Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev

Mathematics (Maths) Class 9

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Class 9 : Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev

The document Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9

Question 1. Construct a triangle ABC in which BC = 7 cm, ∠B = 75º and AB + AC = 13 cm.
 Solution: Steps of construction:

I . Draw a ray BX.
II. From Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev cut off Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev = 7 cm.
III. At B, construct ∠CBY = 75º.
IV. From Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev cut off BD = 13 cm (= AB + BC)

Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev

V. Join D and C.
VI. Bisect DC such that the bisector of DC meets BD at A.
VII.Join AC.
Thus, ΔABC is the required triangle.


Question 2. Construct a triangle ABC in which BC = 8 cm, ∠ B = 45º and AB ∠ AC = 3.5 cm.
 Solution: Steps of construction: 

I . Draw a ray Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev
II. From Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev , cut off BC = 8 cm.
III. Construct ∠CBY = 45º.
IV. From Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev cut off Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev = 3.5 cm.

Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev

V. Join D and C.
VI. Draw PQ, perpendicular bisector of DC, which intersects Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev at A.
VII. Join AC.
Thus, ABC is the required triangle.


Question 3. Construct a triangle PQR in which QR = 6 cm, ∠Q = 60º and PR ∠ PQ = 2cm.
 Solution: Steps of construction:

I . Draw a ray Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev .
II. From Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev, cut off QR = 6 cm.
III. Construct a line YQY' such that ∠RQY = 60º.

Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev

IV. Cut off QS = 2 cm (from QY').
V. Join ‘S’ and ‘R’.
VI. Draw MN, perpendicular bisector of SR, which intersects QY at P.
VII. Join P and R.
Thus, PQR is the required triangle.


Question 4. Construct a triangle XYZ in which ∠ Y = 30º , ∠ Z = 90º and XY + YZ + ZX = 11 cm.
 Solution: Steps of construction:

Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev

I . Draw a line segment AB = 11 cm = (XY + YZ + ZX)
II. Construct ∠BAP = 30º = ∠Y
III. Construct ∠ABQ = 90º = ∠Z
IV. Draw Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev , the bisector of ∠BAP.
V. Draw Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev , the bisector of ∠ABQ, such that Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev and Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev intersect each other at X.
VI. Draw perpendicular bisector of AX, which intersects AB at Y.
VII. Draw perpendicular bisector of XB, which intersects AB at Z.
VIII. Join XY and XZ.
Thus, 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:

I . Draw Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev = 12 cm.
II. Construct ∠CBY = 90º.

Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev

III. From Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev cut off BX = 18 cm.
IV. Join CX.
V. Draw PQ, the perpendicular bisector of CX, such that PQ meets BX at A.
VI. Join AC.
Thus, ABC is the required triangle.

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