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

## Mathematics (Maths) Class 9

Created by: Full Circle

## 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 cut off = 7 cm.
III. At B, construct ∠CBY = 75º.
IV. From cut off BD = 13 cm (= AB + BC) 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 II. From , cut off BC = 8 cm.
III. Construct ∠CBY = 45º.
IV. From cut off = 3.5 cm. V. Join D and C.
VI. Draw PQ, perpendicular bisector of DC, which intersects 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 .
II. From , cut off QR = 6 cm.
III. Construct a line YQY' such that ∠RQY = 60º. 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: I . Draw a line segment AB = 11 cm = (XY + YZ + ZX)
II. Construct ∠BAP = 30º = ∠Y
III. Construct ∠ABQ = 90º = ∠Z
IV. Draw , the bisector of ∠BAP.
V. Draw , the bisector of ∠ABQ, such that and 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 = 12 cm.
II. Construct ∠CBY = 90º. III. From 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.

## Mathematics (Maths) Class 9

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