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Practical Geometry - Construction of Triangles (SAS criterion)

Grade 7CBSE

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

🔑Concepts

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Definition of SAS Criterion: SAS stands for Side-Angle-Side. This criterion states that a unique triangle can be constructed if we are given the lengths of two sides and the measure of the angle included between them. Visually, think of this as two arms of a specific length joined at a fixed pivot point (the angle).

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The Importance of the Included Angle: For the SAS construction to work, the given angle must be the 'included angle', which is the angle located between the two known sides. If the angle is not between the two sides, the triangle might not be unique or even possible to construct. Visually, the angle is the 'hinge' where the two known side lengths meet.

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The Rough Sketch: Before starting the actual construction with tools, always draw a freehand rough sketch. Label the vertices (e.g., AA, BB, CC) and mark the given measurements. This provides a mental map of how the triangle will look, with the base at the bottom and the angle at one of the base vertices.

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Base Construction: The first physical step is drawing one of the given sides (usually the longer one) as a horizontal line segment using a ruler. This segment serves as the foundation of the triangle. For example, if BC=6 cmBC = 6 \text{ cm} is given, draw a straight line segment and label the endpoints BB and CC.

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Angle Ray Construction: At one endpoint of the base, use a protractor (or compass for standard angles) to mark the given angle. Draw a ray (a line starting from the vertex and extending indefinitely) through this mark. This ray represents the path along which the second side of the triangle will lie.

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Marking the Third Vertex: To find the exact location of the third vertex, set your compass to the length of the second given side. Place the metal pointer on the vertex where the angle was drawn and swing an arc to intersect the ray. The intersection point is the third vertex. Visually, the arc acts as a precise distance marker along the angular path.

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Closing the Triangle: Once the third vertex is located, use a ruler to draw a straight line segment connecting it to the other endpoint of the base. This completes the three-sided closed figure. The final shape should clearly show the two specified side lengths and the specific angle degree at the correct vertex.

📐Formulae

SAS Congruence Criterion: ΔABC≅ΔPQR\Delta ABC \cong \Delta PQR if AB=PQAB = PQ, ∠B=∠Q\angle B = \angle Q, and BC=QRBC = QR

Sum of Interior Angles: ∠A+∠B+∠C=180∘\angle A + \angle B + \angle C = 180^{\circ}

Triangle Inequality (Requirement for existence): Side1+Side2>Side3Side_1 + Side_2 > Side_3

💡Examples

Problem 1:

Construct a triangle ΔABC\Delta ABC given AB=5 cmAB = 5 \text{ cm}, BC=7 cmBC = 7 \text{ cm}, and ∠B=60∘\angle B = 60^{\circ}.

Solution:

Step 1: Draw a rough sketch of ΔABC\Delta ABC and label the given parts. Step 2: Draw a line segment BCBC of length 7 cm7 \text{ cm} using a ruler. Step 3: At point BB, draw a ray BXBX making an angle of 60∘60^{\circ} with BCBC using a protractor. Step 4: With BB as center and a radius of 5 cm5 \text{ cm} (the length of ABAB), draw an arc using a compass to cut the ray BXBX at point AA. Step 5: Join ACAC using a ruler to complete the triangle.

Explanation:

We start with the longest side BCBC as the base. Since ∠B\angle B is given, we must construct the angle at vertex BB. The arc of 5 cm5 \text{ cm} ensures that the side ABAB is exactly the required length before we close the triangle by joining AA to CC.

Problem 2:

Construct an isosceles triangle ΔPQR\Delta PQR where the two equal sides PQPQ and PRPR are 6 cm6 \text{ cm} each and the angle between them is 110∘110^{\circ}.

Solution:

Step 1: Draw a line segment PQ=6 cmPQ = 6 \text{ cm}. Step 2: At point PP, use a protractor to draw a ray PYPY such that ∠QPY=110∘\angle QPY = 110^{\circ}. Step 3: Use a compass set to 6 cm6 \text{ cm} width. With PP as center, draw an arc cutting ray PYPY at point RR. Step 4: Join QRQR. Step 5: ΔPQR\Delta PQR is the required isosceles triangle with PQ=PR=6 cmPQ = PR = 6 \text{ cm} and ∠P=110∘\angle P = 110^{\circ}.

Explanation:

In an isosceles triangle with a given included angle, we treat the two equal sides as the two sides of the SAS criterion. Since ∠P\angle P is the angle between PQPQ and PRPR, it must be constructed at the shared vertex PP.