# Geometric Constructions Maths Notes

## Geometric Constructions Maths Notes

Let’s revise certain constructions studied in the previous standards:

1. To construct the perpendicular bisector of a given line segment.
Question: Draw seg AB of length 4.2 cm. Construct its perpendicular bisector.  2. To construct the bisector of any given angle. Question : Draw ∠ABC = 115°, construct its bisector. 3. To construct perpendicular to a line at a given point on it.
Question: Draw AB = 9.7 cm. Take a point P on it such that A-P-B and AP 3.5 cm. Through P draw a line MN perpendicular to seg AB. 4. To construct perpendicular to a line from a point P outside it.
Question: Draw line KL such that KL = 4.5 cm. Consider point P outside It. Through P, draw a line perpendicular to line KL.  5. To construct an angle congruent to the given angle.
Question: Construct ∠PQR congruent to given ∠LMN.  6. To construct a line parallel to a given line and passing through a given point outside the line.
Question: Draw a line 1, take a point P outside it. Draw a line m line 1 passing through point P.
Method 1: Method 1: 7. To divide a given Iin segment into given number of equal parts.
Question: Draw seg PQ of length 5 cm. Divide it into 4 equal parts. Here PA = AB = BC = CQ 8. To divide a line segment in the given ratio.
Question: Draw segment PQ of length 5 cm. Divide it in the ratio 3 : 2. Here M divides PQ in the ratio 3: 2.

9. To construct a triangle whose sides are given.
Question: Construct ∆ ABC such that
AB = 4.2 cm, BC = 5.3 cm and AC = 3.7 cm. Construction of similar triangle:

If two triangles are similar then corresponding sides of similar triangles are in proportion and corresponding angles are congruent.

Using this concept, a triangle similar to the given triangle can be constructed.

Construction of a tangent to a circle at a point on the circle

I. Using the centre of the circle:

Let’s understand, how to construct a tangent at a given point on the circle using the centre of the circle. Consider the following example.

Example:
Draw a circle of radius 3 cm. Mark a point P on the circle. Draw tangent to the circle through point P using the centre of the circle.
Analysis: A circle of radius 3 cm can be drawn.

Let the centre of the given circle be O and line l be the required tangent.

We know, converse of tangent theorem states that, ‘A line perpendicular to radius at its outer end is tangent’.

∴ we construct perpendicular to radius OP at point P, then line l is the required tangent. Steps of construction:

1. Draw a circle with centre O.
2. Take any point P on it.
3. Draw ray OP.
4. Draw line l.⊥ ray OP at point P.

Line l is the required tangent to the circle.

Construction: II. Without using the centre:

Let’s understand how to construct the tangent at a given point on the circle without using the centre of the circle.

Example:
Draw a circle of radius 3 cm. Take any point P on it. Draw tangent to the circle through point P without using the centre of the circle. Analysis:
Analysis: Through P, a chord can be drawn. Let it be PA. Draw any ∠PBA in the alternate segment. Now an ∠APC can be constructed congruent to ∠ ABP, then by converse of tangent secant angle theorem, line PC is the required tangent.

Steps of construction:

1. Draw a circle of radius 3 cm. Take a point P on it.
2. Draw any chord AP and ∠ABP in the alternate segment.
3. With B as the centre and any convenient radius, draw an arc intersecting the sides BA and BP at points M end N respectively.
4. Using the same radius and centre P, draw an arc intersecting chord PA at point R.
5. Taking radius equal to d(M, N) and centre R, draw an arc intersecting the arc drawn in the previous step. Let C be the point of intersection of these arcs. Draw line PC. Line PC is the required tangent to the circle.  To construct tangents to a circle from al point outside the circle:

Let’s understand, how to construct tangents from a given point in the exterior of the circle. Consider the following example.
Example:
Draw a circle of radius 3.5 cm and centre O. Mark a point P at a distance of 6 cm from the centre. Draw tangents to the circle from point P.
Analysis:
A circle of radius 3.5 cm can be drawn and point P at a distance of 6 cm can be located. Consider the given analytical figure. Suppose tangents through P touch the circle at point A and B, then ∠ OAP = ∠ OBP = 90°… (Tangent theorem) we know, ‘Angle inscribed in a semicircle is a right angle.’

∴ A and B lie on the semicircular arcs whose diameter is OP.
A and B, therefore, would be the points of intersection of those semicircular arcs with the circle.
∴ on drawing the perpendiQular bisector of seg OP we can obtain the centre and the radius of the semicircular arcs.
Points of intersection of semIcircular arcs and the circle are points A and B.
∴ tangents PA and PB can be drawn.

Steps of construction :

1. Draw a circle of radius 3.5 cm.
2. Take a point P in the exterior of the circle such that d(O, P) = 6 cm.
3. Draw seg OP. Draw perpendicular bisector of seg OP to get its midpoint M.
4. Draw an arc with radius OM ‘and centre M.
5. Mark the points of intersection of the arc with the circle as A and B.
6. Draw line PA and PB. 