Q.1Â Â Â Â Â Construct an angle of 90Âº at theÂ initialÂ point of a given ray and justify the construction.
Sol.Â Â Â Â Â Â Â Â Steps of construction : Â Â Â
1.Â DrawÂ a ray OA.
2. With its initial pointÂ O asÂ centre and any radius,Â draw an arc CDE, cutting OA at C.
3. WithÂ centreÂ C and sameÂ radiusÂ (as in step 2), drawÂ an arc, cuttingÂ the arc CDE at D.
4. With D as centre and the same radius, drawÂ an arc cutting the arc CDE at E.
5. With D and E as centres, and any convenient radius , draw two arcs intersecting at P.
6. Join OP. Then Â Â
Justification : -
ByÂ construction , OC = CD = OD
Therefore OCD is anÂ equilateral triangle. So,
Again ODÂ = DE = EO
Therefore ODE is also an equilateral triangle. So
Since OP bisects .
Now,
Q.2Â Â Â Â Â Construct an angle of 45Âº at the initialÂ point of aÂ given ray and justify the construction.
Sol.Â
Steps of Construction :
1.Â Draw a ray OA.
2. With O as centre and any suitableÂ radiusÂ 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 centreÂ and same radiusÂ cut the arc at D.
4. With C as centre and radiusÂ moreÂ than half CD draw an arc.
5. With D as centreÂ and sameÂ radiusÂ draw anotherÂ arc to cut theÂ previous arc at E.
6. JoinÂ OE. Then
7. Draw the bisector OF ofÂ
Justification :
By construction and OF is the bisector of
Therefore,Â
Q.3Â Â Â Â Â Â Construct the angles of the followingÂ measurements :
Â Â Â Â Â Â Â Â Â Â Â (i) 30ÂºÂ Â Â Â Â Â Â (ii) Â Â Â Â Â Â Â (iii) 15Âº
Sol.Â Â Â Â Â Â Â Â Â (i) Steps of Construction :
1. DrawÂ a ray OA.
2. With its initialÂ point O as centre and any radius, draw anÂ arc,cuttingÂ OA at C.
3. WithÂ 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 ,draw two arcs intersecting at B.
5. JoinÂ OB. Then
(ii)Â Steps of Construction :
1. Draw an angle AOBÂ = 90Âº
2. Draw the bisector OC ofÂ
3. Bisect , suchÂ that
Thus
(iii) Steps of Construction :
1. Construct an
2. Bisect so that .
3. Bisect , soÂ that
Thus
Q.4Â Â Â Â Â Â Construct the followingÂ angles and verify byÂ measuringÂ them by a protractor :
Â Â Â Â Â Â Â Â Â Â (i)Â 75ÂºÂ Â Â Â Â Â Â (ii) 105ÂºÂ Â Â Â Â Â Â (iii) 135Âº
Sol.Â Â Â Â Â Â Â Â Â (i)Â Steps of Construction :
1.Â Draw a ray OA.
2.Â Construct
3.Â Construct
4.Â Bisect so thatÂ
So, we obtain
= 60Âº + 15Âº = 75Âº
Verification :
OnÂ measuringÂ , withÂ the protractor, we find
(ii) Steps of Construction :
1. Draw a lineÂ segmentÂ XY.
2. Construct and , so that
= 120Âº â€“ 90Âº
= 30Âº
3.Â Bisect angle SYT, byÂ drawing its bisector YZ.
ThenÂ is theÂ Â required angleÂ of 105Âº
Â
(iii) StepsÂ of Construction :
1. Draw
Then Â
2. DrawÂ Â the bisector ofÂ .
Q.5Â Â Â Â Â Â Construct an equilateral triangle, given its sideÂ and justify the construction.
Sol.
Let us draw an equilateralÂ triangleÂ of side 4.6 cm (say).
StepsÂ of Construction :
1. DrawÂ BC = 4.6 cm
2. WithÂ B and C as centres and radiiÂ equal to BC = 4.6 cm, draw twoÂ arcs onÂ the sameÂ side of BC,Â Â Â intersecting each-other at A.
3. Join AB and AC.
Then, ABC is the required equilateralÂ triangle.
Justification : Since by construction :
AB = BC = CA = 4.6 cm
ThereforeÂ ABC is an equilateralÂ triangle.
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