### Constructions - Solutions

CBSE Class 9 Mathematics

NCERT Solutions
CHAPTER 11
Constructions(Ex. 11.1)

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 (initial point of a ray OA) 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 POL 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 =

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) With P and Q as centres and radius greater than ED, draw two arcs intersecting each other at R and let OR intersect the arc PM at S. Then, we get $\mathrm{\angle }$ROA = ${90}^{\circ }$

(f) Now we have to bisect ROA. With L as centre and radius greater than LS, draw an arc.

(g) 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.

(h) 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 =

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) With P and Q as centres and radius greater than PQ, draw two arcs intersecting each other at B and let OB intersect the arc PQ at S. Then we get $\mathrm{\angle }$AOB = ${90}^{\circ }$.

(f) Now we have to bisect BOA. For this, with L as centre and radius greater than LS draw an arc.

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

(h) Join O and T and draw ray OE. Then AOE is the required angle of ${45}^{\circ }$.

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

(j) Now with PS as centre and the same radius as in step 6, draw another arc cutting the arc draw in step 7 at I.

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

(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 =

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

(i)     (ii)      (iii)

Ans. (i) Step of construction o

(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

(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

(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 =

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

Ans. Steps of construction:

(a) Draw a line segment BC of length 5 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:

In right triangle ADB and right triangle ADC,

BD = CD [By construction]