# CLASS 9 MATHS CHAPTER-11 CONSTRUCTIONS

### Exercise 11.1

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

Solution:
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 =

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

Solution:
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 =

Question 3. Construct the angles of the following measurements :

(i)

(ii)

(iii)

Solution:
(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 =

Question 4. Construct the following angles and verify by measuring them by a protractor.
(i) 75°
(ii) 105°
(iii) 135°
Solution:
Step I : Draw .
Step II : With O as centre and having a suitable radius, draw an arc which cuts at B.
Step III : With centre B and keeping the same radius, mark a point C on the previous arc.
Step IV : With centre C and having the same radius, mark another point D on the arc of step II.
Step V : Join and , which gives ∠COD = 60° = ∠BOC.
Step VI : Draw , the bisector of ∠COD, such that
∠COP = ∠COD = (60°) = 30°.
Step VII: Draw , the bisector of ∠COP, such that
∠COQ = ∠COP = (30°) = 15°.

Thus, ∠BOQ = 60° + 15° = 75°∠AOQ = 75°

(ii) Steps of Construction:
Step I : Draw .
Step II : With centre O and having a suitable radius, draw an arc which cuts at B.
Step III : With centre B and keeping the same radius, mark a point C on the previous arc.
Step IV : With centre C and having the same radius, mark another point D on the arc drawn in step II.
Step V : Draw OP, the bisector of CD which cuts CD at E such that ∠BOP = 90°.
Step VI : Draw , the bisector of such that ∠POQ = 15°

Thus, ∠AOQ = 90° + 15° = 105°

(iii) Steps of Construction:
Step I : Draw .
Step II : With centre O and having a suitable radius, draw an arc which cuts at A
Step III : Keeping the same radius and starting from A, mark points Q, R and S on the arc of step II such that .
StepIV :Draw , thebisector of which cuts the arc at T.
Step V : Draw , the bisector of .

Thus, ∠POQ = 135°

Question 5. Construct an equilateral triangle, given its side and justify the construction.
Solution:
pt us construct an equilateral triangle, each of whose side = 3 cm(say).
Steps of Construction:
Step I : Draw .
Step II : Taking O as centre and radius equal to 3 cm, draw an arc to cut at B such that OB = 3 cm
Step III : Taking B as centre and radius equal to OB, draw an arc to intersect the previous arc at C.
Step IV : Join OC and BC.

Thus, ∆OBC is the required equilateral triangle.

Justification:
∵ The arcs and are drawn with the same radius.
∴
⇒ OC = BC [Chords corresponding to equal arcs are equal]
∵ OC = OB = BC
∴ OBC is an equilateral triangle.

### NCERT Solutions for Class 9 Maths Exercise 11.2

Question 1. Construct a ∆ ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm.
Solution:
Steps of Construction:
Step I : Draw .
Step II : Along , cut off a line segment BC = 7 cm.
Step III : At B, construct ∠CBY = 75°
Step IV : From , cut off BD = 13 cm (= AB + AC)
Step V : Join DC.
Step VI : Draw a perpendicular bisector of CD which meets BD at A.
Step VII: Join AC.

Thus, ∆ABC is the required triangle.

Question 2. Construct a ABC in which BC = 8 cm, ∠B = 45° and AB – AC = 35 cm.
Solution:
Steps of Construction:
Step I : Draw .
Step II : Along , cut off a line segment BC = 8 cm.
Step III : At B, construct ∠CBY = 45°
Step IV : From , cut off BD = 3.5 cm (= AB – AC)
Step V : Join DC.
Step VI : Draw PQ, perpendicular bisector of DC, which intersects at A.
Step VII: Join AC.

Thus, ∆ABC is the required triangle.

Question 3. Construct a ∆ ABC in which QR = 6 cm, ∠Q = 60° and PR – PQ = 2 cm.
Solution:
Steps of Construction:
Step I : Draw .
Step II : Along , cut off a line segment QR = 6 cm.
Step III : Construct a line YQY’ such that ∠RQY = 60°.
Step IV : Cut off QS = 2 cm (= PR – PQ) on QY’.
Step V : Join SR.
Step VI : Draw MN, perpendicular bisector of SR, which intersects QY at P.
Step VII: Join PR.

Thus, ∆PQR is the required triangle.

Question 4. Construct a ∆ XYZ in which ∠Y = 30°, ∠Y = 90° and XY + YZ + ZX = 11 cm.
Solution:
Steps of Construction:
Step I : Draw a line segment AB = 11 cm = (XY+YZ + ZX)
Step II : Construct ∠BAP = 30°
Step III : Construct ∠ABQ = 90°
Step IV : Draw AR, the bisector of ∠BAP.
Step V : Draw BS, the bisector of ∠ABQ. Let AR and BS intersect at X.
Step VI : Draw perpendicular bisector of , which intersects AB at Y.
Step VII: Draw perpendicular bisector of , which intersects AB at Z.
Step 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:
Step I : Draw BC = 12 cm.
Step II : At B, construct ∠CBY = 90°.
Step III : Along , cut off a line segment BX = 18 cm.
Step IV : Join CX.
Step V : Draw PQ, perpendicular bisector of CX, which meets BX at A.
Step VI : Join AC.

Thus, ∆ABC is the required triangle.

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