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Practical Geometry - Construction of Triangles (ASA criterion)

Grade 7CBSE

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

🔑Concepts

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The ASA (Angle-Side-Angle) criterion states that a unique triangle can be constructed if two angles and the length of the side included between them are given.

Diagram showing a triangle with two angles and the included side highlighted for ASA construction.
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Before starting the construction, always check if the sum of the two given angles is less than 180∘180^{\circ}. If ∠1+∠2≥180∘\angle_{1} + \angle_{2} \ge 180^{\circ}, a triangle cannot be formed.

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If two angles are given but the side provided is not the included side, use the Angle Sum Property (180∘−(sum of given angles)180^{\circ} - (\text{sum of given angles})) to find the angle adjacent to the given side.

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Steps of Construction: 1. Draw the given side as a base line segment. 2. Use a protractor or compass to draw the two given angles at the endpoints of the segment. 3. Extend the rays until they intersect at the third vertex.

📐Formulae

Angle Sum Property: ∠A+∠B+∠C=180∘\text{Angle Sum Property: } \angle A + \angle B + \angle C = 180^{\circ}

Condition for construction: ∠1+∠2<180∘\text{Condition for construction: } \angle_{1} + \angle_{2} < 180^{\circ}

Calculation of third angle: ∠unknown=180∘−(∠given1+∠given2)\text{Calculation of third angle: } \angle_{unknown} = 180^{\circ} - (\angle_{given1} + \angle_{given2})

💡Examples

Problem 1:

Construct △XYZ\triangle XYZ given m∠X=30∘m\angle X = 30^{\circ}, m∠Y=100∘m\angle Y = 100^{\circ} and XY=6 cmXY = 6\text{ cm}.

Solution:

Step 1: Draw a line segment XYXY of length 6 cm6\text{ cm} using a ruler. Step 2: At point XX, use a protractor to draw a ray XPXP making an angle of 30∘30^{\circ} with XYXY. Step 3: At point YY, use a protractor to draw a ray YQYQ making an angle of 100∘100^{\circ} with YXYX. Step 4: The point where rays XPXP and YQYQ intersect is the vertex ZZ. Step 5: Label the triangle XYZXYZ with the given dimensions.

Explanation:

Since XYXY is the side included between ∠X\angle X and ∠Y\angle Y, we can directly use the ASA criterion. We verify that 30∘+100∘=130∘30^{\circ} + 100^{\circ} = 130^{\circ}, which is less than 180∘180^{\circ}, so the construction is possible.

Problem 2:

Construct △ABC\triangle ABC where BC=5 cmBC = 5\text{ cm}, m∠B=60∘m\angle B = 60^{\circ}, and m∠A=80∘m\angle A = 80^{\circ}.

Solution:

Step 1: Calculate ∠C\angle C using the Angle Sum Property: ∠C=180∘−(60∘+80∘)=180∘−140∘=40∘\angle C = 180^{\circ} - (60^{\circ} + 80^{\circ}) = 180^{\circ} - 140^{\circ} = 40^{\circ}. Step 2: Draw the base BC=5 cmBC = 5\text{ cm} using a ruler. Step 3: At BB, draw a ray making 60∘60^{\circ} with BCBC. Step 4: At CC, draw a ray making 40∘40^{\circ} with CBCB. Step 5: Mark the intersection of these two rays as AA.

Explanation:

In this problem, the given side BCBC is not included between ∠B\angle B and ∠A\angle A. To use the ASA criterion, we must find ∠C\angle C (the angle adjacent to side BCBC along with ∠B\angle B) before drawing.

Problem 3:

Construct △PQR\triangle PQR where PQ=7 cmPQ = 7\text{ cm}, m∠P=45∘m\angle P = 45^{\circ}, and m∠Q=75∘m\angle Q = 75^{\circ}.

Construction of triangle PQR with base 7cm and base angles 45 and 75 degrees.

Solution:

  1. Draw a line segment PQ=7 cmPQ = 7\text{ cm}.
  2. At point PP, draw a ray making an angle of 45∘45^{\circ} with PQPQ.
  3. At point QQ, draw a ray making an angle of 75∘75^{\circ} with QPQP.
  4. Mark the point where the two rays intersect as RR. △PQR\triangle PQR is the required triangle.

Explanation:

This is a direct application of the ASA criterion because the side PQPQ lies between the given angles PP and QQ.

Problem 4:

Construct △LMN\triangle LMN given LM=5.5 cmLM = 5.5\text{ cm}, m∠M=60∘m\angle M = 60^{\circ}, and m∠N=50∘m\angle N = 50^{\circ}.

Construction of triangle LMN showing calculated angle L of 70 degrees and given angle M of 60 degrees.

Solution:

  1. First, find ∠L\angle L using the Angle Sum Property: m∠L=180∘−(60∘+50∘)=70∘m\angle L = 180^{\circ} - (60^{\circ} + 50^{\circ}) = 70^{\circ}.
  2. Draw LM=5.5 cmLM = 5.5\text{ cm}.
  3. Construct ∠M=60∘\angle M = 60^{\circ} and ∠L=70∘\angle L = 70^{\circ} at the respective endpoints.
  4. Their intersection point is NN.

Explanation:

Since the given side LMLM is not the included side for angles MM and NN, we must first calculate ∠L\angle L so that we have two angles adjacent to the known side LMLM.