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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 triangle can be uniquely constructed if the measures of two angles and the length of the side included between them are known. Visually, this looks like a horizontal base line with two diagonal rays extending from its ends to meet at a single point above it.

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The 'Included Side' is a crucial requirement; it must be the line segment that connects the vertices of the two given angles. For instance, in △ABC\triangle ABC, if ∠B\angle B and ∠C\angle C are given, the included side must be BCBC.

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Before starting the construction, always verify the Angle Sum Property. The sum of the two given angles must be strictly less than 180∘180^{\circ} for a triangle to exist. If ∠1+∠2≥180∘\angle 1 + \angle 2 \geq 180^{\circ}, the two rays will either be parallel or move away from each other, never meeting to form a third vertex.

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The construction begins by drawing the given included side as a base line using a ruler. Imagine this as the foundation of the triangle on a flat plane.

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Using a protractor or a compass, rays are drawn from each endpoint of the base line at the specified angle measures. The direction of these rays should be inward toward each other so they can eventually cross.

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The point where the two rays intersect is the third vertex of the triangle. Visually, this intersection point marks the 'peak' of the triangle, and the segments from the base to this point form the remaining two sides.

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If the two given angles are not adjacent to the given side, you must first calculate the third angle using the formula 180∘−(sum of given angles)180^{\circ} - (\text{sum of given angles}) to identify the correct angle for the ASA setup. This ensures the side becomes the 'included' side between two known angles.

📐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.