Chapter 11 - Geometrical Constructions, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9 PDF Download

Geometrical Constructions 

INTRODUCTION
In the chapter "Lines and Angles" and "Triangles" we have proved many theorems and properties by using diagrams in which angles and sides of triangles were drawn in approximate measurement. The diagrams were drawn to have the idea of the situations according to the given conditions. In this chapter, we shall construct some angles and triangles in precise measurement by using only two geometrical instruments. These two instruments' are, a graduated ruler and a compass.

CONSTRUCTION OF PERPENDICULAR BISECTOR OF A LINE SEGMENT
We want to construct the perpendicular bisector of the given line segment AB.

Steps of Construction: 
1. Taking A and B as centres and radius   draw, arcs on both sides of the line segment AB.

2. The arcs are drawn in such a way that on both sides of AB, we get intersection points P and Q.

3. Join PQ. PQ intersects AB at M. Here, PMQ is the required perpendicular bisector of AB.
 

Justification. In figure, Join AP, AQ, BP and BQ.
Here, AP = BP = BQ = AQ (Each = radius of the arc)

Hence, PQ is perpendicular bisector of the chord AB.

 CONSTRUCTION OF BISECTOR OF A GIVEN ANGLE

We want to construct the bisector of given angle ABC.

Steps of Construction:

1. Taking B as centre, we draw an arc of a circle which meets BC at P and BA at Q.

2. Now, taking P and Q as centres and radius  draw two arcs so that they intersect at a point M.

3. Join BM.
Here, the ray BM is the required bisector of ABC.

Justification. In figure, Join PM and QM.
In ΔBPM and ΔBQM, we have

Construction of 60° angle

We will construct 60° angle at the initial point A of the given ray AB.

Steps of Construction : 

1. Taking A as centre and radius = r (say), we draw an arc of a circle. The arc intersects AB at P.

2. Now, taking P as centre and same radius r, we again draw an arc of a circle which intersects the previous arc at Q.

3. Join AQ and produce this as ray AC.

Construction of 30° Angle

We will construct 30° and at the initial point A of the given ray AB.

Steps of construction :
1. Taking A as centre and radius = r (say), we draw an arc of a circle. The arc intersects AB at P ..

2. Now, taking P as centre and same radius r, we again draw an arc of a circle which intersects the previous  arc at Q.

3. Join AQ and produce the ray AC. Here BAC = 60°.

4. Now, taking P and Q as centres and some radius r' > r, we draw arcs which intersect at R.

5. Join AR and produce the ray AD along AR.

6. Here, AD is bisector of BAC = 60°.
Therefore, we have BAD = 30°.

Construction of 45° Angle

We will construct 45° angle at the initial point A of the given ray AB.

Steps of construction :

Construction of 90° Angle

Steps of Construction. :

Ex.1 Take a line segment AB = 6·3 cm. Find the right bisector (perpendicular bisector) of AB.

Sol. Steps of Construction :
1. We take AB = 6·3 cm.
2. Taking A and B as centres and radius   we draw arcs on both sides of AB.

3. Two arcs intersect at P on one side and other two arcs intersect at Q on the second side of AB. Q

4. We join PQ ; it meets AB at M.
Now PMQ is the required right bisector of AB .

Ex.2 Construct an angle of 75° at the initial point of a given ray. Justify the construction.

Sol.

Hence, BAF = BAC + CAF = 60° + 15° = 75°.

CONSTRUCTION OF TRIANGLES

In this section, we shall construct some triangles with given data by using a graduated ruler and a compass.

To construct a triangle, given its base, a base angle and sum of other two sides.

We have to construct  ΔABC. When base BC, B and AB + AC is given.

Here, AB + AC > BC.

Steps of Construction :

1. Draw the given base BC.

2. Construct the base angle CBX = B as given.

3. Cut the line segment BD = AB + AC along BX.

4. Join CD.

5. Draw perpendicular bisector PQ of CD. PQ meets CD at L and BD at A.

6. Join AC.

Here, ΔABC is the required triangle.

Justification. In ΔACD, AL is perpendicular bisector of CD.

⇒ AD = AC.
Now, in ΔABC we have

AB + AC = AB + AD = BD

Note: If sum of two sides is less than the given base, then triangle is not possible.

To construct a triangle, given its base, a base angle and the difference of the other two sides.

Case (a) : We shall construct ΔABC when base BC and B are given. Also, we are given AB – AC. Here, AB > AC.

Steps of Construction :

1. Draw the given base BC.

2. Construct CBX = B as given,

3. Along BX, cut BD = AB – AC, (Here, BD and BA are in same direction) .

4. Join CD.

5. Construct the perpendicular bisector PQ of CD and PQ intersects CD at L.

6. QP (produced) meets BX at A.

7. Join AC.

Here, AC = AD and hence the ΔABC is required triangle.

Justification. In ΔACD, AL is perpendicular bisector of CD
⇒ AC = AD
Now, in ΔABC, we have AB – AC = AB – AD = BD

Case (b) : Let us construct ΔABC, when BC and B are given.
Also, AC > AB, i.e., AB < AC and AC – AB is given.

