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A Tale of Three Intersecting Lines - Construction of Triangles When Some Sides and Angles are Known

Grade 7CBSE

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

🔑Concepts

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The SSS (Side-Side-Side) Criterion: A triangle can be constructed if the lengths of all three sides are known, provided the sum of any two sides is greater than the third side (a+b>ca + b > c).

A triangle ABC showing sides a, b, and c used for SSS construction.
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The SAS (Side-Angle-Side) Criterion: Construction is possible when two sides and the angle included between them are given. The angle must be formed by the two known sides.

Triangle illustrating the SAS criterion with two sides and an included angle.
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The ASA (Angle-Side-Angle) Criterion: Construction requires one side and the two angles at its endpoints. If two angles are given but not the included side, use the Angle Sum Property (∠A+∠B+∠C=180∘\angle A + \angle B + \angle C = 180^\circ) to find the third angle.

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The RHS (Right-angle Hypotenuse Side) Criterion: Specifically for right-angled triangles, you need the length of the hypotenuse and one other side. The angle between the base and perpendicular is always 90∘90^\circ.

📐Formulae

a+b>ca + b > c

∠A+∠B+∠C=180∘\angle A + \angle B + \angle C = 180^\circ

a2+b2=c2 (For RHS construction using Pythagoras Theorem)a^2 + b^2 = c^2 \text{ (For RHS construction using Pythagoras Theorem)}

💡Examples

Problem 1:

Construct a triangle △ABC\triangle ABC where AB=5 cmAB = 5\text{ cm}, BC=6 cmBC = 6\text{ cm}, and AC=7 cmAC = 7\text{ cm}.

Solution:

  1. Draw a line segment BC=6 cmBC = 6\text{ cm}. 2. With BB as center and radius 5 cm5\text{ cm}, draw an arc. 3. With CC as center and radius 7 cm7\text{ cm}, draw another arc intersecting the previous arc at point AA. 4. Join ABAB and ACAC.

Explanation:

This follows the SSS (Side-Side-Side) criterion. Since 5+6>75 + 6 > 7, the triangle construction is possible.

Problem 2:

Construct △PQR\triangle PQR given PQ=3 cmPQ = 3\text{ cm}, QR=5.5 cmQR = 5.5\text{ cm}, and ∠PQR=60∘\angle PQR = 60^\circ.

Solution:

  1. Draw a line segment QR=5.5 cmQR = 5.5\text{ cm}. 2. At point QQ, use a protractor or compass to draw an angle of 60∘60^\circ. 3. From QQ, cut an arc of length 3 cm3\text{ cm} on the angle ray to mark point PP. 4. Join PRPR.

Explanation:

This is an SAS construction because we are given two sides and the angle included between them.

Problem 3:

In △XYZ\triangle XYZ, ∠X=30∘\angle X = 30^\circ and ∠Y=100∘\angle Y = 100^\circ. If the side XY=6 cmXY = 6\text{ cm}, find the third angle and describe the construction.

Solution:

First, find ∠Z\angle Z: 180∘−130∘50∘\begin{array}{r} 180^\circ \\- 130^\circ \\ \hline 50^\circ \end{array}

  1. Draw XY=6 cmXY = 6\text{ cm}. 2. At XX, draw a ray at 30∘30^\circ. 3. At YY, draw a ray at 100∘100^\circ. 4. The intersection point is ZZ.

Explanation:

This uses the ASA criterion. We must ensure the sum of the two given angles is less than 180∘180^\circ for the triangle to exist.

Problem 4:

Construct a right-angled triangle LMNLMN, where ∠M=90∘\angle M = 90^\circ, MN=4 cmMN = 4\text{ cm}, and hypotenuse LN=5 cmLN = 5\text{ cm}.

A right-angled triangle LMN with hypotenuse 5cm and base 4cm.

Solution:

  1. Draw a line segment MN=4 cmMN = 4\text{ cm}.
  2. At point MM, draw a ray MXMX perpendicular to MNMN (making an angle of 90∘90^\circ).
  3. With NN as the center and a radius of 5 cm5\text{ cm} (the hypotenuse), draw an arc that intersects the ray MXMX at point LL.
  4. Join LNLN to complete the triangle LMNLMN.

Explanation:

This construction uses the RHS (Right-angle Hypotenuse Side) property. By drawing the 90∘90^\circ angle first, we establish the direction of the perpendicular side, and the hypotenuse arc fixes the height of the triangle.

Problem 5:

Construct an isosceles triangle DEFDEF where the equal sides DEDE and DFDF are each 5 cm5\text{ cm} and the angle between them is ∠D=40∘\angle D = 40^\circ.

Isosceles triangle DEF with side lengths 5cm and vertex angle 40 degrees.

Solution:

  1. Draw a ray DXDX.
  2. Mark a point EE on DXDX such that DE=5 cmDE = 5\text{ cm}.
  3. At point DD, use a protractor to draw an angle of 40∘40^\circ with respect to DEDE. Label this ray DYDY.
  4. On ray DYDY, mark a point FF such that DF=5 cmDF = 5\text{ cm}.
  5. Join EFEF to form the triangle DEFDEF.

Explanation:

Since two sides and the included angle are given, we use the SAS criterion. Because DE=DFDE = DF, the triangle is isosceles.