The following figures are parallelograms.

Question:

The following figures are parallelograms. Find the degree values of the unknowns x, y, z.

Solution:

(i) $O$ pposite angles of a parallelogram are same.

$\therefore \mathrm{x}=\mathrm{z}$ and $\mathrm{y}=100^{\circ}$

Also, $\mathrm{y}+\mathrm{z}=180^{\circ} \quad\left(s\right.$ um of adjacent angles of a quadrilateral is $\left.180^{\circ}\right)$

$\mathrm{z}+100^{\circ}=180^{\circ}$

$\mathrm{x}=180^{\circ}-100^{\circ}$

$\mathrm{x}=80^{\circ}$

$\therefore \mathrm{x}=80^{\circ}, \mathrm{y}=100^{\circ}$ and $\mathrm{z}=80^{\circ}$

(ii) $O$ pposite angles of a parallelogram are same.

$\therefore \mathrm{x}=\mathrm{y}$ and $\angle \mathrm{RQP}=100^{\circ}$

$\angle \mathrm{PSR}+\angle \mathrm{SRQ}=180^{\circ}$

$\mathrm{y}+50^{\circ}=180^{\circ}$

$\mathrm{x}=180^{\circ}-50^{\circ}$

$\mathrm{x}=130^{\circ}$

$\therefore \mathrm{x}=130^{\circ}, \mathrm{y}=130^{\circ}$

Since $\mathrm{y}$ and $\mathrm{z}$ are alternate angles, $\mathrm{z}=130^{\circ}$.

(iii) $S$ um of all angles in a triangle is $180^{\circ}$.

$\therefore 30^{\circ}+90^{\circ}+\mathrm{z}=180^{\circ}$

$\mathrm{z}=60^{\circ}$

$O$ pposite angles are equal in parallelogram.

$\therefore \mathrm{y}=\mathrm{z}=60^{\circ}$

and $\mathrm{x}=30^{\circ}$ (alternate angles)

(iv) $\mathrm{x}=90^{\circ}$ (vertically opposite angle)

$S$ um of all angles in a triangle is $180^{\circ}$.

$\therefore \mathrm{y}+90^{\circ}+30^{\circ}=180^{\circ}$

$\mathrm{y}=180^{\circ}-\left(90^{\circ}+30^{\circ}\right)=60^{\circ}$

$\mathrm{y}=\mathrm{z}=60^{\circ} \quad(a$ lternate angles $)$

(v) $O$ pposite angles are equal in $a$ parallelogram.

$\therefore \mathrm{y}=80^{\circ}$

$\mathrm{y}+\mathrm{x}=180^{\circ}$

$\mathrm{x}=180^{\circ}-100^{\circ}=80^{\circ}$

$\mathrm{z}=\mathrm{y}=80^{\circ} \quad$ (alternate angles)

(vi) $\mathrm{y}=112^{\circ}$ (opposite angles are equal in a parallelogram)

In $\Delta$ UTW :

$\mathrm{x}+\mathrm{y}+40^{\circ}=180^{\circ}(a$ ngle sum property of $a$ triangle $)$

$\mathrm{x}=180^{\circ}-\left(112^{\circ}-40^{\circ}\right)=28^{\circ}$

$B$ ottom left vertex $=180^{\circ}-112^{\circ}=68^{\circ}$

$\therefore \mathrm{z}=\mathrm{x}=28^{\circ}$ (alternate angles)

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