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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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The SAS (Side-Angle-Side) criterion states that a unique triangle can be constructed when the lengths of two sides and the measure of the angle included between them are known.

Diagram showing two sides and the included angle forming a triangle.
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To construct an SAS triangle, start by drawing the longest known side as the base, then use a protractor or compass to mark the included angle at one endpoint.

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The order of components is vital; the angle must be 'sandwiched' between the two given sides. If the angle is not between the sides (SSA), a unique triangle may not be formed.

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Before starting construction, always draw a rough sketch to visualize the positions of the vertices A,B,CA, B, C and the given dimensions.

📐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.

Problem 3:

Construct ΔXYZ\Delta XYZ where XY=6 cmXY = 6\text{ cm}, YZ=4.5 cmYZ = 4.5\text{ cm}, and ∠Y=45∘\angle Y = 45^{\circ}.

Construction of triangle XYZ with sides 6cm, 4.5cm and angle 45 degrees.

Solution:

  1. Draw a line segment XY=6 cmXY = 6\text{ cm}.
  2. At point YY, use a protractor to draw a ray YQYQ making an angle of 45∘45^{\circ} with YXYX.
  3. Using YY as center and a radius of 4.5 cm4.5\text{ cm}, draw an arc to intersect ray YQYQ at point ZZ.
  4. Join XX and ZZ to complete the triangle ΔXYZ\Delta XYZ.

Explanation:

This follows the SAS criterion because the known angle ∠Y\angle Y is located between the two known sides XYXY and YZYZ.

Problem 4:

Construct an isosceles triangle ΔDEF\Delta DEF where DE=EF=5.5 cmDE = EF = 5.5\text{ cm} and ∠E=90∘\angle E = 90^{\circ}.

Isosceles right triangle DEF with legs of 5.5cm.

Solution:

  1. Draw base EF=5.5 cmEF = 5.5\text{ cm}.
  2. At point EE, construct a perpendicular ray (90∘90^{\circ}).
  3. From EE, mark a point DD on the ray such that ED=5.5 cmED = 5.5\text{ cm}.
  4. Join DD and FF.

Explanation:

Since DE=EFDE = EF, this is an isosceles right-angled triangle. The SAS criterion is satisfied using the two equal sides and the 90∘90^{\circ} angle between them.