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A Tale of Three Intersecting Lines - Constructing a Triangle When its Sides are Given

Grade 7CBSE

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

🔑Concepts

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A triangle can only be constructed if the sum of the lengths of any two sides is strictly greater than the length of the third side. This is known as the Triangle Inequality Property: a+b>ca + b > c, b+c>ab + c > a, and c+a>bc + a > b.

A triangle ABC with sides labeled a, b, and c to illustrate the triangle inequality property.
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The SSS (Side-Side-Side) Criterion: If the lengths of all three sides of a triangle are given and they satisfy the triangle inequality, a unique triangle can be constructed using a ruler and a compass.

Construction step showing a base line and two intersecting arcs from the endpoints.
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To construct a triangle using SSS: 1. Draw the longest side as the base. 2. From one end, draw an arc with a radius equal to the second side. 3. From the other end, draw an arc with a radius equal to the third side. 4. Connect the point of intersection of the arcs to the endpoints of the base.

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If the sum of two shorter sides is equal to the third side (a+b=ca + b = c), the 'arcs' will meet exactly on the base line, resulting in a straight line rather than a triangle. If a+b<ca + b < c, the arcs will never meet.

📐Formulae

a+b>ca + b > c

b+c>ab + c > a

c+a>bc + a > b

P=a+b+cP = a + b + c

💡Examples

Problem 1:

Construct a triangle △ABC\triangle ABC with side lengths AB=5 cmAB = 5\text{ cm}, BC=6 cmBC = 6\text{ cm}, and AC=7 cmAC = 7\text{ cm}. Verify if the construction is possible.

Solution:

First, check the Triangle Inequality:

  1. 5+6=115 + 6 = 11, and 11>711 > 7
  2. 6+7=136 + 7 = 13, and 13>513 > 5
  3. 5+7=125 + 7 = 12, and 12>612 > 6 Since all conditions are met, construction is possible. Steps:
  4. Draw a line segment BC=6 cmBC = 6\text{ cm} using a ruler.
  5. With BB as center and radius 5 cm5\text{ cm}, draw an arc using a compass.
  6. With CC as center and radius 7 cm7\text{ cm}, draw another arc intersecting the previous arc at point AA.
  7. Join ABAB and ACAC. △ABC\triangle ABC is the required triangle.

Explanation:

We use the SSS criterion. By checking the sum of the sides, we ensure the arcs will actually intersect to form a vertex.

Problem 2:

Can you construct a triangle with sides 3 cm3\text{ cm}, 4 cm4\text{ cm}, and 8 cm8\text{ cm}?

Solution:

Check the sum of the two smaller sides: 3+47\begin{array}{r} 3 \\ + 4 \\ \hline 7 \end{array} Here, 3+4=73 + 4 = 7. We compare this to the third side: 7<87 < 8. Since the sum of two sides (7 cm7\text{ cm}) is not greater than the third side (8 cm8\text{ cm}), the triangle cannot be formed.

Explanation:

According to the triangle inequality property, a+ba + b must be greater than cc. Because 3+43 + 4 is less than 88, the arcs drawn from the endpoints of the 8 cm8\text{ cm} base would never meet.

Problem 3:

Construct a triangle △PQR\triangle PQR where PQ=4 cmPQ = 4\text{ cm}, QR=3.5 cmQR = 3.5\text{ cm}, and PR=4 cmPR = 4\text{ cm}. What kind of triangle is this?

Isosceles triangle PQR with base 3.5 cm and two sides of 4 cm.

Solution:

  1. Check the property: 4+3.5=7.5>44 + 3.5 = 7.5 > 4; 4+4=8>3.54 + 4 = 8 > 3.5. Construction is possible.
  2. Draw QR=3.5 cmQR = 3.5\text{ cm} as the base.
  3. Using a compass, draw an arc of radius 4 cm4\text{ cm} from point QQ.
  4. Draw another arc of radius 4 cm4\text{ cm} from point RR.
  5. Mark the intersection as PP and join PQPQ and PRPR. Since PQ=PR=4 cmPQ = PR = 4\text{ cm}, the triangle is an Isosceles Triangle.

Explanation:

Because two sides are of equal length, the triangle formed is isosceles. The intersection of equal arcs from the base endpoints ensures the top vertex is equidistant from both.

Problem 4:

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

Equilateral triangle LMN with all sides 5 cm.

Solution:

  1. Draw base LM=5 cmLM = 5\text{ cm}.
  2. Open the compass to 5 cm5\text{ cm}.
  3. Draw an arc from LL and another arc from MM.
  4. Mark the intersection as NN.
  5. Join LNLN and MNMN. All angles will be 60∘60^{\circ}.

Explanation:

In an equilateral triangle, all three sides are equal. The construction involves setting the compass once and using that same radius for the base and both arcs.