Question

$ABCD$ is a quadrilateral and $BD$ is one of it's diagonals, length of $AB=CD=2.5cm$ and $BD=4cm$ and $\angle ABD=\angle CDB={90}^{\circ },$ then show that $ABCD$ is a $//gm$ and find it's area.

Answer 1

Ref . Image.

AB = CD = 2.5 cm , BD = 4 cm & $\angle ABD=\angle CDB={90}^{\circ }$

we know "if a Transvarsal intersect two lines then that alternate interior angles are equal , then two lines are parallel.

$\angle =ABD=\angle CDB={90}^{\circ }$ given

AB || CD & AB = CD. _____ (1)

In $\Delta ABD$ & $\Delta CDB$

AB = CD

$\angle ABD=\angle CDB.$

BD = BD (common Rule)

so $\Delta ABD\cong \Delta CDB$ ( by R.H.L rule)

so DA = BC

From eq (1) we can say that one pair of sides of quadrilateral ABCD is equal & parallel.

$\therefore$ ABCD is a parallelogram.

Area of 11 gm = base $\times$ corresponding altitude

$=2.5\times 4=10$ sq unit