**Introduction to Construction of Triangles **

there are six elements associated with it, namely, three sides and three angles. Do we need all of these to create a triangle? If we are known, it is well and good.

In variety of practical situations, we may not know all these.

If we know only two, we cannot construct a triangle. Even if three of these are known, we may not be able to construct a triangle. For example,

For example, if two sides and an angle is given we can construct such triangle.

**Perpendicular bisector: **The line which is perpendicular to the given line segment and also bisects the line segment.

**Angle bisector: **The line which divides a complete angle into two equal angles.

**Perimeter: **Sum of all the sides of given geometrical figure or the length of the boundary of any plane figure.

**Altitude**: A perpendicular drawn from the vertex to the opposite side.

**Arc: **Part of the circle.

**Base angle: **Any of the angles formed by the base of a triangle, with the other sides.

**Vertex angle: **The angle at the top of an isosceles triangle.

**Median: **the line drawn from the vertex to the midpoint of the opposite side.

**Note**: 1) At least three parameters are needed to construct a triangle

### 3.4.2** Construction of Triangles – When three sides are given**

**Example 1: **Construct a triangle ABC in which AB = 5cm, BC = 4.3cm and AC = 4cm.

Solution:

(i) Draw a line segment which is sufficiently long using ruler.

(ii) Locate points A and B on it such that AB = 5cm.

(iii) With A as centre and radius 4cm, draw an arc(see figure)

(iv) With B as centre and radius 4.3cm, draw another arc cutting the previous arc at C.

(v) Join AC and BC

Then ABC is the required triangle:

**Construction of Triangles – Exercise 3.4.2**

**Construct a triangle ABC in which AB = 5cm and BC = 4.6 cm and AC = 3.7cm**

Solution:

(i) Draw a line segment which is sufficiently long using ruler.

(ii) Locate points A and B on it such that AB = 5cm.

(iii) With A as centre and radius 4.6cm, draw an arc(see figure)

(iv) With B as centre and radius 3.7cm, draw another arc cutting the previous arc at C.

(v) Join AC and BC

Then ABC is the required triangle

**Construct an equilateral triangle of side 4.8cm**

Solution:

(i) Draw a line segment which is sufficiently long using ruler.

(ii) Locate points A and B on it such that AB = 4.8 cm.

(iii) With A as centre and radius 4.8cm, draw an arc(see figure)

(iv) With B as centre and radius 4.8 cm, draw another arc cutting the previous arc at C.

(v) Join AC and BC

Then ABC is the required triangle:

**Construct a triangle PQR, given that PQ = 5.6cm , PR = 7cm and QR = 4.5cm**

Solution:

(i) Draw a line segment which is sufficiently long using ruler.

(ii) Locate points P and Q on it such that PQ = 5.6cm.

(iii) With P as centre and radius 7cm, draw an arc(see figure)

(iv) With Q as centre and radius 4.5cm, draw another arc cutting the previous arc at C.

(v) Join PR and QR

Then ABC is the required triangle:

**Construct a triangle XYZ in which XY = 7.8cm, YZ = 4.5 cm and XZ = 9.5cm**

Solution:

(i) Draw a line segment which is sufficiently long using ruler.

(ii) Locate points X and Z on it such that XZ = 9.5cm.

(iii) With X as centre and radius 7.8 cm, draw an arc(see figure)

(iv) With Z as centre and radius 4.5cm, draw another arc cutting the previous arc at C.

(v) Join XY and YZ

Then ABC is the required triangle:

**Construct a triangle whose perimeter is 12 cm nd the ratio of their sides is 3:4:5**

Solution:

Let the triangle be ABC with base BC and sides AB and AC

Data: Perimeter of a triangle is 12cm

ratio is 3:4:5

We know, Perimeter = sum of three sides of a triangle

= AB + BC + AC (in triangle ABC)

Therefore, 12 = 3 + 4 + 5

AB = 3/12 * 12 = 4cm

BC = 4/12 * 12 = 3cm

AC = 5/12 * 12 = 5cm

(i) Draw a line segment which is sufficiently long using ruler.

(ii) Locate points B and C on it such that BC = 3cm.

(iii) With B as centre and radius 4cm, draw an arc(see figure)

(iv) With C as centre and radius 5cm, draw another arc cutting the previous arc at C.

(v) Join AC and AB

Then ABC is the required triangle

**Construction of Triangles – When two sides and their included angle are given**

__Example 2: __Construct a triangle PQR, given that PQ = 4cm, QR = 5.2cm and ∠Q = 60˚

Solution:

Steps of Construction:

(i) Draw a line segment which is sufficiently long using ruler.

(ii) Locate points Q and R on it such that QR = 5.2cm.

(iii) At Q, construct a line segment QM, sufficiently large, such that ∠MQR at 60˚ ; use protractor to measure 60˚

(iv) With Q as centre and radius 4cm, draw the line cutting QM at P; join PR.

Then PQR is the required triangle:

**Construction of Triangles – Exercise 3.4.3**

**Construct a triangle ABC, in which AB = 4.5cm, AC = 5.5cm and ∠BAC = 75˚**

Solution:

Steps of Construction:

(i) Draw a line segment which is sufficiently long using ruler.

