## Geometry Chapter 5 Drawing Of Different Geometrical Figures

**Geometry Chapter 5 To draw a perpendicular line to a given line at a point on it:**

**1. Method 1. (paper folding process):**

**Drawing process:**

- Draw a line segment AB on tracing paper.
- Take point O on segment AB.
- Now folds the paper along point O such that the line segments OB coincides with OA.
- Then open the folding and draw vertical line segments OB folding line.
- The line segment PQ is the required perpendicular line segment at O on AB.

**Class 6 Math Solution WBBSE In English**

**Method – 2 (With the help of scale and set square):**

**Drawing Process:**

- Draw a straight line PQ with the help of scale and take a point A on PQ as shown in the.
- A scale is placed on PQ such that one edge of the scale coincides with PQ.
- Now a set square is placed on the scale such that the right-angled point of the set square coincides with point A.
- Then draw a line segment AB at point A along the vertical side of the set square.
- AB is the required perpendicular at A on PQ
- i.e., \(\overline{\mathrm{AB}}\) ⊥ \(\stackrel{\leftrightarrow}{\mathbf{P Q}}\)

**Method – 3. (with the help of scale and pencil compass)**

This can be done in 3 ways as described and drawn below

**Process 1:**

**Drawing process:**

- Draw a straight line PQ with the help of a scale and take a point A on it.
- With the help of a pencil compass, draw a circular arc taking any radius centered at A, so that the arc intersects the line PQ at points C and D.
- Now with the centers at G and D taking any radius greater than CA on the same side of the straight line PQ, draw two arcs and let them meet at point B.
- Join points A and B with the help of a scale.
- AB is the required perpendicular at A on PQ.

**Class 6 Math Solution WBBSE In English**

**Process 2:**

**Drawing Process :**

- Draw a straight line PQ with the help of a scale and take a point A on PQ.
- With the help of a pencil compass, draw a circular arc taking any radius centered at A so that the arc intersects the straight line PQ at points E and F.
- Now with the center at F taking the same radius as before, draw an arc that intersects the previous arc at C.
- With the center at C and taking the same radius draw an arc that intersects the previous arc at point D.
- Now with the center at D and taking the same radius, draw an arc with the help of a pencil compass and let this arc intersects the arc drawn with the center at C at point B as shown in the.
- Join points A and B with the help of a scale and produce the joining line to M as shown in the.
- AB is the required perpendicular line to PQ.
- \(\overline{\mathrm{AM}}\) ⊥ \(\stackrel{\leftrightarrow}{\mathbf{P Q}}\)

**Class 6 Math Solution WBBSE In English**

**Process 3:**

**Drawing Process:**

- Draw a straight line PQ with the help of a scale and take a point on PQ.
- Take point C outside the straight line PQ.
- With the center at C and taking the radius CA, draw a semi-circular arc that intersects the straight line PQ at points A and D respectively.
- With the help of a scale, join D and C and produce DC which intersects the semicircular arc at B as shown in the.
- With the help of a scale, join A and B.
- AB is the required perpendicular line to PQ.
- i.e., \(\overline{\mathrm{AB}}\) ⊥ \(\stackrel{\leftrightarrow}{\mathbf{P Q}}\)

## Geometry Chapter 5 Draw a perpendicular line to a given line from a point lying outside the given line

**Class 6 Math Solutions WBBSE English Medium**

**Method – 1 (Paper folding process):**

**Drawing Process:**

- Take a rectangular tracing paper.
- Draw a straight line AB on this paper and take point O outside the straight line AB.
- Now fold the paper along point O such that the mark of folding the paper will lie on both sides of line AB and the straight line on both sides of the folding should coincide.
- Now open the folding and draw a straight line along the mark of folding by a scale so that this drawing straight line intersects AB at M.
- OM is the required perpendicular straight line to AB
- i.e., \(\overline{\mathrm{OM}}\) ⊥ \(\stackrel{\leftrightarrow}{\mathbf{A B}}\)

**Method – 2 (with the help of scale and pencil compass):**

There are 2 processes that are described and drawn below:

**Class 6 Math Solutions WBBSE English Medium**

**Process 1:**

**Drawing Process :**

- With the help of a scale, draw a straight line AB and take a point O outside the line AB.
- Take a point P on that side of the straight line AB opposite to that of O.
- With the help of a pencil compass, taking a radius equal to OP and centered at O, draw a circular arc that intersects AB at points C and D respectively.
- With the center at C and D, taking the radius greater than half of the length CD draw two arcs on the side of AB where the point P lies.
- Let these two arcs intersect at N.
- Join the points O and N with the scale and let this straight line ON intersect AB at M.
- OM is the required perpendicular from O on AB
- i.e. \(\overline{\mathrm{OM}}\) ⊥ \(\stackrel{\leftrightarrow}{\mathbf{A B}}\)

**Process – 2:**

**Drawing process :**

- With the help of a scale, draw a line AB and take a point O outside the straight line AB.
- We take any two points C and D on the straight line AB.
- With the center at C and taking the radius equal to CO, draw a circular arc.
- With the center at D and taking the radius equal to DO, draw another circular arc that intersects the previous arc at point P.
- Obviously, these two arcs will intersect at O also.
- Now join the points O and P with the scale.
- Let the line segment OP intersect AB at point M.
- OM is the required perpendicular on AB.
- i.e., \(\overline{\mathrm{OM}}\) ⊥ \(\stackrel{\leftrightarrow}{\mathbf{A B}}\).

