**Line segment**

A line segment is a part of a line with two endpoints.

**A line perpendicular to a line segment**

Any line which is perpendicular to a line segment makes an angle of 90°.

**Construction of a line parallel to a given line, through a point not on the line**

We need to construct it using ruler and compass only.

**Step 1:** Draw a line PQ and take a point R outside it.

**Step 2:** Take a point J on the line PQ and join it with R.

**Step 3:** Take J as a centre and draw an arc with any radius which cuts PQ at C and JR at B.

**Step 4:** Now with the same radius, draw an arc taking R as a centre.

**Step 5:** Take the measurement of BC with compass and mark an arc of the same measurement from R to cut the arc at S.

**Step 6:** Now join RS to make a line parallel to PR.

∠ARS = ∠BJC, hence RS ∥ PQ because of equal corresponding angles.

This concept is based on the fact that a transversal between two parallel lines creates a pair of equal corresponding angles.

**Remark:** This can be done by taking alternate interior angles instead of corresponding angles.

**Construction of triangles**

The construction of triangles is based on the rules of congruent triangles. A triangle can be drawn if-

- Three sides are given (SSS criterion).
- Two sides and an included angle are given (SAS criterion).
- Two angles and an included side are given. (ASA criterion).
- A hypotenuse and a side are given for right angle triangle (RHS criterion).

**Construction of a triangle with three given sides (SSS criterion)**

**Example**

Draw a triangle ABC with the sides AB = 6 cm, BC = 5 cm and AC = 9 cm.

**Solution**

**Step 1:** First of all draw a rough sketch of a triangle, so that we can understand how to go ahead.

**Step 2:** Draw a line segment AB = 6 cm.

**Step 3:** From point A, C is 9 cm away so take A as a centre and draw an arc of 9 cm.

**Step 4:** From point B, C is 5 cm away so take B as centre and draw an arc of 5 cm in such a way that both the arcs intersect with each other.

**Step 5:** This point of intersection of arcs is the required point C. Now join AC and BC.

ABC is the required triangle.

**Construction of a triangle if two sides and one included angle is given (SAS criterion)**

**Example**

Construct a triangle LMN with LM = 8 am, LN = 5 cm and ∠NLM = 60°.

**Solution**

**Step 1:** Draw a rough sketch of the triangle according to the given information.

**Step 2:** Draw a line segment LM = 8 cm.

**Step 3:** draw an angle of 60° at L and make a line LO.

**Step 4:** Take L as a centre and draw an arc of 5 cm on LO.

**Step 5:** Now join NM to make a required triangle LMN.

**Construction of a triangle if two angles and one included side is given (ASA criterion)**

**Example**

Draw a triangle ABC if BC = 8 cm, ∠B = 60°and ∠C = 70°.

**Solution**

**Step 1:** Draw a rough sketch of the triangle.

**Step 2:** Draw a line segment BC = 8 cm.

**Step 3:** Take B as a centre and make an angle of 60° with BC and join BP.

**Step 4:** Now take C as a centre and draw an angle of 70° using a protractor and join CQ. The point where BP and QC intersects is the required vertex A of the triangle ABC.

ABC is the required triangle ABC.

**Construction of a right angle triangle if the length of the hypotenuse and one side is given (RHS****criterion)**

**Example**

Draw a triangle PQR which is right angled at P, with QR =7 cm and PQ = 4.5 cm.

**Solution**

**Step 1:** Draw a rough sketch of the triangle.

**Step 2:** Draw a line segment PQ = 4.5 cm.

**Step 3:** At P, draw PS ⊥ PQ. This shows that R must be somewhere on this perpendicular.

**Step 4:** Take Q as a centre and draw an arc of 7 cm which intersects PS at R.

PQR is the required triangle.

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