# NCERT Solutions for Class 9 Maths Exercise 11.1

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NCERT solutions for Class 9 Maths Constructions

## NCERT Solutions for Class 9 Mathematics Constructions

###### 1. Construct an angle of at the initial point of a given ray and justify the construction.

Ans. Steps of construction:

(a) Draw a ray OA.

(b) With O as centre and convenient radius, draw an arc LM cutting OA at L.

(c) Now with L as centre and radius OL, draw an arc cutting the arc LM at P.

(d) Then taking P as centre and radius OL, draw an arc cutting arc PM at the point Q.

(e) Join OP to draw the ray OB. Also join O and Q to draw the OC. We observe that:

AOB = BOC =

(f) Now we have to bisect BOC. For this, with P as centre and radius greater than PQ draw an arc.

(g) Now with Q as centre and the same radius as in step 6, draw another arc cutting the arc drawn in step 6 at R.

(h) Join O and R and draw ray OD.

Then AOD is the required angle of

###### Justification:

Join PL, then OL = OP = PL [by construction]

Therefore OQP is an equilateral triangle and POL which is same as BOA is equal to

Now join QP, then OP = OQ = PQ [ by construction]

Therefore OQP is an equilateral triangle.

POQ which is same as BOC is equal to

By construction OD is bisector of BOC.

DOC = DOB = BOC =

Now, DOA = BOA + DOB

DOA =

DOA =

NCERT Solutions for Class 9 Maths Exercise 11.1

###### 2. Construct an angle of at the initial point of a given ray and justify the construction.

Ans. Steps of construction:

(a) Draw a ray OA.

(b) With O as centre and convenient radius, draw an arc LM cutting OA at L.

(c) Now with L as centre and radius OL, draw an arc cutting the arc LM at P.

(d) Then taking P as centre and radius OL, draw an arc cutting arc PM at the point Q.

(e) Join OP to draw the ray OB. Also join O and Q to draw the OC. We observe that:AOB = BOC =

(f) Now we have to bisect BOC. For this, with P as centre and radius greater than PQ draw an arc.

(g) Now with Q as centre and the same radius as in step 6, draw another arc cutting the arc drawn in step 6 at R.

(h) Join O and R and draw ray OD. Then AOD is the required angle of

(i) With L as centre and radius greater than LS, draw an arc.

(j) Now with S as centre and the same radius as in step 2, draw another arc cutting the arc draw in step 2 at T.

(k) Join O and T and draw ray OE.

Thus OE bisects AOD and therefore AOE = DOE =

###### Justification:

Join LS then OLS is isosceles right triangle, right angled at O.

OL = OS

Therefore, O lies on the perpendicular bisector of SL.

SF = FL

And OFS = OFL [Each ]

Now in OFS and OFL,

OF = OF [ Common]

OS = OL [By construction]

SF = FL [Proved]

OFS OFL [By SSS rule]

SOF = LOF [By CPCT]

Now SOF + LOF = SOL

SOF + LOF =

2LOF =

LOF =

And AOE =

NCERT Solutions for Class 9 Maths Exercise 11.1

###### 3. Construct the angles of the following measurements:

(i)

(ii)

(iii)

Ans. (i) Steps of construction:

(a) Draw a ray OA.

(b) With O as centre and a suitable radius, draw an arc LM that cuts OA at L.

(c) With L as centre and radius OL, draw an arc to cut LM at N.

(d) Join O and N draw ray OB. Then AOB =

(e) With L as centre and radius greater than LN, draw an arc.

(f) Now with N as centre and same radius as in step 5, draw another arc cutting the arc drawn in step 5 at P.

(g) Join O and P and draw ray OC. Thus OC bisects AOB and therefore AOC = BOC =

###### (ii) Steps of construction:

(a) Draw a ray OA.

(b) With O as centre and convenient radius, draw an arc LM cutting OA at L.

(c) Now with L as centre and radius OL, draw an arc cutting the arc LM at P.

(d) Then taking P as centre and radius OL, draw an arc cutting arc PM at the point Q.

(e) Join OP to draw the ray OB. Also join O and Q to draw the OC. We observe that:

AOB = BOC =

(f) Now we have to bisect BOC. For this, with P as centre and radius greater than PQ draw an arc.

(g) Now with Q as centre and the same radius as in step 6, draw another arc cutting the arc drawn in step 6 at R.

(h) Join O and R and draw ray OD. Then AOD is the required angle of

(i) With L as centre and radius greater than LS, draw an arc.

