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Question-1

Construct a triangle ABC in which BC = 4.6 cm, B = 45° and AB + CA = 8.2 cm.

Solution:
                          
Steps of Construction

1. Draw a ray BX and cut off a line segment BC = 4.6 cm from it.
2. Construct XBY = 45°.
3. From BY cut off a line segment BD = 8.2 cm.
4. Join CD.
5. Draw the perpendicular bisector of CD, intersecting BY at a point A.
6. Join AC.
Then, ABC is the required triangle.

Question-2

Construct a right triangle when one side is 3.5 cm and the sum of the other side and hypotenuse is 5.5 cm.

Solution:
                                                     

Steps of Construction

1. Draw a ray BX and cut off a line segment BD = 5.5 cm from it.
2. Construct XBY = 90°

3. From BY cut off a line segment BD = 5.5 cm
4. Join CD.
5. Draw the perpendicular bisector of CD, intersecting BD at a point A.
6. Join AC.

Then, ABC is the required triangle.

Question-3

Construct a triangle ABC where base BC = 6.5 cm, CA + AB = 10 cm and B = 60°.

Solution:
                                                 

Steps of Construction

1. Draw a ray BX and cut off a line segment BC = 6.5 cm from it.
2. Construct XBY = 60°

3. From BY cut off a line segment BD = 10 cm.
4. Join CD.
5. Draw the perpendicular bisector of CD, intersecting BD at a point A.
6. Join AC.

Then, ABC is the required triangle.

Question-4

 Construct a triangle ABC in which BC = 4.5 cm, B = 45° and AB – AC = 2.5 cm.
 


Solution:
  
                                                                                         
Steps of Construction: 

1. Draw a ray BX and cut off a line segment BC = 4.5 cm.
2. Construct YBC = 45° 

3. Cut off a line segment BD = 2.5 cm from BY.
4. Join CD.
5. Draw perpendicular bisector RS of CD intersecting BY at a point A.
6. Join AC.

Then ABC is the required triangle.

Question-5

Construct a triangle ABC where BC = 5 cm, B = 30° and AC - AB = 2 cm.

Solution:
                                              
Steps of Construction

1. Draw a ray BX and cut off a line segment BC = 5 cm from it.
2. Construct a ray BY making an angle of 30o with BC and produce YB to form a line YBY'.
3. Cut off a line segment BD' = 2 cm from BY'.
4. Join CD'.
5. Draw perpendicular bisector RS of CD' intersecting BY at a point A.
6. Join AC.

Then, ABC is the required triangle.

Question-6

Construct a triangle ABC whose perimeter is 12 cm, B = 60°  and C = 45°.

Solution:
                                                     
Steps of Construction

1. Draw a ray PX and cut off a line segment PQ = 12 cm from it.
2. At P, construct YPQ = 30°(60°)
3. At Q, construct ZQP = 22(45°). 
Let the rays PY and QZ intersect at the point A.
4. Draw the perpendicular bisector of AP intersecting PQ at a point B.
5. Draw the perpendicular bisector of AQ intersecting PQ at a point C.
6. Join AB and AC.

Then, ABC is the required triangle.

Question-7

Construct a triangle ABC with perimeter 10 cm and each base angle is of 45°.

Solution:
                                               
Steps of Construction

1. Draw a ray PX and cut off a line segment PQ = 10 cm from it.
2. At P, construct YPQ = 22° (45° )
3. At Q, construct ZQP = 22°(45° )
4. Draw the perpendicular bisector of AP intersecting PQ at a point B.
5. Draw the perpendicular bisector of AQ intersecting PQ at a point C.
6. Join AB and AC.

Then, ABC is the required triangle.

Question-8

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

Solution:
                                          
Steps of Construction

1. Draw a line segment BC = 6 cm.
2. Bisect BC. Let D be its mid-point.
3. With D as centre and radius 4 cm, draw an arc.
4. With B as centre and radius 6 cm, draw another arc intersecting the above arc at a point A.
5. Join AB and AC.

Then, ABC is the required triangle.

Question-9

Construct a right triangle when one side is 2 cm and the sum of the other side and hypotenuse is 6 cm.

Solution:
Given: In Δ ABC, BC = 2 cm, BD = 6cm and B = 90°

Required: To construct Δ ABC.

                                                                       
Steps of Construction

1. Draw a ray BX and cut off a line segment BC = 2 cm from it.

2. Construct XBY = 90
°.


3. From BY cut off a line segment BD = 6 cm

4. Join CD.

5. Draw the perpendicular bisector of CD, intersecting BD at a point A.

6. Join AC.

Then, ABC is the required triangle.

Question-10

Construct a triangle ABC where base BC = 8 cm, CA + AB = 15 cm and B = 60°.

Solution:
Given: In Δ ABC, base BC = 8 cm, CA + AB = 15 cm and B = 60°.

Required: To construct Δ ABC.


Steps of Construction

1. Draw a ray BX and cut off a line segment BC = 8 cm from it.

2. Construct
XBY = 60°.

3. From BY cut off a line segment BD = 15 cm.

4. Join CD.

5. Draw the perpendicular bisector of CD, intersecting BD at a point A.

6. Join AC.

Question-11

Construct a triangle ABC in which BC = 6 cm, B = 45° and AB - AC = 2.5 cm.

Solution:
Given: In Δ ABC, BC = 6 cm, B = 45° and AB - AC = 2.5 cm.

Required: To construct Δ ABC.
 



Steps of Construction

1. Draw a ray BX and cut off a line segment BC = 6 cm.

2. Construct
YBC = 45°.

3. Cut off a line segment BD = 2.5 cm from BY.

4. Join CD.

5. Draw perpendicular bisector RS of CD intersecting BY at a point A.

6. Join AC.

Then ABC is the required triangle.

Question-12

Construct a triangle ABC in which base, BC = 6 cm; B = 60° and sum of other two sides = 9 cm.

Solution:
                              
Steps of Construction

(i) Draw BC = 6 cm.
(ii) Construct CBX = 60°
(iii) With B as centre and radius 9 cm cut an arc on the ray BX at P.
(iv) Join CP.
(v) Draw the perpendicular bisector of CP to intersect BP at A.
(vi) Join AC.

Then, ABC is the required triangle.

Question-13

Construct a triangle ABC in which base BC = 5.7 cm, B = 30° and difference between other two sides is 3 cm.

Solution:
                                              
Steps of Construction

(i) Draw BC = 5.7 cm.
(ii) Construct CBX = 30°
(iii) With B as centre and radius 3 cm cut an arc on the ray BX at P.
(iv) Join PC.
(v) Draw the perpendicular bisector of PC, meeting BP produced at A.
(vi) Join AC.

Then, ABC is the required triangle.




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