NCERT Solutions for Class 9 Maths Chapter 11 - Constructions
Learn to draw triangles and bisectors of line segments with TopperLearning’s NCERT Solutions for CBSE Class 9 Mathematics Chapter 11 Constructions. Get step-by-step instructions to construct angles, bisectors and line segments. Revise the chapter solutions to understand how to give an appropriate justification for your construction.
Practise drawing arcs, triangles etc. according to the given data using the CBSE Class 9 Maths textbook solutions created by our subject experts. You can access these solutions any time at our online learning portal to revisit the topic of constructions and thus do well in your Maths exam.
Chapter 11 - Constructions Exercise Ex. 11.1
(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersect PQ at R.
(iii) Taking S as centre and with the same radius as before, drawn an arc intersecting the arc at T (see figure)
(iv) Taking S and T as centre, draw arc of same radius to intersect each other at U.
(v) Join PU, which is the required ray making 90o with given ray PQ.
We can justify the construction, if we can prove

For this let us join PS and PT



(i)Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersect PQ at R.
(iv)Taking S and T as centre draw arc of same radius to intersect each other at U.
(vi) Now from R and V draw arcs with other at W with radius more than

To justify the construction, we have to prove











= 60o + 30o
= 90o
In step (vi) of this construction, we constructed PW as the bisector of




Step I: Draw the given ray PQ. Now taking P as centre and with some radius, draw an arc of a circle which intersects PQ at R.


(i) Take the given ray PQ. Draw an arc of some radius, taking point P as its centre, which intersect PQ at R.
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(v) Join PU. Let it intersect arc at point V.

(vii) Let it intersects the arc at X. Taking X and R as centre and radius more than


Step I: Draw the given ray PQ. Now taking P as centre and with some radius, draw an arc of a circle which intersects PQ at R.
Step II: Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point S.
Step III: Now taking R and S as centre and with radius more than

Step IV: Let is intersects the arc at U. Now taking U and R as centre and with

The steps of construction are as follows:
(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersect PQ at R.
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(iv) Taking S and T as centre draw arc of same radius to intersect each other at U.

(B) 105o
The steps of construction are as follows:
(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersect PQ at R.
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(iii) Taking S as centre and with the same radius as before, drawn an arc intersecting the arc at T (see figure)
(iv) Taking S and T as centre draw arc of same radius to intersect each other at U.
(v) Join PU. Let it intersects the arc at V. Now taking T and V as centre draw arcs with radius more than

(C) 135o
The steps of construction are as follows:
(i) Take the given ray PQ. Extend PQ on opposite side of Q. Draw a semicircle of some radius taking point P as its centre, which intersect PQ at R and W.
(ii) Taking R as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(iii)Taking S as centre and with the same radius as before, drawn an arc intersecting the arc at T (see figure)
(iv) Taking S and T as centre, draw arc of same radius to intersect each other at U.
(v) Join PU. Let it intersect the arc at V. Now taking V and W as centre and with radius more than

Also, each angle of an equilateral triangle is 60.
The steps of construction are as follows:
Step I: Draw a line segment AB of 5 cm length. Draw an arc of some radius, while taking A as its centre. Let it intersect AB at P.
Step II: Now taking P as centre draw an arc to intersect the previous arc at E. Join AE.
Step III: Taking A as centre draw an arc of 5 cm radius, which intersects extended line segment AE at C. Join AC and BC.

To justify the construction, we have to prove that ABC is an equilateral triangle i.e.AB = BC = AC = 5 cm and



Now, in


Since, AC = AB, we have


Now, in

























Chapter 11 - Constructions Exercise Ex. 11.2



Step I: Draw the line segment BC = 8 cm and at point B make an angle of 45o say

Step II: Cut the line segment BD = 3.5 cm (equal to AB - AC) on ray BX.
Step III: Join DC and draw the perpendicular bisector PQ of DC.
Step IV: Let it intersect BX at point A. Join AC.

Step I: Draw line segment QR of 6 cm. At point Q draw an angle of 60o say

Step II: Cut a line segment QS of 2 cm from the line segment QT extended an opposite side of line segment XQ. (As PR> PQ and PR - PQ = 2cm). Join SR.
Step III: Draw perpendicular bisector AB of line segment SR. Let it intersect QX at point P. Join PQ, PR.






Step I: Draw line segment AB of 12 cm. Draw a ray AX making 90o with AB.

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