(j) Now with S as centre and the same radius as in step 2, draw another arc cutting the arc draw in step 2 at T.

(k) Join O and T and draw ray OE. Thus OE bisects AOD and therefore AOE = DOE = .

(l) Let ray OE intersect the arc of circle at N.

(m) Now with L as centre and radius greater than LN, draw an arc.

(n) With N as centre and same radius as in above step and draw another arc cutting arc drawn in above step at I.

(o) Join O and I and draw ray OF. Thus OF bisects AOE and therefore AOF = EOF = .

###### (iii) Steps of construction:

(a) Draw a ray OA.

(b) With O as centre and a suitable radius, draw an arc LM that cuts OA at L.

(c) With L as centre and radius OL, draw an arc to cut LM at N.

(d) Join O and N draw ray OB. Then AOB =

(e) With L as centre and radius greater than LN, draw an arc.

(f) Now with N as centre and same radius as in step 5, draw another arc cutting the arc drawn in step 5 at P.

(g) Join O and P and draw ray OC. Thus OC bisects AOB and therefore AOC = BOC = .

(h) Let ray OC intersects the arc of circle at point Q.

(i) Now with L as centre and radius greater than LQ; draw an arc.

(j) With Q as centre and same radius as in above step, draw another arc cutting the arc shown in above step at R.

(k) Join O and R and draw ray OS. Thus OS bisects AOC and therefore COS = AOS =

NCERT Solutions for Class 9 Maths Exercise 11.1

###### 4. Construct the following angles and verify by measuring them by a protactor:

(i)

(ii)

(iii)

Ans. (i) Step of construction of

(a) Draw ABE = and ABF = . [ Follow the same steps as done in Question 1 and Question 3 (i)]

(b) Let ray BF intersects the arc of circle at G.

(c) Now with M as centre and radius greater than MG draw an arc.

(d) With G as centre and with same radius as in step (c), draw an arc which intersects the previous arc at point H.

(e) Draw a ray BC passing through H which bisects EBF.

Thus ABC = is the required angle.

Justification:

EBF = ABF – ABE =

Now EBF = CBF = EBF = [ BC is the bisector of EBF]

ABC = ABE + EBC =

(ii) Steps of construction of

NCERT Solutions for Class 9 Maths Exercise 11.1

(a) Draw ABE = and ABF =

(b) Let ray BE intersects the arc of circle at M and ray BF intersects the arc of circle N.

(c) With point M as centre and radius greater than MN, draw an arc.

(d) With N as centre and with same radius as in step (c), draw another arc which intersects the previous arc at P.

(e) Draw a ray BC passing through P which bisects EBF.

Thus ABC = is the required angle.

###### Justification:

EBF = ABF – ABE=

Now EBC = CBF = EBF = [ BC is the bisector of EBF]

ABC = ABE + EBC =

(iii) Steps of construction of

NCERT Solutions for Class 9 Maths Exercise 11.1

(a) Draw a ray OA.

(b) With O as centre and convenient radius, draw an arc LM (having length more than the semicircle) cutting OA at L.

(c) Now with L as centre and radius = OL; draw an arc cutting the arc LM at P.

(d) Then taking P as centre and radius OL, draw an arc cutting arc PM at Q.

(e) Now bisect POQ by ray OB, we get AOB =

(f) Now taking Q as centre and radius OL, draw an arc cutting QM at N.

(g) Join O and N to draw the ray OC.

Thus we get AOC = BOC = AOB =

(h) Now bisect BOC by ray OD.

Then AOD is the required angle of

AOD = AOB + BOD =

NCERT Solutions for Class 9 Maths Exercise 11.1

###### 5. Construct an equilateral triangle, given its side and justify the construction.

Ans. Steps of construction:

(a) Draw a line segment BC of length 6 cm.

(b) At B draw XBC =

(c) Draw perpendicular bisector PQ of line segment BC.

(d) Let A and D be the points where PQ intersects the ray BX and side BC respectively.

(e) Join AC.

Thus ABC is the required equilateral triangle.

###### Justification:

BD = CD [By construction]

B = C = [By CPCT]

A = (B + C)

= = =

A = B = C =

ABC is an equilateral triangle.

## NCERT Solutions for Class 9 Maths Exercise 11.1

NCERT Solutions for Class 9 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 9 Maths includes text book solutions from Mathematics Book. NCERT Solutions for CBSE Class 9 Maths have total 15 chapters. 9 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 9 solutions PDF and Maths ncert class 9 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.

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