Class 9 Some Constructions of Triangles For CBSCE-NCERT

### Topic covered

color{red} ♦ Some Constructions of Triangles

### Some Constructions of Triangles

Some Constructions So far, some basic constructions have been considered.
Next, some constructions of triangles will be done by using the constructions given in earlier classes and given above. Recall that SAS, SSS, ASA and RHS rules give the congruency of two triangles.

Therefore, a triangle is unique if : (i) two sides and the included angle is given, (ii) three sides are given, (iii) two angles and the included side is given and, (iv) in a right triangle, hypotenuse and one side is given.

Now, let us consider some more constructions of triangles. You may have noted that at least three parts of a triangle have to be given for constructing it but not all combinations of three parts are sufficient for the purpose.

For example, if two sides and an angle (not the included angle) are given, then it is not always possible to construct such a triangle uniquely.

 color{ blue} text(Construction 11.4 :)
"To construct a triangle, given its base, a base angle and sum of other two sides."

Given the base BC, a base angle, say ∠B and the sum AB + AC of the other two sides of a triangle ABC, you are required to construct it.

color{blue}text(Steps of Construction :)

1. Draw the base BC and at the point B make an angle, say XBC  equal to the given angle.
2. Cut a line segment BD equal to AB + AC from the ray BX.
3. Join DC and make an angle DCY equal to ∠BDC.
4. Let CY intersect BX at A (see Fig. 11.4).

Then, ABC is the required triangle. Let us see how you get the required triangle.

Base BC and ∠B are drawn as given. Next in triangle ACD,

∠ACD = ∠ ADC (By construction)

Therefore, AC = AD and then

AB = BD – AD = BD – AC

color{purple} {AB + AC = BD}

 color{blue} "Alternative method :"

Follow the first two steps as above. Then draw perpendicular bisector PQ of CD to intersect BD at a point A (see Fig 11.5). Join AC. Then ABC is the required triangle. Note that A lies on the perpendicular bisector of CD, therefore AD = AC.

 color{blue} text(Remark :)

The construction of the triangle is not possible if the sum AB + AC ≤ BC.

 color{blue} {"Construction 11.5 :"}

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

Given the base BC, a base angle, say ∠B and the difference of other two sides AB – AC or AC – AB, you have to construct the triangle ABC. Clearly there are following two cases:

 color{blue} text( Case (i) :) Let AB > AC that is AB – AC is given.

 color{blue} text( Steps of Construction :)

1. Draw the base BC and at point B make an angle say XBC equal to the given angle.

2. Cut the line segment BD equal to AB – AC from ray BX.

3. Join DC and draw the perpendicular bisector, say PQ of DC.

4. Let it intersect BX at a point A. Join AC (see Fig. 11.6).

Then ABC is the required triangle.

Let us now see how you have obtained the required triangle ABC.

Base BC and ∠B are drawn as given. The point A lies on the perpendicular bisector of DC. Therefore,

AD = AC

So, BD = AB – AD = AB – AC.

color{blue} text(Case (ii) :)

Let AB < AC that is AC – AB is given.

color{blue} text( Steps of Construction :)

1. Same as in case (i).

2. Cut line segment BD equal to AC – AB from the line BX extended on opposite side of line segment BC.

3. Join DC and draw the perpendicular bisector, say PQ of DC.

4. Let PQ intersect BX at A. Join AC (see Fig. 11.7).

Then, ABC is the required triangle.

You can justify the construction as in case (i).

color{blue} text( Construction 11.6 :)

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

Given the base angles, say ∠ B and ∠ C and BC + CA + AB, you have to construct the triangle ABC.

Steps of Construction :

1. Draw a line segment, say XY equal to BC + CA + AB.

2. Make angles LXY equal to ∠B and MYX equal to ∠C.

3. Bisect ∠ LXY and ∠ MYX. Let these bisectors intersect at a point A [see Fig. 11.8(i)].

4. Draw perpendicular bisectors PQ of AX and RS of AY.

5. Let PQ intersect XY at  B and RS intersect XY at C. Join AB and AC [see Fig 11.8(ii)].

Then ABC is the required triangle. For the justification of the construction, you

observe that, B lies on the perpendicular bisector  PQ of AX.

Therefore, XB = AB and similarly, CY = AC.

This gives BC + CA + AB = BC + XB + CY = XY.

Again ∠BAX = ∠AXB (As in Delta AXB, AB = XB) and

color{green} {∠ABC = ∠BAX + ∠AXB = 2 ∠AXB = ∠LXY}

Similarly, color{orange} {∠ACB = ∠MYX} as required.
Q 3210478319

Construct a triangle ABC, in which ∠B = 60°, ∠ C = 45° and AB + BC + CA = 11 cm.

Class 9 Chapter 11 Example 1
Solution:

1. Draw a line segment PQ = 11 cm.( = AB + BC + CA).

2. At P construct an angle of 60° and at Q, an angle of 45°.

3. Bisect these angles. Let the bisectors of these angles intersect at a point A.

4. Draw perpendicular bisectors DE of AP to intersect PQ at B and FG of AQ to intersect PQ at C.

5. Join AB and AC (see Fig. 11.9). Then, ABC is the required triangle.