(ii) Locate points A and B on it such that AB = 4.5cm.

(iii) At A, construct a line segment AE, sufficiently large, such that ∠BAC at 60˚ ; use protractor to measure 60˚

(iv) With A as centre and radius 5.5cm, draw the line cutting AE at C; join BC.

Then ABC is the required triangle:

**2. Construct a triangle PQR in which PQ=5.4cm, QR = 5.5cm and PQR=55°**

Solution:

Steps of Construction:

(i) Draw a line segment which is sufficiently long using ruler.

(ii) Locate points P and Q on it such that PQ = 5.4cm.

(iii) At Q, construct a line segment QS, sufficiently large, such that ∠PQR = 55˚ ; use protractor to measure 55˚

(iv) With Q as centre and radius 5.5 cm, draw the line cutting QS at R; join PR.

Then PQR is the required triangle

**3. Construct a triangle XYZ in which XY = 5cm, YZ = 5.5cm and ∠XYZ = 100°**

Solution:

Steps of Construction:

(i) Draw a line segment which is sufficiently long using ruler.

(ii) Locate points X and Y on it such that XY = 5cm.

(iii) At Y, construct a line segment YW, sufficiently large, such that ∠XYZ = 100˚ ; use protractor to measure 100˚

(iv) With Y as centre and radius 5.5 cm, draw the line cutting YW at Z; join XZ.

Then XYZ is the required triangle:

**4. Construct a triangle LMN in which LM = 7.8 cm, MN = 6.3cm and ∠LMN = 45°**

Solution:

Steps of Construction:

(i) Draw a line segment which is sufficiently long using ruler.

(ii) Locate points M and N on it such that MN = 6.3 cm.

(iii) At M, construct a line segment MO, sufficiently large, such that ∠LMN = 45˚ ; use protractor to measure 45˚

(iv) With M as centre and radius 7.8 cm, draw the line cutting MO at N; join LN.

Then LMN is the required triangle:

**Construction of Triangles – When two angles and included side are**** give****n:**

**Example 3: **Construct a triangle XYZ in which XY = 4.5cm and ∠X = 100° and

∠Y = 50°

Solution:

Steps of Construction:

(i) Draw a line segment which is sufficiently long using ruler.

(ii) Locate points X and Y on it such that XY = 4.5 cm.

(iii) Construct a line segment XP, sufficiently large, such that ∠PXY= 100˚ ;Construct a line segment YZ, sufficiently large, such that ∠XYZ = 50˚, use protractor to measure

(iv) Extend XP and YQ to intersect at Z.

Then XYZ is the required triangle:

## Construction of Triangles – Exercise 3.4.4

1.** Construct a triangle ABC in which AB = 6.5cm. ∠A = 45° and ∠B = 60°**

Solution:

Steps of Construction:

(i) Draw a line segment which is sufficiently long using ruler.

(ii) Locate points A and B on it such that AB = 6.5 cm.

(iii) Construct a line segment AD, sufficiently large, such that ∠A = 45˚ ;Construct a line segment BE, sufficiently large, such that ∠B = 60˚, use protractor to measure

(iv) Extend AD and BE to intersect at C.

Then ABC is the required triangle:

**2. Construct a triangle PQR in which QR = 4.8 cm. ∠Q = 45° and ∠R = 55°**

Solution:

Steps of Construction:

(i) Draw a line segment which is sufficiently long using ruler.

(ii) Locate points Q and R on it such that QR = 4.8 cm.

(iii) Construct a line segment QA, sufficiently large, such that ∠Q = 45˚ ;Construct a line segment RB, sufficiently large, such that ∠R = 55˚, use protractor to measure

(iv) Extend QA and RB to intersect at P.

Then PQR is the required triangle:

**3. Construct a triangle ABC in which BC = 5.2 cm. ∠B = 35° and ∠C = 80°**

Solution:

Steps of Construction:

(i) Draw a line segment which is sufficiently long using ruler.

(ii) Locate points B and C on it such that BC = 5.2 cm.

(iii) Construct a line segment BD, sufficiently large, such that ∠B = 35˚ ;Construct a line segment CE, sufficiently large, such that ∠C = 80˚, use protractor to measure

(iv) Extend BD and CE to intersect at A.

Then ABC is the required triangle:

4. Construct a triangle ABC in which BC = 6 cm. ∠B = 30° and ∠C = 125°

Solution:

Steps of Construction:

(i) Draw a line segment which is sufficiently long using ruler.

(ii) Locate points B and C on it such that BC = 6 cm.

(iii) Construct a line segment BD, sufficiently large, such that ∠B = 30˚ ;Construct a line segment CE, sufficiently large, such that ∠R = 125˚, use protractor to measure

(iv) Extend BD and CE to intersect at A.

