Question

ABCD is a trapezium such that AB parallel to DC and AB=3cm,if the diagonals AC and BD intersect at O such that AO/OC=BO/OD=1/2. Then calculate DC

Answer 1

Given:

ABCD is a trapezium such that AB parallel to DC and AB=3cm.

In a trapezium consider the angles AOB and DOC.

AO/OC=BO/OD=1/2 which is given, Angle AOB=Angle DOC these are vertically opposite angles.

Therefore, Triangle AOB is similar to that of DOC. The SAS similarity test. Thus AB/DC=AO/OC=BO/OD=1/2. Therefore,AB/DC=1/2 3/DC=1/2 Hence DC=6

Answer 2

The value of DC is 6 cm.

Step-by-step explanation:

• Given, in ΔAOB and ΔCOD,

$\frac{AO}{OC}=\frac{BO}{OD}=\frac{1}{2}$

∠AOB=∠COD

∴ΔAOB≈ΔCOD

• Since,ΔAOB≈ΔCOD

∴   $\frac{AO}{OC}=\frac{BO}{OD}=\frac{AB}{CD}=\frac{1}{2}$

=>$\frac{AB}{CD}=\frac{1}{2}$

Given,AB=3cm

=>$\frac{3}{CD}=\frac{1}{2}$

=>CD=6cm

• The value  of DC=6cm.