Class 9 NCERT Solutions 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.
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Ex. 11.1
Ex. 11.2
Constructions Exercise Ex. 11.1
Solution 5
We know that all sides of an equilateral triangle are equal. So, all sides of this equilateral triangle will be 5 cm.
Also, each angle of an equilateral triangle is 60.
The steps of construction are as follows:
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.
ABC is the required equilateral triangle of side 5 cm.
Justification of Construction:
To justify the construction, we have to prove that ABC is an equilateral triangle i.e.AB = BC = AC = 5 cm andA = B = C = 60o.
Now, inABC, we have AC = AB = 5 cm and 
A = 60o
Since, AC = AB, we have
B = C (angles opposite to equal sides of a triangle)
Now, inABC
A + B + C = 180o (angle sum property of a triangle)
60o + C + C = 180o
60o + 2 C = 180o
2 C = 180o - 60o = 120o
C = 60o
To justify the construction, we have to prove that ABC is an equilateral triangle i.e.AB = BC = AC = 5 cm and
A = B = C = 60o.Now, in
ABC, we have AC = AB = 5 cm and A = 60oSince, AC = AB, we have
B = C (angles opposite to equal sides of a triangle)Now, in
ABC A + B + C = 180o (angle sum property of a triangle)60o + C + C = 180o60o + 2 C = 180o 2 C = 180o - 60o = 120o C = 60o
Now, we have A = B = C = 60o ... (1)
A = B = C = 60o ... (1)A = B and A = CBC = AC and BC = AB (sides opposite to equal angles of a triangle)AB = BC = AC = 5 cm ... (2)Equations (1) and (2) show that the ABC is an equilateral triangle.
ABC is an equilateral triangle.Solution 1
Following are the steps of construction:
(i) Take the given ray PQ. Draw an arc of some radius taking point P as its centre, which intersect 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.
(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.
(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.
Justification of Construction:
We can justify the construction, if we can prove
UPQ = 90o.
For this let us join PS and PT
We can justify the construction, if we can prove
UPQ = 90o.For this let us join PS and PT
We have SPQ = TPS = 60o. In (iii) and (iv) steps of this construction, we have drawn PU as the bisector of TPS.
SPQ = TPS = 60o. In (iii) and (iv) steps of this construction, we have drawn PU as the bisector of TPS.
Solution 2
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.
(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.
(iv)Taking S and T as centre draw arc of same radius to intersect each other at U.
(v) Join PU. Let it intersect arc at point V.
(vi) Now from R and V draw arcs with other at W with radius more than
RV to intersect each other. PW is the required ray making 45o with PQ.
(vi) Now from R and V draw arcs with other at W with radius more than
RV to intersect each other. PW is the required ray making 45o with PQ.
Justification of Construction:
To justify the construction, we have to prove
WPQ = 45o. Join PS and PT
To justify the construction, we have to prove
WPQ = 45o. Join PS and PT
We have SPQ = TPS = 60o. In (iii) and (iv) steps of this construction, we have drawn PU as the bisector of TPS.
SPQ = TPS = 60o. In (iii) and (iv) steps of this construction, we have drawn PU as the bisector of TPS.
UPS = TPS= 
Now, UPQ = SPQ + UPS
= 60o + 30o
= 90o
In step (vi) of this construction, we constructed PW as the bisector ofUPQ
UPQ = SPQ + UPS= 60o + 30o
= 90o
In step (vi) of this construction, we constructed PW as the bisector of
UPQWPQ = 
UPQ = 
= 45oSolution 3
(i) 30o
The steps of construction are as follows:
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 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 
RS draw arcs to intersect each other at T. Join PT which is the required ray making 30o with the given ray PQ.
RS draw arcs to intersect each other at T. Join PT which is the required ray making 30o with the given ray PQ.
(ii) 22
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.
(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 intersect arc at point V.
(v) Join PU. Let it intersect arc at point V.
(vi) Now from R and V draw arcs with radius more than RV to intersect each other at W. Join PW.
