- Construct an angle of 90˚ at the initial point of a given ray and justify the construction.
Solution:
Construction:
- Draw a line ray AB with initial point A
- Taking A as a centre draw an arc XY with any radius intersecting the ray AB at P.
- Draw another arc with same radius by taking C as centre which intersects the arc XY at Q. Draw a ray joining AQ.
- Taking Q as centre with same radius draw another arc which intersects the arc XY at R. Draw a ray joining AR.
- Now, taking Q and R as centres with any radius draw two arc intersecting one another at S.
- Join AS. Then angle SAB = 90˚
Justification:
By construction, we have AP = AQ = PQ, therefore, triangle APQ is an equilateral triangle. Thus, ∠APQ = ∠AQP = ∠QAP = 60˚
Also, we have, AQ = AR =RQ , therefore, triangle ARQ is also an equilateral triangle. Thus, ∠ARQ = ∠AQR = ∠QAR = 60˚
Since, AS bisects ∠QAR then we have ∠QAS = ∠SAR = 1/2∠QAR = 1/2 x 60˚ = 30˚
Thus, ∠PAS = ∠PAQ + ∠QAS = 60˚ + 30˚ = 90˚
- Construct an angle of 45˚ at the initial point of a given ray and justify the construction.
Solution:
Construction:
Draw a line ray AB with initial point A
- Taking A as a centre draw an arc XY with any radius intersecting the ray AB at P.
- Draw another arc with same radius by taking C as centre which intersects the arc XY at Q. Draw a ray joining AQ.
- Taking Q as centre with same radius draw another arc which intersects the arc XY at R. Draw a ray joining AR.
- Now, taking Q and R as centres with any radius draw two arc intersecting one another at S.
- Join AS. Then angle SAB = 90˚
- Again taking S and P as centres with any radius draw two arcs intersecting each other at L.
- Join AL . Therefore, ∠PAL = 45 ˚
Justification:
Join GH and CH
In ∆AHG and ∆AHC, we have [arcs of equal radii]
HG = HC [radii of the same arc]
AG = AC [common]
⸫∆AHG ≅ ∆AHC [SSS congruence]
⇒∠HAG = ∠HAC [CPCT] ……………(i)
But ∠HAG + ∠HAC = 90˚ [By construction] ……………….(ii)
⇒∠HAG = ∠HAC = 45˚ [from (i) and (ii)]
- Construct the angles of the following measurements:
(i) 30° (ii) 22 1/2° (iii) 15°
Solution:
(i) 30° = 60 ˚/2
Construction:
Draw a ray AB with initial point A
- Draw an arc XY with any radius by taking A as centre which cuts the ray AB at C.
- Draw an arc by taking C as centre with same radius which intersects arc XY at D. Since AC = AD = DC then, ∆ADC is an equilateral triangle.
- ∠ADC = ∠ACD = ∠DAC = 60°
- Taking C and D as centres with any radius draw two arcs intersecting one another at E. Join AE.
- Thus, 1/2∠EAC = ∠DAC ; 1/2∠EAC = 60˚ ⇒∠EAC = 30˚
(ii)221/2˚ = 90˚/2 = 45˚
Construction:
Draw a line ray AB with initial point A
- Taking A as a centre draw an arc XY with any radius intersecting the ray AB at P.
- Draw another arc with same radius by taking C as centre which intersects the arc XY at Q. Draw a ray joining AQ.
- Taking Q as centre with same radius draw another arc which intersects the arc XY at R. Draw a ray joining AR.
- Now, taking Q and R as centres with any radius draw two arc intersecting one another at S.
- Join AS. Then angle SAB = 90˚
- Taking S and P as centres draw two arcs intersecting each other at T. Join AT. i.e., ∠TAP = 1/2 ∠SAP
∠TAP = 1/2 x 90 ˚ = 45 ˚ = 221/2˚
(iii) 15°
Construction:
Draw a ray AB with initial point A
- Draw an arc XY with any radius by taking A as centre which cuts the ray AB at C.
