Question 6

We will use the properties of parallelograms and rhombuses to find the unknown angles in each quadrilateral.

Step 1 — Finding angles in parallelogram PEAR (i)

A parallelogram is a quadrilateral with two pairs of parallel sides. In quadrilateral PEAR, the arrows show that PR is parallel to EA, and PE is parallel to RA. So, PEAR is a parallelogram. We are given that $\angle P = 40^\circ$. In a parallelogram, consecutive angles are supplementary, meaning they add up to $180^\circ$. So, $\angle P + \angle E = 180^\circ$.

$40^\circ + \angle E = 180^\circ$ $\angle E = 180^\circ - 40^\circ$

$\boxed{\angle E = 140^\circ}$

Also, $\angle P + \angle R = 180^\circ$.

$40^\circ + \angle R = 180^\circ$ $\angle R = 180^\circ - 40^\circ$

$\boxed{\angle R = 140^\circ}$

Opposite angles in a parallelogram are equal. So, $\angle A = \angle P$.

$\boxed{\angle A = 40^\circ}$

The angles of parallelogram PEAR are $\angle P = 40^\circ$, $\angle E = 140^\circ$, $\angle A = 40^\circ$, and $\angle R = 140^\circ$.

Step 2 — Finding angles in parallelogram PQRS (ii)

Quadrilateral PQRS is a parallelogram because its opposite sides are marked as parallel. We are given that $\angle P = 110^\circ$. Consecutive angles in a parallelogram are supplementary. So, $\angle P + \angle Q = 180^\circ$.

$110^\circ + \angle Q = 180^\circ$ $\angle Q = 180^\circ - 110^\circ$

$\boxed{\angle Q = 70^\circ}$

Also, $\angle P + \angle S = 180^\circ$.

$110^\circ + \angle S = 180^\circ$ $\angle S = 180^\circ - 110^\circ$

$\boxed{\angle S = 70^\circ}$

Opposite angles in a parallelogram are equal. So, $\angle R = \angle P$.

$\boxed{\angle R = 110^\circ}$

The angles of parallelogram PQRS are $\angle P = 110^\circ$, $\angle Q = 70^\circ$, $\angle R = 110^\circ$, and $\angle S = 70^\circ$.

Step 3 — Finding angles in rhombus UVWX (iii)

A rhombus is a quadrilateral where all four sides are equal in length. Quadrilateral UVWX is a rhombus because all its sides are marked with tick marks, meaning they are equal. We are given that $\angle XVU = 30^\circ$. The diagonal XV divides the rhombus into two triangles, $\triangle UVX$ and $\triangle VWX$. In a rhombus, the diagonals bisect (cut into two equal parts) the angles at the vertices they connect. So, diagonal XV bisects $\angle V$ and $\angle X$. This means $\angle XVW$ is equal to $\angle XVU$.

$\angle XVW = \angle XVU$

$\boxed{\angle XVW = 30^\circ}$

Now consider $\triangle UVX$. Since UV = UX (all sides of a rhombus are equal), $\triangle UVX$ is an isosceles triangle. In an isosceles triangle, the angles opposite the equal sides are equal. So, $\angle UXV$ is equal to $\angle XVU$.

$\angle UXV = \angle XVU$

$\boxed{\angle UXV = 30^\circ}$

Since diagonal XV bisects $\angle X$, $\angle WXV$ is equal to $\angle UXV$.

$\angle WXV = \angle UXV$

$\boxed{\angle WXV = 30^\circ}$

Now let us find $\angle VUX$, which is the angle $\angle U$ of the rhombus. The sum of angles in any triangle is $180^\circ$. In $\triangle UVX$, we have $\angle VUX + \angle XVU + \angle UXV = 180^\circ$.

