The document RD Sharma Solutions -Ex-17.3, Constructions, Class 9, Maths Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.

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**Q1. Construct a △ABC in which BC=3.6cm, AB+AC=4.8cm and ∠B=60 ^{0}.**

__Steps of Construction:__

- Construct a line segment BC of 3.6cm.
- At the point B, draw ∠XBC=60
^{0}. - Keeping B as center and radius 4.8cm draw an arc which intersects XB at D.
- Join DC.
- Draw the perpendicular bisector of DC which intersects DB at A.
- Join AC.

Hence △ABC is the required triangle.

** **

**Q2. Construct a △ABC in which AB+AC=5.6cm ,BC=4.5cm and ∠B=45 ^{0}.**

__Steps of Construction:__

- Construct a line segment BC of 4.5cm.
- At the point B, draw ∠XBC=45
^{0}. - Keeping B as centre and radius 5.6cm draw an arc which intersects XB at D.
- Join DC.
- Draw the perpendicular bisector of DC which intersects DB at A.
- Join AC.

Hence △ABC is the required triangle

** **** **

**Q3. Construct a △ABC in which BC=3.4cm, AB-AC=1.5cm and ∠B=45 ^{0}.**

__Steps of Construction:__

- Construct a line segment BC of 3.4cm.
- At the point B, draw ∠XBC=45
^{0}. - Keeping B as centre and radius 1.5cm draw an arc which intersects XB at D.
- Join DC.
- Draw the perpendicular bisector of DC which intersects DB at A.
- Join AC.

Hence △ABC is the required triangle.

** **

**Q4. Using rulers and compasses only, construct a △ABC, given base BC=7cm, ****∠ABC=60 ^{0} **

__Steps of Construction:__

- Construct a line segment BC of 7cm
- At the point B, draw ∠XBC=60
^{0}. - Keeping B as center and radius 12cm draw an arc which intersects XB at D
- Join DC
- Draw the perpendicular bisector of DC which intersects DB at A
- Join AC

Hence △ABC is the required triangle.

**Q5. Construct a triangle whose perimeter is 6.4cm, and angles at the base are 60 ^{0} 45^{0}.**

__Steps of Construction:__

- Draw a line segment XY of 6.4cm.
- Draw ∠DXY=60
^{0}and ∠EYX=45^{0}. - Draw the angle bisectors of ∠DXY and ∠EYX which intersect each other at A.
- Draw the perpendicular bisector of AX and AY which intersect XY at B and C respectively.
- Join AB and AC.

Hence △ABC is the required triangle.

** **

**Q6. Using rulers and compasses only, construct a △ABC from the following data:**

**AB+BC+CA=12cm, ∠B=45 ^{0 }and ∠C=60^{0}**

__Steps of Construction:__

- Draw a line segment XY of 12cm.
- Draw ∠DXY=45
^{0}and ∠EYX=60^{0}. - Draw the angle bisectors of ∠DXY and ∠EYX which intersect each other at A.
- Draw the perpendicular bisector of AX and AY which intersect XY at B and C respectively.
- Join AB and AC.

Hence △ABC is the required triangle.

**Q7. Construct a right-angled triangle whose perimet****er is equal to 10cm and one acute angle equal to 600.**

__Steps of Construction:__

- Draw a line segment XY of 10cm.
- Draw ∠DXY=900 and ∠EYX=60
^{0}. - Draw the angle bisectors of ∠DXY and ∠EYX which intersect each other at A.
- Draw the perpendicular bisector of AX and AY which intersect XY at B and C respectively.
- Join AB and AC.

Hence △ABC is the required triangle.

**Q8. Construct a triangle ABC such that BC= 6cm, AB= 6cm and median AD=4cm.**

__Steps of construction:__

- Draw a line segment BC of 6cm.
- Take mid-point O of side BC.
- With center B and D and radii 6cm and 4cm, draw two arcs which intersect each other at A.
- Join AB, AD and AC.

Hence △ABC is the required triangle.

**Q9. Construct a right-angled triangle ABC whose base BC is 6cm and the sum of the hypotenuse AC and other side AB is 10cm.**

__Steps of Construction:__

- Construct a line segment BC of 6cm.
- At the point B, draw ∠XBC=90
^{0}. - Keeping B as center and radius 10cm draw an arc which intersects XB at D.
- Join DC.
- Draw the perpendicular bisector of DC which intersects DB at A.
- Join AC.

Hence △ABC is the required triangle.

** **

**Q10. Construct a triangle XYZ in which ∠Y=30 ^{0}, ∠Z=90^{0} and XY+YZ+ZX=11cm.**

__Steps of construction:__

- Draw a line segment AB of 11cm.
- Draw ∠DAB=30
^{0}and ∠FBA=90^{0}. - Draw the angle bisectors of ∠DAB and ∠EBA which intersect each other at X.
- Draw the perpendicular bisector of XA and XB which intersect AB at Y and Z respectively
- Join XY and XZ.

Hence △XYZ is the required triangle.

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