NCERT Solutions Class 9 Maths
Chapter – 11 (Constructions)
The NCERT Solutions in English Language for Class 9 Mathematics Chapter – 11 Constructions Exercise 11.2 has been provided here to help the students in solving the questions from this exercise.
Chapter 11: Constructions
Exercise – 11.2
1. Construct a triangle ABC in which BC = 7 cm, ∠B = 75° and AB+AC = 13 cm.
Answer – Construction Procedure
1. Draw a line segment of base BC = 7 cm
2. Measure and draw ∠B = 75° and draw the ray BX
3. Take a compass and measure AB + AC = 13 cm.
4. With B as the centre, draw an arc at the point be D
5. Join DC
6. Now, draw the perpendicular bisector of the line DC, and the intersection point is taken as A
7. Then, join AC
8. Therefore, ABC is the required triangle.
2. Construct a triangle ABC in which BC = 8 cm, ∠B = 45° and AB–AC = 3.5 cm.
Answer – Construction Procedure
1. Draw a line segment of base BC = 8 cm
2. Measure and draw ∠B = 45° and draw the ray BX
3. Take a compass and measure AB-AC = 3.5 cm
4. With B as the centre, draw an arc at point D on the ray BX
5. Join DC
6. Now, draw the perpendicular bisector of the line CD, and the intersection point is taken as A
7. Then, join AC
8. Therefore, ABC is the required triangle.
3. Construct a triangle PQR in which QR = 6 cm, ∠Q = 60° and PR–PQ = 2cm.
Answer – Construction Procedure
1. Draw a line segment of base QR = 6 cm
2. Measure and draw ∠Q = 60°, and let the ray be QX
3. Take a compass and measure PR – PQ = 2 cm.
4. Since PR – PQ is negative, QZ will be below the line QR.
5. With Q as the centre, draw an arc at the point Z on the ray QX
6. Join ZR
7. Now, draw the perpendicular bisector of the line DR and the intersection point is taken as P.
8. Then, join PR
9. Therefore, PQR is the required triangle.
4. Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.
Answer – Construction Procedure
1. Draw a line segment AB which is equal to XY + YZ + ZX = 11 cm.
2. Make an angle ∠LAB = 30° from the point A.
3. Make an angle ∠MBA = 90° from the point B.
4. Bisect ∠LAB and ∠MBA at point X.
5. Now, take the perpendicular bisector of the lines as XA and XB and the intersection point as Y and Z, respectively.
6. Join XY and XZ
7. Therefore, XYZ is the required triangle
5. Construct a right triangle whose base is 12 cm, and the sum of its hypotenuse and other side is 18 cm.
Answer – Construction Procedure:
1. Draw a line segment of base BC = 12 cm
2. Measure and draw ∠B = 90°, and draw the ray BX
3. Take a compass and measure AB + AC = 18 cm
4. With B as the centre, draw an arc at the point D on the ray BX
5. Join DC
6. Now, draw the perpendicular bisector of the line CD, and the intersection point is taken as A.
7. Then, join AC
8. Therefore, ABC is the required triangle.
