CLASS 6TH | UNIT 4: GEOMETRY | TRIANGLES (INCLUDING TYPES,PROPERTIES AND CONSTRUCTION) | REVISION NOTES

Construction of geometrical objects is an important part of geometry. This is nothing but practical geometry. Every object is constructed using certain instruments following certain rules and methods. In this chapter, you will learn about how to construct line segments, angles, triangles and circles by using tools such as ruler and compass.

Things To Remember

• Constructing a circle given its radius:
  1. Open the compasses and measure the required radius, say 3 cm.
  2. Mark a point O with a sharp pencil where we want the centre of the circle to be.
  3. Place the pointer of the compasses on O.

• Rotate the compasses to draw the circle. Take care to complete the movement around in one instant.

• Constructing a line segment of given length:
  1. Draw a line l and mark a point A on it.
  2. Place the compasses pointer on zero mark of the ruler. Open it to place the pencil point upto the required length.
  3. Taking caution that the opening of the compasses has not changed, place the pointer of the compasses at A and make an arc to cut l at B.
  4. AB is the line segment of the required length.
• Constructing a copy of a given line segment:
  1. Given \ov{AB} whose length is not known.
  2. Fix the compasses pointer on A and the pencil end on B. The opening of the instrument now gives the length of \ov {AB}.
  3. Draw any line l. Choose a point C on l. Without changing the compasses setting, place the pointer on C.
  4. Swing an arc that cuts l at a point, say, D. Now \ov{CD} is a copy of \ov{AB}.
• Constructing perpendicular to a line through a point on it:
  1. Given a point P on a line l.
  2. With P as the centre and a convenient radius, construct an arc intersecting the line l at two points A and B.
  3. With A and B as centres and a radius greater than AP construct two arcs, which cut each other at Q.
  4. Join PQ. Then PQ is perpendicular to l. We write {PQ}↖{↔} ⊥ l.
• Constructing perpendicular to a line through a point not on it:
  1. Given a line l and a point P not on it.
  2. With P as the centre, draw an arc which intersects line l at two points A and B.
  3. Using the same radius and with A and B as centres, construct two arcs that intersect at a point, say Q, on the other side.
  4. Join PQ. Then PQ is perpendicular to l. We write {PQ}↖{↔} ⊥ l.
• Constructing the perpendicular bisector of a line segment:
  1. Draw a line segment \ov{AB} of any length.
  2. With A as the centre, using compasses, draw a circle. The radius of your circle should be more than half the length of \ov{AB}.
  3. With the same radius and with B as the centre, draw another circle using compasses. Let it cut the previous circle at C and D.
  4. Join \ov{CD}. It cuts \ov{AB} at O. Use your divider to verify that O is the midpoint of \ov{AB}. Also, verify that ∠COA and ∠COB are right angles. Therefore, \ov{CD} is the perpendicular bisector of \ov{AB}.
• Constructing an angle of a given measure:
  1. Draw \ov{AB} of any length.
  2. Place the centre of the protractor at A and the zero edge along \ov{AB}.
  3. Start with zero near B. Mark point C at the required angle measure, say 40°.
  4. Join AC. ∠BAC is the required angle.
• Constructing a copy of an angle of unknown measure:
  Given ∠A, whose measure is not known.
  1. Draw a line l and choose a point P on it.
  2. Place the compasses at A and draw an arc to cut the rays of ∠A at B and C.
  3. Use the same compasses setting to draw an arc with P as the centre, cutting l in Q.
  4. Set your compasses to the length BC with the same radius.
  5. Place the compasses pointer at Q and draw the arc to cut the arc drawn earlier in R.
  6. Join PR. This gives us ∠P. It has the same measure as ∠A. This means ∠QPR has the same measure as ∠BAC.

• Constructing bisector of an angle:
  Let an angle, say, ∠A be given.
  1. With A as the centre and using compasses, draw an arc that cuts both rays of ∠A. Label the points of intersection as B and C.
  2. With B as the centre, draw (in the interior of ∠A) an arc whose radius is more than half the length BC.
  3. With the same radius and with C as the centre, draw another arc in the interior of ∠A. Let the two arcs intersect at D. Then AD is the required bisector of ∠A.
• Constructing a 60° angle:
  1. Draw a line l and mark a point O on it.
  2. Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line at a point say, A.
  3. With the pointer at A (as centre), now draw an arc that passes through O.
  4. Let the two arcs intersect at B. Join OB. We get ∠BOA whose measure is 60°.
• Constructing a 30° angle:
  1. Construct an angle of 60° as shown earlier. Now, bisect this angle. Each angle is 30°.
• Constructing a 120° angle:
  1. Draw any line PQ and take a point O on it.
  2. Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line at A.
  3. Without disturbing the radius on the compasses, draw an arc with A as the centre which cuts the first arc at B.
  4. Again without disturbing the radius on the compasses and with B as the centre, draw an arc which cuts the first arc at C.
  5. Join OC, ∠COA is the required angle whose measure is 120°.
• Constructing a 90° angle:
  1. Construct a perpendicular to a line from a point lying on it, as discussed earlier. This is the required 90° angle.
• Constructing a triangle given all three sides:
  Constructing a ΔABC such that AB = 6 cm, BC = 5 cm and CA = 7 cm.
  1. Draw one of the sides say AB = 6 cm.
  2. Using compasses and taking A as the centre, draw an arc of radius 7 cm.
  3. With B as the centre, draw an arc of radius 5 cm, that cuts the first arc at the point C.
  4. Join AC and BC. ΔABC is the required triangle.
• Constructing a triangle when two sides and the included angle is given:
  Constructing a ΔABC such that AB = 5 cm, ∠A = 60° and CA = 8 cm.
  1. Draw AB = 5 cm.
  2. With the help of a compass construct &angPAB = 60°.
  3. With A as centre and radius 5 cm cut AP at point C.
  4. Join BC. ΔABC is the required triangle.
• Constructing a triangle when two angles and included side is given:
  Constructing ΔABC such that AB = 4 cm, ∠A = 30°, ∠B = 60°.
  1. Draw AB = 4 cm.
  2. At A construct ∠QAB = 30°.
  3. At B construct ∠PBA = 60°.
  4. AQ and BP intersect each other at C. ΔABC is the required triangle.
• Constructing the circumcircle of a triangle:

