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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: This criterion states that a triangle can be uniquely constructed if the lengths of all three of its sides are given. It is one of the fundamental conditions for triangle congruence and construction.

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Triangle Inequality Property: Before beginning construction, verify that the sum of the lengths of any two sides is greater than the third side (e.g., a+b>ca + b > c). If this condition is not met, the arcs drawn during construction will not intersect, and no triangle can be formed.

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The Rough Sketch: Always start by drawing a small, freehand labeled triangle. This visual aid helps in planning the placement of the vertices and choosing which side to draw as the horizontal base on the paper.

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Drawing the Base: Use a ruler to draw the longest given side as a horizontal line segment. For example, if BC=6 cmBC = 6\text{ cm} is the base, mark point BB at 00 and point CC at 6 cm6\text{ cm} on the ruler to create a straight foundation for the triangle.

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Locating the Third Vertex using Arcs: To find the top vertex (e.g., point AA), use a compass. Set the compass width to the length of the second side (ABAB) using a ruler, place the metal pointer on BB, and draw a curved arc above the base. Then, set the compass to the length of the third side (ACAC), place the pointer on CC, and draw another arc.

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Point of Intersection: The point where the two curved arcs cross each other is the third vertex of the triangle. This intersection point is exactly at the required distances from both endpoints of the base line.

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Completing the Figure: Use a ruler to join the intersection point to the endpoints of the base line segment. This creates two straight sloping sides, resulting in a closed three-sided polygon that satisfies the given measurements.

📐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.