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Practical Geometry - Construction of Quadrilaterals given different conditions (sides, diagonals, angles)

Grade 8ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Fundamental Requirement for Unique Construction: To construct a unique quadrilateral, exactly five independent measurements (elements) are required. These can be combinations of sides, diagonals, and angles. For example, knowing only four sides is insufficient because the shape can 'flex' or 'tilt' into different parallelograms unless a fifth measurement, like a diagonal, is provided to lock the structure.

The Role of the Rough Sketch: Before starting the actual construction with a compass and ruler, always draw a freehand rough sketch. Label the vertices in cyclic order (e.g., A,B,C,DA, B, C, D going clockwise). This visual map helps you identify which triangles can be constructed first using the SSS (Side-Side-Side) or SAS (Side-Angle-Side) criteria.

Construction using Four Sides and One Diagonal: This is the most common condition where the diagonal divides the quadrilateral into two distinct triangles. Visualize the quadrilateral as two triangles, ABC\triangle ABC and ADC\triangle ADC, sharing a common base ACAC. You first construct ABC\triangle ABC using the SSS criterion and then use the remaining two sides to find vertex DD relative to the fixed points AA and CC.

Construction using Three Sides and Two Diagonals: When two diagonals and three sides are given, the strategy shifts to constructing a triangle formed by two diagonals and one side. For instance, if you know sides AB,BC,CDAB, BC, CD and diagonals AC,BDAC, BD, you first construct ABC\triangle ABC. Then, using BB and CC as centers, you draw arcs with radii equal to BDBD and CDCD respectively to find the intersection point DD.

Construction using Two Adjacent Sides and Three Angles: In this case, you must start by drawing the side that has two known angles at its endpoints. For example, if ABAB is the base and A\angle A and B\angle B are known, draw ABAB first, then construct the rays for both angles. Use the second side length to mark a point on one of these rays, and from that new vertex, construct the third given angle to find the final intersection point.

Properties of Special Quadrilaterals: For specific shapes like Parallelograms, Rhombuses, and Squares, you may be given fewer than five measurements because their properties provide the hidden values. For a Parallelogram, opposite sides and angles are equal. For a Rhombus, all sides are equal and diagonals bisect each other at 9090^{\circ} (right angles). For a Square, all sides are equal and every interior angle is 9090^{\circ}.

Using the Angle Sum Property: If you are given three angles of a quadrilateral, you can always find the fourth angle using the formula Fourth Angle=360(Sum of three given angles)\text{Fourth Angle} = 360^{\circ} - (\text{Sum of three given angles}). This is particularly useful when the side you want to use as a base doesn't have its corresponding angles explicitly stated in the problem.

📐Formulae

Sum of interior angles of a quadrilateral: A+B+C+D=360\angle A + \angle B + \angle C + \angle D = 360^{\circ}

Area of a general quadrilateral given a diagonal dd and perpendiculars h1,h2h_1, h_2 from opposite vertices: Area=12×d×(h1+h2)Area = \frac{1}{2} \times d \times (h_1 + h_2)

Area of a Rhombus using diagonals d1d_1 and d2d_2: Area=12×d1×d2Area = \frac{1}{2} \times d_1 \times d_2

Area of a Parallelogram: Area=base×heightArea = \text{base} \times \text{height}

Property of Rhombus Diagonals: Diagonal1Diagonal2\text{Diagonal}_1 \perp \text{Diagonal}_2 (They are perpendicular bisectors of each other)

💡Examples

Problem 1:

Construct a quadrilateral ABCDABCD where AB=4 cm,BC=6 cm,CD=5 cm,DA=5.5 cmAB = 4 \text{ cm}, BC = 6 \text{ cm}, CD = 5 \text{ cm}, DA = 5.5 \text{ cm} and diagonal AC=7 cmAC = 7 \text{ cm}.

Solution:

  1. Draw a line segment AB=4 cmAB = 4 \text{ cm}.
  2. With BB as center and radius 6 cm6 \text{ cm}, draw an arc.
  3. With AA as center and radius 7 cm7 \text{ cm} (the diagonal), draw another arc to intersect the previous arc at point CC.
  4. Join BCBC and ACAC. This completes ABC\triangle ABC.
  5. Now, with AA as center and radius 5.5 cm5.5 \text{ cm}, draw an arc on the side opposite to BB.
  6. With CC as center and radius 5 cm5 \text{ cm}, draw an arc to intersect the previous arc at point DD.
  7. Join ADAD and CDCD.
  8. ABCDABCD is the required quadrilateral.

Explanation:

This construction uses the 'Four Sides and One Diagonal' condition. The diagonal ACAC splits the quadrilateral into two triangles, ABC\triangle ABC and ADC\triangle ADC, both of which are constructed using the SSS (Side-Side-Side) rule.

Problem 2:

Construct a quadrilateral PQRSPQRS where PQ=3.5 cm,QR=6.5 cm,P=75,Q=105PQ = 3.5 \text{ cm}, QR = 6.5 \text{ cm}, \angle P = 75^{\circ}, \angle Q = 105^{\circ} and R=120\angle R = 120^{\circ}.

Solution:

  1. Draw line segment PQ=3.5 cmPQ = 3.5 \text{ cm}.
  2. At point QQ, construct an angle PQR=105\angle PQR = 105^{\circ} using a protractor or compass.
  3. From the ray starting at QQ, cut off a length QR=6.5 cmQR = 6.5 \text{ cm} using a compass.
  4. At point RR, construct an angle of 120120^{\circ} relative to the segment QRQR.
  5. At point PP, construct an angle of 7575^{\circ} relative to the segment PQPQ.
  6. Extend the rays from point PP and point RR until they intersect. Label the point of intersection as SS.
  7. PQRSPQRS is the required quadrilateral.

Explanation:

This construction follows the 'Two Adjacent Sides and Three Angles' condition. By starting with side PQPQ and its adjacent angles, we establish the positions of P,Q,P, Q, and RR. The final vertex SS is found by the intersection of the rays formed by the remaining angles.