## 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.

Answer:

2. To construct the bisector of any given angle. Question : Draw ∠ABC = 115°, construct its bisector.

Answer:

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.

Answer:

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.

Answer:

5. To construct an angle congruent to the given angle.

Question: Construct ∠PQR congruent to given ∠LMN.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

**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:**

- Draw a circle with centre O.
- Take any point P on it.
- Draw ray OP.
- 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:

- Draw a circle of radius 3 cm. Take a point P on it.
- Draw any chord AP and ∠ABP in the alternate segment.
- With B as the centre and any convenient radius, draw an arc intersecting the sides BA and BP at points M end N respectively.
- Using the same radius and centre P, draw an arc intersecting chord PA at point R.
- 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 :**

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

Answer: