Construction of Quadrilaterals

• Five independent measurements are required.
• Four sides are not sufficient because a four-bar linkage is mechanically flexible (non-rigid). A quadrilateral with four fixed side lengths can flex into infinitely many different parallelograms or irregular shapes with varying angles. A fifth measurement (such as a diagonal or an interior angle) is strictly required to lock the figure into a rigid triangle pair.

• Steps of Construction:
1. Draw a line segment $AC = 7\text{ cm}$ (the diagonal).
2. Draw the perpendicular bisector of $AC$, intersecting $AC$ at point $O$ ($OA = OC = 3.5\text{ cm}$).
3. Since the diagonals of a square are equal and bisect each other perpendicularly at $90^\circ$, half the diagonal is $OB = OD = 3.5\text{ cm}$.
4. With center $O$ and radius $3.5\text{ cm}$, draw arcs on both sides of the perpendicular bisector to cut it at points $B$ and $D$.
5. Join $AB, BC, CD$, and $DA$.
6. $ABCD$ is the required square.

• Check if $\triangle ABC$ can be constructed with sides $AB = 3\text{ cm}, BC = 4\text{ cm}$, and $AC = 8\text{ cm}$.
• By the Triangle Inequality Theorem, the sum of any two sides must exceed the third side:

$$AB + BC = 3 + 4 = 7\text{ cm}$$

$$\text{Third side } AC = 8\text{ cm}$$

• Since $3 + 4 = 7 < 8\text{ cm}$, $\triangle ABC$ cannot exist.
• Consequently, the quadrilateral $ABCD$ cannot be constructed.

• Minimum number: $3$ independent measurements.
• Valid Sets of Data:
1. Two adjacent sides and the included angle (e.g., $AB, BC, \angle B$).
2. Two adjacent sides and one diagonal (e.g., $AB, BC, AC$).
3. One side and both diagonals (e.g., $AB, AC, BD$).

• An angle of $75^\circ$ lies exactly halfway between $60^\circ$ and $90^\circ$:

$$\frac{60^\circ + 90^\circ}{2} = \frac{150^\circ}{2} = 75^\circ$$

• Therefore, first construct a $60^\circ$ arc and a $90^\circ$ ray, then bisect the $30^\circ$ angle between the $60^\circ$ and $90^\circ$ rays.

• Steps:
1. Draw base segment $AB = 6\text{ cm}$.
2. At vertex $B$, construct a $90^\circ$ right angle ray $BX$ using compass.
3. With center $A$ and radius $7.5\text{ cm}$ (diagonal length), draw an arc intersecting ray $BX$ at point $C$.
4. With center $C$ and radius $6\text{ cm}$, draw an arc to the left.
5. With center $A$ and radius equal to $BC$, draw an arc intersecting the previous arc at $D$.
6. Join $AD$ and $CD$. $ABCD$ is the required rectangle.

• In ICSE board examinations, marks are specifically allocated for the demonstration of correct geometric construction technique (such as intersecting arcs for angle bisection, perpendicular bisectors, and point triangulation).
• Erasing construction arcs makes it impossible for the examiner to verify whether the drawing was drafted using a compass or improperly measured using a protractor/ruler, leading to loss of method marks.

• No. Knowing only that all four sides are $5\text{ cm}$ is insufficient because a rhombus requires two independent measurements.
• Four equal sides can form an infinite family of rhombuses with varying interior angles (from needle-thin rhombuses with angles near $0^\circ$ to a square at $90^\circ$). An angle or a diagonal length is mandatory.