Question

Let ABCD be a parallelogram. The diagonals bisect in E.
(i) If AB = 6 cm and AD = 4 cm, find CD and BC.
(ii) If DE = 5 cm and AE = 7 cm, find BD and AC.
(iii) If $\angle DAB={72}^o$, find the measure of $\angle CBA$.
(iv) If $AD=(x+2y),BC=(2x+3),DC=(x+7)$ and $AB=(3y+2)$, find AB and BC.

Answer 1

(i) Given $AB=6$ and $AD=4$, we have to find CD and BC. We know that in a parallelogram opposite sides are equal. Hence $CD=AB=6cm$ and $CB=AD=4cm$.
(ii) Given : $DE=5,AE=7$
To find BD and AC.
We have
$AC=2\left(AE\right)=2\times \left(7\right)=14cm$ (E is the midpoint of AC)
$BD=2\left(DE\right)=2\times \left(5\right)=10cm$ (E is the midpoint of BD)
(iii) Given: $\angle DAB={72}^o$
To Find $\angle CBA$.
Observe $\angle CDA+\angle DAB={180}^o$
(Sum of two angles on the same side of a parallelogram $={180}^o$.)
$\angle CBA={180}^o$.
(iv) Given : $AD=(x+2y),DC=(x+7),BC=(2x+3),AB=(3y+2)$
To find AB and BC.
We know that $AD=BC$. Hence $x+2y=2x+3$. We get
$x-2y=-3$ .....(1)
Again $AB=DC$ and hence $3y+2=x+7$. we obtain
$x-3y=-5$ ...(2)
Solving (1) and (2) for x and y, we get $x=1,y=2$. Hence
$AB=3y+2=3\times 2+2=8cm$
$BC=2x+3=2+3=5cm$