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 is 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ABC is required triangle.

Justification.

InACD. AL is perpendicular bisector of CD

 ^{AC = AD}

Now, inABC, we have AB - AC = AB - AD = BD

Case (b) :

Let us constructABC, 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.

Justification.

InACD, Al perpendicular bisector of CD.

 AC = AD.

Now, inABC, 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ABC if its base angles B and C are given and its perimeter AB + BC + CA is given equal to p.

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 respectively, Also PQ intersects GA at L and RS intersects HA at M.

6. Join AB and AB.

Here,ABC is the required triangle.

Justification

BL is right bisector of GA.

 BA = BG ...(i)

Sol. Steps of Construction : 1. Draw BC = 8 cm

2. Construct^{CBX = 30}⁰ 3. Along BX, cut BD = 12 cm 4. Join CD

5. Draw PQ right bisector of CD.

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

7. Join CA; CA = AD because AL is perpendicular bisector of CD.

Now,ABC is the required triangle.

Ex.4 ConstructABC such that B = 5.8 cm, BC + CA = 7 cm andB = 60⁰ Sol. Steps of Construction :

1. Draw AB = 5.8 cm andABX = 60⁰ 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CAD, CL is perpendicular bisector of AD.

 CA = CD

 BC + CA = BC + CD = BD = 7 cm.

Hence,ABC is the required triangle.

Ex.5 ConstructABC such that BC = 6 cm,B = 45⁰and AC - AB = 2 cm.

Sol. Steps of Construction:

1. Draw BC = 6 cm andCBX = 45⁰

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.

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

6. Join AC.

ABC it the required triangle.

Ex.6 ConstructABC such that BC = 6 cm,B = 45⁰and AB - AC = 3 cm.

Sol. Steps of Construction :

1. Draw BC = 6 cm and^{CBX = 45}⁰ 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 AB.

Here,ABC is the required triangle.

Ex.7 ConstructABC such that^{B = 60}⁰^,^{C = 75}⁰and AB + BC + CA = 13 cm.

Sol. Steps of Construction : 1. Draw GH = 13 cm.

2. DrawHGX = 60⁰andGHY = 75⁰

3. Bisector 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ABC is the required triangle.

Ex.6 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 = 90⁰ 3. Cut BD = 10 m along BX.

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

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

6. Join AC.

We observe that isACD, AL is right bisector of CD.

Therefore AC = AD.

Thus, inABC, AC is hypotenuse and AB + AC = AB + AD = BC = 10 cm.

i.e., AB + AC = 10 cm. Hence, theABC is the required right triangle.

Ex.9 ConstructABC in which AC = 7 cm,C = 60⁰and AB + BC = 12 c m.

Sol. Steps of Construction:

1. Draw AC = 7 cm 2. DrawACX = 60⁰

3. Cut CD = 12 cm along CX.

4. Join AD and draw right bisector PQ and AD.

5. PQ intersects AD at L and CD at B.

6. Join AB.

Now,ABC is the required triangle.

Justification

InABC, 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.

EXERCISE

1. Take AB = 7.4 cm and find its perpendicular bisector.

2. Construct the following angles at the initial point of a given ray : (i) 15⁰

13. Construct a right triangle whose base is 8 cm and the sum of its hypotenuse and the other side is 16 cm.

14. ConstructABC in which AC = 5 cm,^{C = 60}⁰and AB + BC = 9 cm.

1. Construct an angle of 90⁰at the initial point of a given ray and justify the construction.

2. Construct an angle of 45⁰at the initial point of a given ray and justify the construction.

3. Construct the angles of the following measurements : (i) 30⁰ (ii) 

221 (iii) 15⁰

4. Construct the following angles and verify by measuring them by a protractor : (i) 75⁰ (ii) 105⁰(iii) 135⁰

5. Construct an equilateral triangle, given its side and justify the construction.

6. Construct a triangle ABC in which BC = 7 cm,B = 75⁰and AB + AC = 13 cm.

7. Construct a triangle ABC in which BC = 8 cm,B = 45⁰and AB - AC = 3.5 cm.

8. Construct a triangle PQR in which QR = 6 cm,Q = 60⁰and PR - PQ = 2 cm.

9. Construct a triangle XYZ in which^{Y = 30}⁰^,^{Z = 90}⁰and XY + YZ + ZX = 11 cm.

In document Circle, Geometrical Constructions Statistics & Probability (Page 58-63)