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# 2.Consider the following parallelograms. Find the values of the unknowns x, y, z.

(i)Here, ABCD is a parallelogram.

$$∠B = ∠D\quad$$ [Opposite angles of a parallelogram are equal]

$$∠D = 100^{\circ}$$

$$\Rightarrow y = 100^{\circ}$$

$$∠A + ∠B = 180^{\circ}\quad$$ [Adjacent angles of a parallelogram are supplementary]

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

$$\Rightarrow z = 180^{\circ} – 100^{\circ} = 80^{\circ}$$

$$∠A = ∠C\quad$$ [Opposite angles of a parallelogram are equal]

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

$$Hence, x = 80^{\circ}, y = 100^{\circ} and\; z = 80^{\circ}$$

(ii)We have PQRS as a parallelogram.

$$∠P + ∠S = 180^{\circ}\quad$$ [Adjacent angles of parallelogram]

$$\Rightarrow x + 50^{\circ} = 180^{\circ}$$

$$\Rightarrow x = 180^{\circ} – 50^{\circ} = 130^{\circ}$$

$$Now, ∠P = ∠R\quad$$[Opposite angles are equal]

$$\Rightarrow x = y$$

$$\Rightarrow y = 130^{\circ}$$

$$Also, y = z\quad$$ [Alternate angles]

$$\Rightarrow z = 130^{\circ}$$

$$Hence, x = 130^{\circ}, y = 130^{\circ} and z = 130^{\circ}$$

(iii) Here we have, ABCD as a rhombus because diagonals intersect at $$90^{\circ}$$

$$So, x = 90^{\circ}$$

So,now in ∆OCB we have,

$$x + y + 30^{\circ} = 180^{\circ}\quad$$ [Angle sum property]

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

$$\Rightarrow y + 120^{\circ} = 180^{\circ}$$

$$\Rightarrow y = 180^{\circ} – 120^{\circ} = 60^{\circ}$$

$$y = z \quad$$[Alternate angles]

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

$$Hence, x = 90^{\circ}, y = 60^{\circ} and z = 60^{\circ}$$

(iv) ABCD is a parallelogram

∠A + ∠B = 180^{\circ} (Adjacent angles of a parallelogram are supplementary)

$$\Rightarrow x + 80^{\circ} = 180^{\circ}$$

$$\Rightarrow x = 180^{\circ} – 80^{\circ} = 100^{\circ}$$

$$Now, ∠D = ∠B\quad$$ [Opposite angles of a parallelogram are equal]

$$\Rightarrow y = 80^{\circ}$$

$$Also, z = ∠B = 80^{\circ}\quad$$ (Alternate angles)

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

(v) In this case we have ABCD as a parallelogram.

$$∠D = ∠B \quad$$[Opposite angles of a parallelogram are equal]

$$y = 112^{\circ}$$

$$x + y + 40^{\circ} = 180^{\circ}\quad$$ [Angle sum property]

$$\Rightarrow x + 112^{\circ} + 40^{\circ} = 180^{\circ}$$

$$\Rightarrow x + 152^{\circ} = 180^{\circ}$$

$$\Rightarrow x = 180^{\circ} – 152^{\circ} = 28^{\circ}$$

$$z = x = 28^{\circ} \quad$$[Alternate angles]

$$So\; we \;have, x = 28^{\circ}, y = 112^{\circ}, z = 28^{\circ}$$.