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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

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A quadrilateral is uniquely determined if five independent measurements are given. These can be various combinations of sides, diagonals, and angles.

Quadrilateral ABCD with diagonal AC showing a construction requirement.
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Case 1: Construction given four sides and one diagonal (S−S−S−S−DS-S-S-S-D). We treat the quadrilateral as two triangles sharing a common base (the diagonal).

Illustration showing arcs being drawn from the ends of a diagonal to locate a third vertex.
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Case 2: Construction given three sides and two diagonals (S−S−S−D−DS-S-S-D-D). This involves constructing one triangle first and then finding the fourth vertex using the second diagonal.

Quadrilateral construction with two diagonals shown.
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Case 3: Construction given two adjacent sides and three angles (S−S−A−A−AS-S-A-A-A). We first construct the given sides and then use the angles at the endpoints to find the intersection for the final vertex.

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Case 4: Construction given three sides and two included angles (S−S−S−A−AS-S-S-A-A). We draw the central side, construct the two angles, and then mark off the other two sides along these angle arms.

📐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: Diagonal1⊥Diagonal2\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=105∘PQ = 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 120∘120^{\circ} relative to the segment QRQR.
  5. At point PP, construct an angle of 75∘75^{\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.

Problem 3:

Construct a quadrilateral ABCDABCD where AB=5 cm,BC=4.5 cm,CD=4 cmAB = 5 \text{ cm}, BC = 4.5 \text{ cm}, CD = 4 \text{ cm}, ∠B=100∘\angle B = 100^{\circ} and ∠C=75∘\angle C = 75^{\circ}.

Construction of quadrilateral ABCD with given angles and sides.

Solution:

  1. Draw a line segment BC=4.5 cmBC = 4.5 \text{ cm}.
  2. At point BB, construct ∠XBC=100∘\angle XBC = 100^{\circ}.
  3. On the ray BXBX, mark a point AA such that BA=5 cmBA = 5 \text{ cm}.
  4. At point CC, construct ∠YCB=75∘\angle YCB = 75^{\circ}.
  5. On the ray CYCY, mark a point DD such that CD=4 cmCD = 4 \text{ cm}.
  6. Join AA and DD. ABCDABCD is the required quadrilateral.

Explanation:

This is a Case 4 construction (three sides and two included angles). We start with side BCBC because the angles at both its endpoints are known.

Problem 4:

Construct a rhombus PQRSPQRS with diagonals PR=6 cmPR = 6 \text{ cm} and QS=8 cmQS = 8 \text{ cm}.

Rhombus PQRS showing perpendicular bisecting diagonals.

Solution:

  1. Draw a line segment QS=8 cmQS = 8 \text{ cm}.
  2. Construct the perpendicular bisector of QSQS. Let it meet QSQS at point OO.
  3. Since diagonals of a rhombus bisect each other at right angles, mark points PP and RR on the bisector such that OP=OR=62=3 cmOP = OR = \frac{6}{2} = 3 \text{ cm}.
  4. Join PQ,QR,RSPQ, QR, RS and SPSP. PQRSPQRS is the required rhombus.

Explanation:

A rhombus is a special quadrilateral. Given only two diagonals, we utilize the property that they are perpendicular bisectors of each other.