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# Construction of Triangles

3.4.1 Introduction:

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 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: Exercise 3.4.2

1. 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: 1. 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: 1. 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: 1. 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: 1. 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 3.4.3 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: Exercise 3.4.3

1. 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: 3.4.4 When two angles and included side are given:

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: 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: 3.4.7 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:

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

Then, LMN is the required triangle. Exercise 3.4.7

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

Solution:

Steps of construction:

1. Draw a line segment XY.
2. Locate A, B on XY such that AB = 5 cm
3. Construct a line segment BD, sufficiently large, such that ∠B = 90°
4. 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:

1. Draw a line segment.
2. Locate Q, R on XY such that QR = 3 cm
3. Construct a line segment SR, sufficiently large, such that ∠R = 90°
4. 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:

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

Then, ABC is the required triangle. 3.4.8 To construct an isosceles triangle whose base and corresponding altitude are given

Exercise 3.4.8

1. 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:

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

1. Draw a line segment YZ whose length is 5.8cm
2. Draw the perpendicular bisector of YZ; call it AB with B on YZ
3. With B as centre and radius 3.8 cm, draw an arc cutting AB at X; join XY and XZ.
4. 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:

1. Draw a line segment PQ whose length is 7.2cm
2. Draw the perpendicular bisector of ; call it AB with B on YZ
3. With B as centre and radius 3.8 cm, draw an arc cutting AB at X; join XY and XZ.
4. Then XYZ is the required triangle. 3.4.9 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:

1. Draw a line segment XY
2. Take a point M on XY and draw a line MP ⊥ XY
3. With M as centre and radius 4 cm, draw an arc cutting MP at A.
4. Construct B and C on XY such that ∠MAB = 80˚/2 = 40˚ and ∠MAC = 80˚/2 = 40˚
5. Then ABC is the required triangle. Exercise 3.4.9

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

Solution:

Steps of construction:

1. Draw a line segment XY
2. Take a point M on XY and draw a line MP ⊥ XY
3. With M as centre and radius 4.5 cm, draw an arc cutting MP at A.
4. Construct B and C on XY such that ∠MAB = 70˚/2 = 35˚ and ∠MAC = 70˚/2 = 35˚
5. 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:

1. Draw a line segment XY
2. Take a point M on XY and draw a line MP ⊥ XY
3. With M as centre and radius 6.6 cm, draw an arc cutting MP at A.
4. Construct B and C on XY such that ∠MAB = 60˚/2 = 30˚ and ∠MAC = 60˚/2 = 30˚
5. Then ABC is the required triangle. 3. Construct an isosceles triangle whose altitude is 5cm and vertex angle is 90˚

Solution:

Steps of construction:

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