(vii) Let it intersects the arc at X. Taking X and R as centre and radius more than RX draw arcs to intersect each other at Y. Joint PY which is the required ray making 22 with the given ray PQ.
other at W. Join PW.(vii) Let it intersects the arc at X. Taking X and R as centre and radius more than
RX draw arcs to intersect each other at Y. Joint PY which is the required ray making 22 with the given ray PQ.
(iii) 150
The steps of construction are as follows:
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
RS draw arcs to intersect each other at T. Join PTStep IV: Let is intersects the arc at U. Now taking U and R as centre and with
radius more than RU draw arc to intersect each other at V. Join PV which is the required ray making 15o with given ray PQ.
Solution 4
(A) 75o
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.
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.
(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 S and V as centre draw arcs with radius more than
SV. Let those intersect each other at W. Join PW, which is the required ray making 75o with the given ray PQ.
SV. Let those intersect each other at W. Join PW, which is the required ray making 75o with the given ray PQ.
Now, we can measure the angle so formed with the help of a protractor. It comes to be 75o.
(B) 105o
(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
TV. Let these arcs intersect each other at W. Join PW, which is the required ray making 105o with the given ray PQ.
Now, we can measure the angle so formed with the help of a protractor. It comes to be 105o.
(C) 135o
The steps of construction are as follows:
(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
VW draw arcs to intersect each other at X. Join PX which is the required ray making 135o with the given line PQ.
Now, we can measure the angle so formed with the help of a protractor. It comes to be 135o.
Constructions Exercise Ex. 11.2
Solution 1
The steps of construction for the required triangles are as follows:
Step I: Draw a line segment BC of 7 cm. At point B draw an angle of 75o say
XBC.
XBC. Step II: Cut a line segment BD = 13 cm (that is equal to AB + AC) from the ray BX.
Step III: Join DC and make an angle DCY equal to BDC
BDC Step IV: Let CY intersects BX at A. 
ABC is the required triangle. 
ABC is the required triangle. Solution 2
The steps of construction for the required triangles are as follows:
Step I: Draw the line segment BC = 8 cm and at point B make an angle of 45o say
XBC.
Step II: Cut the line segment BD = 3.5 cm (equal to AB - AC) on ray BX.
Step I: Draw the line segment BC = 8 cm and at point B make an angle of 45o say
XBC. 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.
ABC is the required triangle.
Solution 3
The steps of construction for the required triangles are as follows:
Step I: Draw line segment QR of 6 cm. At point Q draw an angle of 60o say
XQR.
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.
PQR is the required triangle.
Step I: Draw line segment QR of 6 cm. At point Q draw an angle of 60o say
XQR.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.
PQR is the required triangle.
Solution 4
The steps of construction for the required triangles are as follows:
Step I: Draw a line segment AB of 11 cm.(As XY + YZ + ZX = 11 cm)
Step II: Construct an angle 
PAB of 30o at point A and an angle QBA of 90o at point B.
PAB of 30o at point A and an angle QBA of 90o at point B.Step III: Bisect PAB and QBA. Let these bisectors intersect each other at point X.
PAB and QBA. Let these bisectors intersect each other at point X.Step IV: Draw perpendicular bisector ST of AX and UV of BX. Step V: Let ST intersects AB at Y and UV intersects AB at Z.Join XY, XZ.
XYZ is the required triangle.
XYZ is the required triangle.
Solution 5
The steps of construction for the required triangles are as follows:
Step I: Draw line segment AB of 12 cm. Draw a ray AX making 90o with AB.
Step I: Draw line segment AB of 12 cm. Draw a ray AX making 90o with AB.
Step II: Cut a line segment AD of 18 cm. (As sum of other two side is 18) from ray AX.
Step III: Join DB and make an angle DBY equal to ADB.
Step IV: Let BY intersects AX at C. Join AC, BC.
ABC is the required triangle.
ABC is the required triangle.
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