Steps of Construction :

1. Draw the given base BC.

2. Construct CBX = B as given.

3. Cut BD = AC – AB along XB (produced),
Here, BD and BA are in opposite directions. 

4. Join CD.
5. Construct the perpendicular bisector PQ of CD.
6. QP meets BX at A and CD at L.
7. Join AC.
Here, AC = AD = AB + BD
i.e., AC – AB = BD.
Hence, ΔABC is the required triangle.

Justification. In ΔACD, AL is perpendicular bisector of CD.
⇒ AC = AD.
Now, in ΔABC, we have
AC – AB = AD – AB = BD.

To construct a triangle, given its two base angles and the perimeter of the triangle.

We shall construct ΔABC if its base angles B and C are given and its perimeter AB + BC + CA is given equal to p.

Steps of Construction:

1. Cut GH = AB + BC + CA = p units.

2. Construct  (as given)

3. Draw bisectors of XGH and YHG and both intersect at A.

4. Construct PQ and RS perpendicular bisectors of GA and HA respectively.

5. PQ and RS intersect GH at B and C respectively. Also, PQ intersects GA at L and RS intersects HA at M.

6. Join AB and AC.

Here, ΔABC is the required triangle.

Justification
BL is right bisector of GA.

Now, BA = BG; Similarly, CA = CH.
Thus, we have AB + BC + CA = BG + BC + CH = GH

Ex.3 Construct ΔABC in which BC = 8 cm, B = 30° and AB + AC = 12 cm.

Sol. Steps of Construction :
1. Draw BC = 8 cm.
2. Construct CBX = 30°

3. Along BX, cut BD = 12 cm
4. Join CD.
5. Draw PQ right bisector of CD. 30°
6. PQ intersects BD at A and CD at L.
7. Join CA; CA = AD because AL is perpendicular bisector of CD.

Now, ΔABC is the required triangle..

Ex4. Construct ΔABC such that AB = 5·8 cm, BC + CA = 7 cm and B = 60°.

Sol. Steps of Construction :
1. Draw AB = 5·8 cm and ABX = 60°.
2. Cut BD = 7 cm along BX
3. Join AD and draw PQ, right bisector of AD.
4. PQ intersects AD at L and BD at C.
5. Join AC.
We observe that in ΔCAD, CL is perpendicular bisector of AD.
⇒ CA = CD
⇒ BC + CA = BC + CD = BD = 7 cm.

Hence, ΔABC is the required triangle.

Ex5. Construct ΔABC such that BC = 6 cm, B = 45° and AC – AB = 2 cm.

Sol. Steps of Construction :
1. Draw BC = 6 cm and CBX = 45°

2. Produce XB to D so that BD = 2 cm, i.e., BD = AC – AB.
Here, BD and BA are to be in opposite directions.

3. Join CD.
4. Draw PQ, right bisector of CD. 45°
5. PQ intersects BX at A and CD at L.
6. Join AC.

ΔABC is the required triangle.

Ex.6 Construct ΔABC such that BC = 6 cm, B = 45° and AB – AC= 3 cm.

Sol. Steps of Construction :
1. Draw BC = 6 cm and CBX = 45°
2. On BX cut BD = 3 cm.
3. Join CD.
4. Draw PQ, right bisector of CD.

5. PQ intersects BX at A and CD at L.
6. Join AC.
Here, ΔABC is the required triangle.

Ex.7 Construct ΔABC such that B = 60°, C = 75° and AB + BC + CA = 13 cm.

Sol. Steps of Construction :
1. Draw GH = 13 cm.
2. Draw HGX = 60° and GHY = 75°.
3. Bisectors of HGX and GHY intersect at A.

4. Draw PQ, right bisector of GA. PQ intersects GH at B and GA at L.

5. Draw RS, right bisector of HA. RS intersects GH at C and HA at M.

6. Join AB and AC.

The ΔABC is the required triangle.

Ex.8 Construct a right triangle whose base is 6 cm and sum of its hypotenuse and the other side is 10 cm.

Sol. Steps of Construction :
1. Draw base BC of the triangle equal to 6 cm.

2. Construct CBX = 90°

3. Cut BD = 10 cm along BX.

4. Join CD and draw PQ, right bisector of CD. L Q

5. PQ intersects BD at A and CD at L.

6. Join AC.
We observe that in ACD, AL is right bisector of CD.
Therefore AC = AD.
Thus, in ΔABC, AC is hypotenuse and
AB + AC = AB + AD = BD = 10 cm.
i.e., AB + AC = 10 cm.

Hence, the ΔABC is the required right triangle.

Ex.9 Construct ΔABC in which AC = 7 cm, C = 60° and AB + BC = 12 cm.
 Sol. Steps of Construction :
1. Draw AC = 7 cm
2. Draw ACX = 60°
3. Cut CD = 12 cm along CX.
4. Join AD and draw right bisector PQ of AD. Q
5. PQ intersects AD at L and CD at B.
6. Join AB.
Now, ΔABC is the required triangle.

Justification
In ΔABC, BL is right bisector of AD.
Therefore, AB = BD.
We have AB + BC = BD + BC = CD = 12 cm.
i.e., AB + BC = 12 cm as required.