Question

in a trapezium ABCD AB line segment is parallel to CD line segment E and F are the mid points AD line segment and BC line segment respectively if AB =6 Cms, CD=8 Cms, then find EF​

Answer 1

Answer:

$ef = \frac{(ab + dc)}{2}$

ABCD is trapezium in which AB∥DC.

EF is parallel to side DC. 

Then we have AB∥DC∥EF.

Hence we have also trapezium ABFE and trapezium EFCD.

Let AP be the perpendicular to DC and this intersects EF at Q. 

AQ will be perpendicular to EF.

For △APD and △AQE we have EAAD=AQAP=2

This gives AP=2AQ  

i.e, AQ=QP

Consider the area  we have area ABCD= area ABFE+ area EFCD                    

(21)AP×(AB+DC)=(21)AQ×(AB+EF)+(21)QP×(EF+DC)

⇒AP(AB+DC)=AP×2AB+AP×2EF+AP×2EF+AP×2

Answer 2

Answer:

Answer:

ef = \frac{(ab + dc)}{2}ef=

2

(ab+dc)

ABCD is trapezium in which AB∥DC.

EF is parallel to side DC.

Then we have AB∥DC∥EF.

Hence we have also trapezium ABFE and trapezium EFCD.

Let AP be the perpendicular to DC and this intersects EF at Q.

AQ will be perpendicular to EF.

For △APD and △AQE we have EAAD=AQAP=2

This gives AP=2AQ

i.e, AQ=QP

Consider the area we have area ABCD= area ABFE+ area EFCD

(21)AP×(AB+DC)=(21)AQ×(AB+EF)+(21)QP×(EF+DC)

⇒AP(AB+DC)=AP×2AB+AP×2EF+AP×2EF