Construction of Special Types of Quadrilaterals

• Step 1: Draw a line segment $AC = 7\text{ cm}$ using a ruler.
• Step 2: Construct the perpendicular bisector of $AC$. Label the point of intersection as $O$ (midpoint of $AC$).
• Step 3: Calculate half of diagonal $BD$: $\frac{6\text{ cm}}{2} = 3\text{ cm}$.
• Step 4: With center $O$ and compass radius $3\text{ cm}$, strike arcs on both sides of the perpendicular bisector to cut it at points $B$ and $D$.
• Step 5: Join $AB, BC, CD,$ and $DA$ using a ruler.
• $ABCD$ is the required rhombus! (All sides will measure exactly $\sqrt{3.5^2 + 3^2} \approx 4.61\text{ cm}$).

• A triangle is a rigid polygon uniquely determined by its three side lengths ($SSS$).
• A quadrilateral with four fixed side lengths is flexible and non-rigid: the angles between the sides can vary continuously without changing the side lengths (it can flex into infinitely many different shapes).
• To make a quadrilateral rigid, the angular freedom must be removed by providing at least one diagonal or one angle.
• Therefore, four sides are insufficient; a minimum of five independent elements is mandatory.

• Step 1: Draw a line segment $AC = 6.4\text{ cm}$.
• Step 2: Draw the perpendicular bisector $XY$ of line segment $AC$, intersecting $AC$ at midpoint $O$.
• Step 3: Since diagonals of a square are equal ($BD = AC = 6.4\text{ cm}$), $OB = OD = \frac{6.4}{2} = 3.2\text{ cm}$.
• Step 4: With center $O$ and compass radius $3.2\text{ cm}$, cut arcs on $XY$ at points $B$ and $D$.
• Step 5: Join $AB, BC, CD,$ and $DA$. $ABCD$ is the required square.

• Notice that side $QR$ is given, but angle $\angle R$ at vertex $R$ is unknown, while angle $\angle S$ is given where side $RS$ is unknown.
• Before starting, we must calculate the missing angle $\angle R$ using the Angle Sum Property of a quadrilateral ($360^{\circ}$):

$$\angle P + \angle Q + \angle R + \angle S = 360^{\circ}$$

$$75^{\circ} + 105^{\circ} + \angle R + 100^{\circ} = 360^{\circ}$$

$$280^{\circ} + \angle R = 360^{\circ}$$

$$\mathbf{\angle R = 360^{\circ} - 280^{\circ} = 80^{\circ}}$$

• Now we have base $PQ$, $\angle P = 75^{\circ}, \angle Q = 105^{\circ}$, side $QR = 5.2\text{ cm}$, and $\angle R = 80^{\circ}$, allowing seamless construction!

• Step 1: Draw base line segment $AB = 5.5\text{ cm}$.
• Step 2: At vertex $B$, construct a perpendicular ray $BX$ at $90^{\circ}$ using compass.
• Step 3: With center $A$ and radius $7\text{ cm}$ (the diagonal), strike an arc cutting ray $BX$ at vertex $C$.
• Step 4: With center $C$ and radius $5.5\text{ cm}$ ($= AB$), draw an arc to the left.
• Step 5: With center $A$ and radius equal to length $BC$, draw an arc cutting the previous arc at vertex $D$.
• Step 6: Join $CD$ and $DA$. $ABCD$ is the required rectangle.

• A rough sketch provides a visual architectural blueprint displaying the relative positions of all vertices cyclically ($A, B, C, D$).
• It reveals which triangle should be constructed first as a rigid base.
• It highlights whether all required dimensions are directly usable or if missing angles must be calculated.
• ICSE examination marking schemes award explicit marks for a properly labeled rough sketch.

• No, it is impossible.
• Two adjacent sides provide only $2$ independent pieces of information.
• Even though opposite sides are equal ($a, b, a, b$), the angle between them is completely free to rotate.
• A parallelogram requires at least three independent elements (e.g., two adjacent sides and an included angle, or two sides and a diagonal) to be uniquely constructed.