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} {"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
Similarly, `color{orange} {∠ACB = ∠MYX}` as required.
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.