**Class 6 Math Solutions WBBSE English Medium**

**Method – 3 (with the help of scale and set square):**

**Drawing Process :**

- With the help of a scale, draw the straight line AB and take a point O outside the line AB. ,
- Place any side other than the hypotenuse of the set square such that this side coincides with AB.
- Now place a scale along the hypotenuse of the set square so that the edge of the scale coincides with the hypotenuse of the set square.
- Now press the scale strongly and ascend the set square and move if necessary so that the vertical edge of the set square coincides with point O.
- In this position, mark point M where the vertical edge of the set square intersects the line AB as shown in the.
- Now join O and M.
- OM is the required perpendicular on the straight line AB.
- i.e., \(\overline{\mathrm{OM}}\) ⊥ \(\stackrel{\leftrightarrow}{\mathbf{A B}}\).

## Geometry Chapter 5 Perpendicular-bisector

**Definition: **

- The perpendicular upon a line segment at its mid-point is called the perpendicular bisector of the line segment.
- In the above AB is a line segment and P is its mid-point.
- OP is perpendicular to the line segment AB at P.
- OP is called the perpendicular bisector, of AB.
- So a perpendicular bisector is

1. Perpendicular to the given line segment.

2. It divides the given line segment into equal parts i.e., a perpendicular bisector upon a line segment bisects the given line segment. - Now we shall discuss how to draw a perpendicular bisector to a given line segment.

**Class 6 Math Solutions WBBSE English Medium**

## Geometry Chapter 5 Draw the perpendicular bisector of a given line segment

The different methods of drawing the perpendicular bisector upon a given line segment are discussed below:

**Method – 1 (Paper folding process):**

**Drawing process:**

- We take a rectangular piece of paper and fold the paper along a horizontal line.
- Then open the folding (in along the CD).
- Along the folding, draw a line segment AB.
- Now the paper is folded vertically (in the along EF) such that point A completely falls D on point B.
- Now, open the folding paper and draw a line segment OP along the vertical folding and it intersects the line segment AB at point P.
- ∴ OP is the required perpendicular bisector of the line segment AB.

**Method – 2 (with the help of scale and pencil compass):**

**Drawing Process:**

- At first, we draw a line segment AB with the help of scale.
- With the centers at A and B respectively, taking the
- radius equal to the length AB, we draw two circular arcs with help of a pencil compass.
- Let the arcs intersect each other at points O and Q.
- Join the points O and Q with a scale.
- Let this line segment OQ intersect AB at P.
- OP is the required perpendicular bisector of the line segment AB.

**Class 6 Math Solution WBBSE**

Method – 3 (with the help of scale and pencil compass):

**Drawing Process:**

- With the help of a scale, draw a line segment AB.
- With the center at point A, taking the radius greater than, half of AB, draw two arcs, one arc on each side of the line segment AB.
- With the center at point B, taking the same radius, draw two arcs, one arc on each side of the line segment AB.
- Let these arcs intersect the previous arcs at points O and Q respectively.
- With the help of a scale, join O and Q.
- Let the line segment OQ intersect the line segment AB at P.
- OP is the required perpendicular bisector of the line segment AB.

## Geometry Chapter 5 To Draw an angle that is equal to a given angle

Let ∠AOB be a given angle. We have to draw an angle that is equal to ∠AOB.

**Drawing Process:**

- With the help of a scale, a line segment QR be drawn.
- With the center O of the angle AOB and taking any radius.
- We draw a circular arc that intersects the side OA at C and the side OB at D.
- Now with the center at Q of the line segment QR and taking a radius equal to \(\overline{\mathrm{OC}}\) or \(\overline{\mathrm{OD}}\) draw a circular arc that intersects the line segment QR at E.
- Now, with the center at E and taking the radius \(\overline{\mathrm{CD}}\), draw another circular arc that intersects the previous arc with the center at Q at F.
- Join the points Q and F by a scale and produce QF to point P.
- Then ∠PQR is the required angle.
- ∴ ∠PQR = AOB.

**Class 6 Math Solution WBBSE**

## Geometry Chapter 5 To bisect a given angle (with the help of a scale and a pencil compass)

Let ∠AOB be a given angle. We have to bisect it.

**Drawin**g Procedure:

- First, with the center at O of the ∠AOB and taking the radius, draw a circular arc.
- Let this arc intersect the side OA and the side OB of the angle ∠AOB at points C and D respectively.
- Now with the centers at points C and D and taking the radius equal to CD (or greater than half of CD) within the angle ∠AOB, we draw consecutively two arcs.
- Let these two arcs intersect each other at point Q.
- Join the points O and Q by a scale and produce OQ to point P.
- Then OP is the required bisector i.e. OP is the bisector of the ∠AOB
- ∴ ∠AOP = ∠BOP.