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# Class 9 NCERT Solutions- Chapter 11 Constructions – Exercise 11.2

• Last Updated : 28 Dec, 2020

### Question 1. Construct a âˆ† ABC in which BC = 7 cm, âˆ B = 75Â° and AB + AC = 13 cm.

Solution:

Steps of construction:

1. Draw a line segment BC base of cm is drawn.
2. At point B draw an angle of 75Â°.
3. Cut BD =13cm from BY.
4. Join âˆ D which intersect BD at A.
5. Join AC. Now triangle ABC is the required triangle

### Question 2. Construct a ABC in which BC = 8 cm, âˆ B = 45Â° and AB â€“ AC = 35 cm.

Solution:

Steps of construction:

1. Draw  a line segment BC=8cm.
2. At point B, draw angle 45Â°.
3. Cut BD=3.5 from BY.
4. Join CD.
5. Draw perpendicular bisector of CD, which construct BY at A.
6. Join AC. NOW, ABC is the required triangle.

### Question 3. Construct a âˆ† ABC in which QR = 6 cm, âˆ Q = 60Â° and PR â€“ PQ = 2 cm.

Solution:

Steps of construction:

1. Draw a line segment QR=6cm.
2. At point Q draw angle 60Â°.
3. Extend PQ to Yâ€™.
4. Cut QS =2cm from QYâ€™.
5. Join RS.
6. Draw perpendicular bisector of RS which intersect QY at P.
7. Join PR. Now, PQR is the required triangle.

### Question 4. Construct a âˆ† XYZ in which âˆ Y = 30Â°, âˆ Y = 90Â° and XY + YZ + ZX = 11 cm.

Solution:

Steps of construction:

1. Draw a line segment AB=11cm.
2. At point A draw âˆ BAP=30Â°.
3. At point B draw angle 90Â°.
4. Draw the bisector of âˆ BAP and âˆ ABR which intersect each other at X.
5. Join AX and BX.
6. Draw perpendicular bisector of AX and BX which intersect AB on Y and Z respectively.
7. Join XY and XZ. Then XYZ is the required triangle.

### Question 5. Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.

Solution:

Steps of construction:

1. Draw a line segment of BC=12cm.
2. At point B draw angle b=90Â°
3. Cut BD =18cm.
4. Join CD.
5. Draw perpendicular bisector of CD which intersect BD at point A.
6. Join AC. Now ABC is the required triangle.

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