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Practical Geometry - Construction of Triangles (SSS criterion)

Grade 7CBSE

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

🔑Concepts

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The SSS (Side-Side-Side) criterion states that a triangle can be uniquely constructed if the lengths of all three of its sides are known.

A triangle ABC with sides labeled a, b, and c.
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Before starting construction, always verify the Triangle Inequality Theorem: the sum of the lengths of any two sides must be strictly greater than the length of the third side. If a+b≤ca + b \le c, no triangle can be formed.

Two short lines placed end-to-end that fail to span the length of a third longer line.
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Construction Step 1: Draw the longest side as a base using a ruler. Label the endpoints (e.g., QQ and RR for side QRQR).

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Construction Step 2: Use a compass. Set the width to the second side length, place the pointer on one endpoint, and draw an arc above the base line.

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Construction Step 3: Adjust the compass to the third side length. Place the pointer on the other endpoint and draw a second arc intersecting the first one. The intersection point is the third vertex.

📐Formulae

Condition for existence: side1+side2>side3side_{1} + side_{2} > side_{3}

Triangle Inequality 1: AB+BC>ACAB + BC > AC

Triangle Inequality 2: BC+AC>ABBC + AC > AB

Triangle Inequality 3: AC+AB>BCAC + AB > BC

Perimeter of the triangle: P=AB+BC+CAP = AB + BC + CA

💡Examples

Problem 1:

Construct a triangle ABCABC such that AB=5 cmAB = 5\text{ cm}, BC=6 cmBC = 6\text{ cm}, and AC=7 cmAC = 7\text{ cm}.

Solution:

Step 1: Check the inequality: 5+6>75 + 6 > 7, 6+7>56 + 7 > 5, and 5+7>65 + 7 > 6. Since all are true, construction is possible. \nStep 2: Draw a line segment BC=6 cmBC = 6\text{ cm} using a ruler. \nStep 3: With BB as the center and a radius of 5 cm5\text{ cm} (length of ABAB), draw an arc using a compass. \nStep 4: With CC as the center and a radius of 7 cm7\text{ cm} (length of ACAC), draw another arc cutting the previous arc at point AA. \nStep 5: Join ABAB and ACAC using a ruler.

Explanation:

This follows the SSS construction method. We start with the base BCBC and use the compass to find the exact point AA that is simultaneously 5 cm5\text{ cm} away from BB and 7 cm7\text{ cm} away from CC.

Problem 2:

Determine if a triangle can be constructed with sides 3 cm3\text{ cm}, 4 cm4\text{ cm}, and 8 cm8\text{ cm}.

Solution:

Step 1: Identify the lengths: a=3a = 3, b=4b = 4, c=8c = 8. \nStep 2: Apply the Triangle Inequality Property: a+b=3+4=7a + b = 3 + 4 = 7. \nStep 3: Compare the sum to the third side: 7<87 < 8. \nStep 4: Since the sum of the two shorter sides is not greater than the third side, the condition a+b>ca + b > c is not satisfied.

Explanation:

In SSS construction, if the sum of two sides is less than or equal to the third side, the arcs drawn from the endpoints of the base will never meet. Therefore, a triangle cannot be formed.

Problem 3:

Construct an equilateral triangle PQRPQR with each side measuring 5 cm5\text{ cm}.

Construction of equilateral triangle PQR with intersecting arcs above the base.

Solution:

  1. Draw a line segment QR=5 cmQR = 5\text{ cm}.
  2. Taking QQ as center and radius 5 cm5\text{ cm}, draw an arc.
  3. Taking RR as center and radius 5 cm5\text{ cm}, draw another arc to intersect the previous arc at PP.
  4. Join PQPQ and PRPR.

Explanation:

In an equilateral triangle, all three sides are equal. Since PQ=QR=RP=5 cmPQ = QR = RP = 5\text{ cm}, we use the SSS construction method with equal radii for both arcs.

Problem 4:

Construct an isosceles triangle XYZXYZ where XY=XZ=6 cmXY = XZ = 6\text{ cm} and YZ=4 cmYZ = 4\text{ cm}.

Isosceles triangle XYZ with base 4cm and equal sides 6cm.

Solution:

  1. Draw the base YZ=4 cmYZ = 4\text{ cm}.
  2. With YY as center and radius 6 cm6\text{ cm}, draw an arc.
  3. With ZZ as center and radius 6 cm6\text{ cm}, draw an arc intersecting the first arc at XX.
  4. Join XYXY and XZXZ.

Explanation:

Since two sides are equal (6 cm6\text{ cm} each), the arcs drawn from YY and ZZ will have the same radius, making the triangle symmetric about the perpendicular bisector of the base.