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Constructions : Exercise 11.1 (Mathematics NCERT Class 9th)

Q.1      Construct an angle of 90º at the  initial  point of a given ray and justify the construction.
Sol.         Steps of construction :     

1.  Draw  a ray OA.
2. With its initial point  O as  centre and any radius,  draw an arc CDE, cutting OA at C.
3. With  centre  C and same  radius  (as in step 2), draw  an arc, cutting  the arc CDE at D.
4. With D as centre and the same radius, draw  an arc cutting the arc CDE at E.
5. With D and E as centres, and any convenient radius \left( {more\,than\,{1 \over 2}DE} \right), draw two arcs intersecting at P.
6. Join OP. Then  \angle AOP = {90^o}  
Justification : -
By  construction , OC = CD = OD
Therefore \Delta OCD is an  equilateral triangle. So, \angle COD ={ 60^o}
Again OD  = DE = EO

Therefore \Delta ODE is also an equilateral triangle. So \angle DOE = {60^o}
Since OP bisects \angle DOE,\,so\,\angle POD = {30^o}.

Q.2      Construct an angle of 45º at the initial  point of a  given ray and justify the construction.

Steps of Construction :
1.  Draw a ray OA.
2. With O as centre and any suitable  radius  draw an arc cutting OA at B.
3. With B as centre and same radius cut the previous drawn arc at C and then with C as centre  and same radius  cut the arc at D.
4. With C as centre and radius  more  than half CD draw an arc.
5. With D as centre  and same  radius  draw another  arc to cut the  previous arc at E.
6. Join  OE. Then \angle AOE = {90^o}
7. Draw the bisector OF of  \angle AOE.\,Then\,\angle AOF = {45^o}

2Justification :
By construction \angle AOE = {90^o} and OF is the bisector of \angle AOE

Q.3       Construct the angles of the following  measurements :
              (i) 30º          (ii) 22{1 \over 2}^\circ            (iii) 15º
Sol.          (i) Steps of Construction :

1. Draw  a ray OA.
2. With its initial  point O as centre and any radius, draw an  arc,cutting  OA at C.

3. With  centre C and Same radius (as in step 2). Draw an arc,cutting the arc of step 2 in D.
4. With C and D as centres, and any convenient radius \left( {more\,than\,{1 \over 2}CD} \right),draw two arcs intersecting at B.
5. Join  OB. Then \angle AOB = {30^o}

(ii)  Steps of Construction :
1. Draw an angle AOB  = 90º
2. Draw the bisector OC of  \angle AOB\,,\,\,then\,\angle AOC = {45^o}
3. Bisect \angle AOC, such  that \angle AOD\, = \,\angle COD = {22.5^o}
Thus \angle AOD\, = {22.5^o}

(iii) Steps of Construction :
1. Construct an \angle AOB\, = {60^o}
2. Bisect \angle AOB\, so that \angle AOC = \angle BOC = {30^o}.
3. Bisect \angle AOC, so  that \angle AOD = \angle COD = {15^o}
Thus \angle AOD = {15^o}

Q.4       Construct the following  angles and verify by  measuring  them by a protractor :
               (i)  75º           (ii) 105º           (iii) 135º
Sol.          (i)  Steps of Construction :

1.  Draw a ray OA.
2.  Construct \angle AOB = {60^o}

3.  Construct \angle AOP = {90^o}
4.  Bisect \angle BOP so that 
\angle BOQ = {1 \over 2}\angle BOP
 = {1 \over 2}\left( {\angle AOP - \angle AOB} \right)
 = {1 \over 2}\left( {90^o - 60^o} \right) = {1 \over 2} \times {30^o} = {15^o}
So, we obtain
\angle AOQ = \angle AOB + \angle BOQ
= 60º + 15º = 75º

Verification :
On  measuring  \angle AOQ, with  the protractor, we find \angle AOQ = {75^o}

(ii) Steps of Construction :
1. Draw a line  segment  XY.
2. Construct \angle XYT = {120^o} and \angle XYS = {90^o} , so that
\angle SYT = \angle XYT - \angle XYS
= 120º – 90º
= 30º
3.  Bisect angle SYT, by  drawing its bisector YZ.
Then  \angle XYZ is the   required angle  of 105º


7(iii) Steps  of Construction :

1. Draw \angle AOE = 90^\circ
Then \angle LOE = 90^\circ  
2. Draw   the bisector of  \angle LOE.
Then\,\,\angle AOF = {135^o}

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

Let us draw an equilateral  triangle  of side 4.6 cm (say).
Steps  of Construction :
1. Draw  BC = 4.6 cm
2. With  B and C as centres and radii  equal to BC = 4.6 cm, draw two  arcs on  the same  side of BC,    intersecting each-other at A.
3. Join AB and AC.

9Then, ABC is the required equilateral  triangle.
Justification : Since by construction :
AB = BC = CA = 4.6 cm
Therefore  \Delta ABC is an equilateral  triangle.


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