# 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 $\angle DOE,\,so\,\angle POD = {30^o}$.
Now, $\angle AOP = {\mkern 1mu} \angle COD + \angle DOP = 60^\circ� + 30^\circ� = {90^o}$

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

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  $\angle AOE.\,Then\,\angle AOF = {45^o}$

By construction $\angle AOE = {90^o}$ and OF is the bisector of $\angle AOE$
Therefore, $\angle AOF = {1 \over 2}\angle AOE = {1 \over 2} \times 90^\circ� = {45^o}$

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 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.
Sol.

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.

Justification : Since by construction :
AB = BC = CA = 4.6 cm
Therefore  $\Delta$ ABC is an equilateral  triangle.

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