- Draw an arc by taking C as centre with same radius which intersects arc XY at D. Since AC = AD = DC then, ∆ADC is an equilateral triangle.
- ∠ADC = ∠ACD = ∠DAC = 60°
- Taking C and D as centres with any radius draw two arcs intersecting one another at E. Join AE.
- Thus, 1/2∠EAC = ∠DAC ; 1/2∠EAC = 60˚ ⇒∠EAC = 30˚
- Taking E and C as centres draw two arcs with any radius intersecting each other at F. Join AF
- Thus, 1/2∠FAC = ∠EAC ; 1/2∠FAC = 30˚ ⇒∠FAC = 15˚
- Construct the following angles and verify by measuring them by a protractor:
(i) 75° (ii) 105° (iii) 135°
Solution:
(i) 75°
Construction:
Draw a line ray AB with initial point A
- Taking A as a centre draw an arc XY with any radius intersecting the ray AB at P.
- Draw another arc with same radius by taking C as centre which intersects the arc XY at Q. Draw a ray joining AQ.
- Since AQ = AP = PQ, ∆APQ is an equilateral triangle. Therefore, ∠PAQ = 60 ˚
- Taking Q as centre with same radius draw another arc which intersects the arc XY at R. Draw a ray joining AR.
- Now, taking Q and R as centres with any radius draw two arc intersecting one another at S.
- Join AS. Then angle SAB = 90˚
- ∠SAB = ∠PAQ + ∠QAS ⇒90 ˚ = 60 ˚+ ∠QAS ⇒∠QAS = 30˚
- Taking S and Q as centres draw two arcs intersecting each other at M.
- Join Then ∠MAQ = 1/2∠QAS = 1/2x 30˚ = 15˚
- Therefore, ∠PAM = ∠MAQ + ∠PAQ = 60˚ + 15˚ = 75˚
(ii) 105°
Construction:
Draw a line ray AB with initial point A
- Taking A as a centre draw an arc XY with any radius intersecting the ray AB at P.
- Draw another arc with same radius by taking C as centre which intersects the arc XY at Q. Draw a ray joining AQ.
- Since AQ = AP = PQ, ∆APQ is an equilateral triangle. Therefore, ∠PAQ = 60 ˚
- Taking Q as centre with same radius draw another arc which intersects the arc XY at R. Draw a ray joining AR.
- Now, taking Q and R as centres with any radius draw two arc intersecting one another at S.
- Join AS. Then angle SAB = 90˚
- Taking S and R as centre with any radius draw two arcs intersecting each other at T.
- Join AT. Then, ∠SAT = 1/2 ∠SAR = 1/2 x 30 ˚ = 15 ˚
- Thus, ∠PAT = ∠PAS + ∠SAT = 90 ˚ + 15 ˚ = 105 ˚
(iii) 135°
Construction:
Draw a line ray AB with a point O.
- Taking O as a centre draw an arc XY with any radius intersecting the ray AB at C.
- Draw another arc with same radius by taking C as centre which intersects the arc XY at D.
- Taking D as centre with same radius draw another arc which intersects the arc XY at E. Draw a ray joining OF.
- Now, taking D and E as centres with any radius draw two arc intersecting one another at F.
- Join OF. Then angle ∟FOC = 90˚
- Taking O as radius draw an arc interesting the ray AB at G. ∟FOG = 90˚
- Now, taking G and F as centres with any radius draw two arc intersecting one another at H.
- Join OH. ∟HOF = 1/2∟FOG =1/2 x 90˚ = 45˚
- ∠HOC = ∠FOC + ∠HOF = 90˚ + 45˚ = 135˚
- Construct an equilateral triangle, given its side and justify the construction.
Solution:
Steps of Construction:
- Draw a line segment AB of the given length
- Taking A and B as centres with radius same as that of the length of AB draw two arcs intersecting each other at C
- Join AC and BC.
- Then triangle ABC is an equilateral triangle.
Justification:
- By construction AB= BC = AC
- Then triangle ABC is an equilateral triangle.
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