A circle that passes through all the three vertices of a triangle is called the circumcircle of the triangle.

• The point where the perpendicular bisectors of the sides of a triangle meet is called the circumcenter. Here O is the circumcenter.
• OA = OB = OC = radius of circumcircle = Circumradius.
• Constructing the in-circle of a triangle:

A circle drawn inside a triangle such that it touches all the three sides of the triangle is called the in-circle of the triangle.

• The point where the bisectors of the angles of a triangle meet, is called incenter. Here I is the incenter.
• The length of the perpendicular, here IP, is called the inradius.

A polygon is a plane figure with a minimum of three straight lines and three angles.

A triangle is a polygon with three sides. A triangle consists 3 vertices and 3 sides encompassing 3 angles. The sum of the interior angles of a triangle is always 180 degrees.

In this chapter, you will learn about types of triangles and their properties.

Triangles: A triangle is a plane figure bounded by three line segments.

Vertex: Vertex of a triangle is a point where any two of its sides meet.

Angles:

• Sum of interior angles of a triangle is always 180°.
• An interior angle of a triangle can be represented by the letter representing the corresponding vertex.
• When any side of a triangle is extended, the angle formed outside the triangle is called an exterior angle.
• An exterior angle of a triangle is an adjacent and supplementary angle to the corresponding interior angle of a triangle.
• An exterior angle is equal to the sum of opposite interior angles.

Types of triangles according to angles

• If each angle of a triangle is acute i.e. less than 90°, it is called acute angled triangle.
• If each angle of a triangle is equal to  90°, it is called right angled triangle.
  • In a right angled triangle, the side opposite to the right angle is called hypotenuse and it is the largest side of the right angled triangle.             
• If each angle of a triangle is obtuse i.e. greater than 90°, it is called the obtuse angled triangle.

Types of triangles according to sides:

• A triangle with at least two sides equal is called isosceles triangle.
• PQ =PR,so ∠R = ∠P.
• A triangle with all sides equal is called equilateral triangle. Each interior angle of the equilateral triangle = (180°/3) = 90°.
• If the three sides of a triangle are unequal then it is called scalene triangle.

Let us consider an example:

Example:

Find the value of x in the figure given  below:

Solution:

∠ACD = ∠A + ∠B.

115° = 2x + 3x.

4x = 115°.

x = 23°.

It is important to learn construction of triangles. For the construction of a triangle, it is not mandatory to have all its dimensions and angles. A triangle can be constructed given any one of the following set of measurements:

• The length of two sides and measurement of the angle associated.
• Measurements of two angles and length of the side associated.
• For a right-angled triangle, the length of one side and the length of the hypotenuse.
• The length of all the three sides of a triangle.

In this chapter, you will learn construction of triangles with examples.

We shall be constructing a triangle when any of the following conditions are given:

• The length of three sides.
• The length of two sides and angle included between these two sides.
• Any two angles and the included side i.e. the side common to both angles.

Here are some examples:

Example 1:

Construct a triangle ABC such that BC = 4cm, AC = 6cm and AB = 7. cm.

Solution:

Steps:

1) Draw line segment AB = 7.6 cm

2) Taking A as the centre, draw an arc of radius 6 cm.

3)  Taking B as the centre, draw an arc of radius 4 cm which cuts the previous arc at C.

4) Join AC and BC.

ABC is the required triangle. (Refer to the figure on the left side).

Example 2: Construct a triangle ABC such that BC = 5 cm, ∠ABC = 60° and AB = 3cm.

Solution:

Steps:

1) Draw line segment BC  = 5 cm

2) Construct  ∠PBC = 60°

3)  Taking B as the centre, draw an arc of radius 3 cm which cuts BP at A such that BA = 3 cm.

4) Join A and C.

ABC is the required triangle. (Refer to the figure on the right side).