NCERT Solutions for Class 9 Maths Chapter 11 - Constructions

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Chapter 11 - Constructions Exercise Ex. 11.1

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.
 
(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.
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions
Justification of Construction:

We can justify the construction, if we can prove Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsUPQ = 90o.
For this let us join PS and PT
Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions
 
We have Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsSPQ = Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsTPS = 60o. In (iii) and (iv) steps of this construction, we have drawn PU as the bisector of Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsTPS.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions
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.
 
(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.

(vi) Now from R and V draw arcs with other at W with radius more than Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions RV to intersect each other. PW is the required ray making 45o with PQ.
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions
 
Justification of Construction:

To justify the construction, we have to prove Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsWPQ = 45o.     Join PS and PT
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions
We have Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsSPQ = Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsTPS = 60o. In (iii) and (iv) steps of this construction, we have drawn PU as the bisector of Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsTPS.
Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsNcert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsUPS = Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions  Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsTPS= Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions
Now, Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsUPQ = Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsSPQ + Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsUPS
                  = 60o + 30o
                  = 90o
    In step (vi) of this construction, we constructed PW as the bisector of Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsUPQ
Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsNcert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsWPQ = Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsUPQ = Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions  = 45o
 
Solution 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 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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions RS draw arcs to intersect each other at T. Join PT which is the required ray making 30o with the given ray PQ.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions
(ii) 22Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions
          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 intersect arc at point V.
 
(vi) Now from R and V draw arcs with radius more than  RV to intersect each Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions other at W. Join PW.
            
(vii) Let it intersects the arc at X. Taking X and R as centre and radius more than Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions RX draw arcs to intersect each other at Y. Joint PY which is the required ray making 22Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions  with the given ray PQ.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions
 
(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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions RS draw arcs to intersect each other at T. Join PT

Step IV: Let is intersects the arc at U. Now taking U and R as centre and with Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions 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.
 
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions
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.
 
(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 S and V as centre draw arcs with radius more thanNcert Solutions Cbse Class 9 Mathematics Chapter - Constructions   SV. Let those intersect each other at W. Join PW, which is the required ray making 75o with the given             ray PQ.
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions
 
Now, we can measure the angle so formed with the help of a protractor. It comes to be 75o.

(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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions TV. Let these arcs intersect each other at W. Join PW, which is the required ray making 105o with the given ray PQ.
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions
 
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:

(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 Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions  VW draw arcs to intersect each other at X. Join PX which is the required ray making 135o with the given line PQ.
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions
 
Now, we can measure the angle so formed with the help of a protractor. It comes to be 135o.                                    

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:

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. Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsABC is the required equilateral triangle of side 5 cm.
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions
 
Justification of Construction:
To justify the construction, we have to prove that ABC is an equilateral triangle i.e.AB = BC = AC = 5 cm and Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsA = Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsB = Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsC = 60o.
 
Now, in Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsABC, we have AC = AB = 5 cm and Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsA = 60o
Since, AC = AB, we have
Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsB = Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsC                         (angles opposite to equal sides of a triangle)
Now, in Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsABC
Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsNcert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsA + Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsB + Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsC = 180o    (angle sum property of a triangle)
 Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions60oNcert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsC + Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsC = 180o
 Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions60o + 2 Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsC = 180o
Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions 2 Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsC = 180o - 60o = 120o
Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsC = 60o
Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions
 
Now, we have Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsA = Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsB = Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsC = 60o              ... (1)
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsNcert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsA = Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsB and Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsA = Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsC
Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsBC = AC and BC = AB          (sides opposite to equal angles of a triangle)
Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsAB = BC = AC = 5 cm                                  ... (2)
Equations (1) and (2) show that the Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsABC is an equilateral triangle.

Chapter 11 - 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 sayNcert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsXBC.  
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 Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsBDC  
Step IV: Let CY intersects BX at A. Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsABC is the required triangle.   Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions
 
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 Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsXBC.     
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. Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsABC is the required triangle.
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions
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 Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsXQR.
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. Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsPQR is the required triangle.
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions
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 Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsPAB of 30o at point A and an angle Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsQBA of 90o at point B.
Step III: Bisect Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsPAB and Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsQBA. 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.Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsXYZ is the required triangle.
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions
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 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.Ncert Solutions Cbse Class 9 Mathematics Chapter - ConstructionsABC is the required triangle.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Constructions