Construction of Triangles - Constructions, Class 9, Mathematics | Extra Documents & Tests for Class 9 PDF Download

CONSTRUCTION OF TRIANGLES

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

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. B

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 ∠XGH = ∠B (as given) and ∠YHG =∠C (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, Δ ABC is the required triangle.