$\angle VUX + 30^\circ + 30^\circ = 180^\circ$ $\angle VUX + 60^\circ = 180^\circ$ $\angle VUX = 180^\circ - 60^\circ$

$\boxed{\angle VUX = 120^\circ}$

Finally, let us find $\angle VWX$, which is the angle $\angle W$ of the rhombus. In a rhombus, opposite angles are equal. So, $\angle VWX$ is equal to $\angle VUX$.

$\angle VWX = \angle VUX$

$\boxed{\angle VWX = 120^\circ}$

The remaining angles are $\angle XVW = 30^\circ$, $\angle UXV = 30^\circ$, $\angle WXV = 30^\circ$, $\angle VUX = 120^\circ$, and $\angle VWX = 120^\circ$. (The given angle is $\angle XVU = 30^\circ$).

Step 4 — Finding angles in rhombus OIAE (iv)

Quadrilateral OIAE is a rhombus because all its sides are marked with tick marks, meaning they are equal. We are given that $\angle AEO = 20^\circ$. The diagonal OE divides the rhombus into two triangles, $\triangle AEO$ and $\triangle OIE$. In a rhombus, the diagonals bisect the angles at the vertices they connect. So, diagonal OE bisects $\angle E$ and $\angle O$. This means $\angle OEI$ is equal to $\angle AEO$.

$\angle OEI = \angle AEO$

$\boxed{\angle OEI = 20^\circ}$

Now consider $\triangle AEO$. Since AE = AO (all sides of a rhombus are equal), $\triangle AEO$ is an isosceles triangle. In an isosceles triangle, the angles opposite the equal sides are equal. So, $\angle AOE$ is equal to $\angle AEO$.

$\angle AOE = \angle AEO$

$\boxed{\angle AOE = 20^\circ}$

Since diagonal OE bisects $\angle O$, $\angle IOE$ is equal to $\angle AOE$.

$\angle IOE = \angle AOE$

$\boxed{\angle IOE = 20^\circ}$

Now let us find $\angle EAO$, which is the angle $\angle A$ of the rhombus. The sum of angles in any triangle is $180^\circ$. In $\triangle AEO$, we have $\angle EAO + \angle AEO + \angle AOE = 180^\circ$.

$\angle EAO + 20^\circ + 20^\circ = 180^\circ$ $\angle EAO + 40^\circ = 180^\circ$ $\angle EAO = 180^\circ - 40^\circ$

$\boxed{\angle EAO = 140^\circ}$

Finally, let us find $\angle OIA$, which is the angle $\angle I$ of the rhombus. In a rhombus, opposite angles are equal. So, $\angle OIA$ is equal to $\angle EAO$.

$\angle OIA = \angle EAO$

$\boxed{\angle OIA = 140^\circ}$

The remaining angles are $\angle AOE = 20^\circ$, $\angle OEI = 20^\circ$, $\angle IOE = 20^\circ$, $\angle EAO = 140^\circ$, and $\angle OIA = 140^\circ$. (The given angle is $\angle AEO = 20^\circ$).

Answer

(i) The angles of parallelogram PEAR are $\angle P = 40^\circ$, $\angle E = 140^\circ$, $\angle A = 40^\circ$, and $\angle R = 140^\circ$. (ii) The angles of parallelogram PQRS are $\angle P = 110^\circ$, $\angle Q = 70^\circ$, $\angle R = 110^\circ$, and $\angle S = 70^\circ$. (iii) The angles formed by the diagonal in rhombus UVWX are $\angle XVU = 30^\circ$, $\angle XVW = 30^\circ$, $\angle UXV = 30^\circ$, $\angle WXV = 30^\circ$, $\angle VUX = 120^\circ$, and $\angle VWX = 120^\circ$. (iv) The angles formed by the diagonal in rhombus OIAE are $\angle AEO = 20^\circ$, $\angle AOE = 20^\circ$, $\angle OEI = 20^\circ$, $\angle IOE = 20^\circ$, $\angle EAO = 140^\circ$, and $\angle OIA = 140^\circ$.