Then ABC is the required triangle:

**Construction of Triangles – To construct a right triangle whose one side and hypotenuse are given**

**Example 6: **Construct a right triangle LMN in which ∠M = 90, MN = 4cm and LN = 6.2cm

Solution:

Steps of construction:

- Draw a line segment XY.
- Locate M, N on XY such that MN = 4cm
- Construct a line segment MP, sufficiently MP, sufficiently large, such that ∠NMP = 40°
- With N as centre and radius 6.2cm draw an arc, cutting MP at L; join NL.

Then, LMN is the required triangle.

**Construction of Triangles – Exercise 3.4.7**

**Construct a right angle triangle ABC in which ∠B = 90° , AB = 5cm and AC = 7cm**

**So**lution:

Steps of construction:

- Draw a line segment XY.
- Locate A, B on XY such that AB = 5 cm
- Construct a line segment BD, sufficiently large, such that ∠B = 90°
- With A as centre and radius 7 cm draw an arc, cutting BD at C; join AC.

Then, ABC is the required triangle.

2.**Construct a right angle triangle PQR in which ∠R = 90° , PQ = 4cm and QR = 3cm**

Solution:

Steps of construction:

- Draw a line segment.
- Locate Q, R on XY such that QR = 3 cm
- Construct a line segment SR, sufficiently large, such that ∠R = 90°
- With Q as centre and radius 4 cm draw an arc, cutting RS at P; join PR.

Then, PQR is the required triangle.

**3.Construct a right angle triangle ABC in which ∠B = 90° , BC = 4cm and AC = 5 cm**

Solution:

Steps of construction:

- Draw a line segment XY.
- Locate B, C on XY such that BC = 4 cm
- Construct a line segment BD, sufficiently large, such that ∠B = 90°
- With C as centre and radius 5 cm draw an arc, cutting BD at A; join AC.

Then, ABC is the required triangle.

### Construction of Triangles – To construct an isosceles triangle whose base and corresponding altitude are given

**Construction of Triangles – Exercise 3.4.8**

**Construct an isosceles triangle ABC in which base BC = 6.5 cm and altitude from A on BC is 4 cm**

Solution:

Steps of construction:

- Draw a line segment BC whose length is 6.5cm
- Draw the perpendicular bisector of BC; call it XY with Y on BC
- With X as centre and radius 4 cm, draw an arc cutting XY at A; join AB and AC.
- Then ABC is the required triangle.

**2. Construct an isosceles triangle XYZ in which base YZ = 5.8 cm and altitude from X on YZ is 3.8cm**

Steps of construction:

- Draw a line segment YZ whose length is 5.8cm
- Draw the perpendicular bisector of YZ; call it AB with B on YZ
- With B as centre and radius 3.8 cm, draw an arc cutting AB at X; join XY and XZ.
- Then XYZ is the required triangle.

**3. Construct an isosceles triangle PQR in which base PQ = 7.2 cm and altitude from R on pq is 5cm**

Solution:

Steps of construction:

- Draw a line segment PQ whose length is 7.2cm
- Draw the perpendicular bisector of ; call it AB with B on YZ
- With B as centre and radius 3.8 cm, draw an arc cutting AB at X; join XY and XZ.
- Then XYZ is the required triangle.

**Construction of Triangles – To construct an isosceles triangle when its altitude and vertex angle are given**

**Example 8**: Construct an isosceles triangle whose altitude is 4cm and vertex angle is 80˚

Solution:

Steps of construction:

- Draw a line segment XY
- Take a point M on XY and draw a line MP ⊥ XY
- With M as centre and radius 4 cm, draw an arc cutting MP at A.
- Construct B and C on XY such that ∠MAB = 80˚/2 = 40˚ and ∠MAC = 80˚/2 = 40˚
- Then ABC is the required triangle.

## Construction of Triangles – Exercise 3.4.9

**Construct an isosceles triangle whose altitude is 4.5 cm and vertex angle is 70˚**

Solution:

Steps of construction:

- Draw a line segment XY
- Take a point M on XY and draw a line MP ⊥ XY
- With M as centre and radius 4.5 cm, draw an arc cutting MP at A.
- Construct B and C on XY such that ∠MAB = 70˚/2 = 35˚ and ∠MAC = 70˚/2 = 35˚
- Then ABC is the required triangle.

**2. Construct an isosceles triangle whose altitude is 6.6 cm and vertex angle is 60˚**

Solution:

Steps of construction:

- Draw a line segment XY
- Take a point M on XY and draw a line MP ⊥ XY
- With M as centre and radius 6.6 cm, draw an arc cutting MP at A.
- Construct B and C on XY such that ∠MAB = 60˚/2 = 30˚ and ∠MAC = 60˚/2 = 30˚
- Therefore, ABC is the required triangle.

**3. Construct an isosceles triangle whose altitude is 5cm and vertex angle is 90˚**

Solution:

Steps of construction:

- Draw a line segment XY
- Take a point M on XY and draw a line MP ⊥ XY
- With M as centre and radius 5 cm, draw an arc cutting MP at A.
- Construct B and C on XY such that ∠MAB = 90˚/2 = 45˚ and ∠MAC = 90˚/2 = 45˚
- Then ABC is the required triangle.

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