/BX = 1

=> QX ║ AB

Hence  P & Q are line PX

ΔBQX ≈ ΔBDC

also QB/BD = QX/CD

=> 1/2 = QX/CD

=> QX = CD/2

PQ = | QX - PX |

= |  CD/2 - AB/2|

= |( CD - AB)/2 |

PQ = (1/2) ( Difference of parallel sides  )

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Answer 2

Step-by-step explanation:

PQ= 1/2 (DIFFERENCE OF PARALLEL SIDES)