Breath Math

1. 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˚

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2. 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)]

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3. 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˚

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4. 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˚

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5. 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.