1. In Fig. 6.28, find the values of ` \x\ ` and ` \y\ ` and then show that AB ` \||\ ` CD.
solution :
Find,
AB ` \||\ ` CD
` \x=?\ `
` \y=?\ `
In the Fig.
` \=>y=130°\ `[ Vertically opposite angles ]
Now,
` \=>x+50°=180°\ `[ Linear pair of angles ]
` \=>x=180°-50°\ `
` \=>x=130°\ `
` \=>x=y\ `[ Alternate interior angles ]
We say,
AB ` \||\ ` CD
So,
The value of ` \x\ ` and ` \y\ ` is 130° respectively.
2. In Fig. 6.29, if AB ` \||\ ` CD, CD ` \||\ ` EF and ` \y:z=3:7\ `, find ` \x\ `.
solution :
Given,
AB ` \||\ ` CD, CD ` \||\ ` EF and ` \y:z=3:7\ `.
Find,
` \x=?\ `
` \y=?\ `
` \z=?\ `
In th Fig.
AB ` \||\ ` CD, CD ` \||\ ` EF
So,
AB ` \||\ ` EF
` \=>x=z\ `----- (i)[ Alternate interior angles ]
` \=>x+y=180°\ `----- (ii)[ Interior angles on the same side of the transversal ]
From (i) and (ii),
` \z+y=180°\ `----- (iii)
Now,
let ` \y=3p\ ` and ` \z=7p\ `[ As ` \y:z=3:7\ ` ]
From (iii),
` \∴3p+7p=180°\ `
` \=>10p=180°\ `
` \=>p=\frac{180°}{10}\ `
` \=>p=18°\ `
Now, ` \y=3×18°=54°\ `
and, ` \z=7×18°=126°\ `
From (ii),
` \=>x+y-180°\ `
` \=>x+54°=180°\ `
` \=>x=180°-54°\ `
` \=>x=126°\ `
So,
The value of ` \x\ ` is 126°.
3. In Fig. 6.30, if AB ` \||\ ` CD, EF ⊥ CD and ∠GDE = 126°, find ∠AGE, ∠GEF and ∠FGE.
solution :
Given,
AB ` \||\ ` CD, EF` \⊥\ `CD and ` \∠GED=126°\ `.
Find,
∠AGE, ∠GEF and ∠FGE.
In the Fig.
AB ` \||\ ` CD
` \=>∠AGE=∠GED\ `[ Alternate interior angles ]
So,
` \∠AGE=126°\ `
Now,
` \∠GED=∠GEF+∠FED\ `
` \=>126°=∠GEF+90°\ `
` \=>∠GEF=126°-90°\ `
` \=>∠GEF=36°\ `
Then,
` \∠AGE+∠FGE=180°\ `[ Linear pair of angles ]
` \=>126°+∠FGE=180°\ `
` \=>∠FGE=180°-126°\ `
` \=>∠FGE=54°\ `
So,
The value of ` \∠AGE=126°\ `, ` \∠GEF=36°\ ` and ` \∠FGE=54°\ `.
4. In Fig. 6.31, if PQ ` \||\ ` ST,
∠PQR=110° and ∠RST=130°, find ∠QRS.
[
Hint : Draw a line parallel to ST
through point R. ]
solution :
Given,
PQ ` \||\ ` ST, ` \∠PQR=110°\ ` and ` \∠RST=130°\ `
Find,
` \∠QRS=?\ `
Let,
Draw a parallel line to ST through point R is MN.
In the Fig.
ST ` \||\ ` MN
So,
` \∠RST=∠MRS\ `[ Alternate interior angles ]
` \=>∠MRS=130°\ `
Similarly,
PQ ` \||\ ` ST, ST ` \||\ ` MN,
Then, PQ ` \||\ ` MN
` \∠PQR=∠QRN\ `[ Alternate interior angles ]
` \∠QRN=110°\ `
Now,
` \∠MRQ+∠QRS+∠SRN=180°\ `[ Linear pair of angles ]
` \=>∠MRQ+∠QRN=180°\ `` \[ ∴∠QRS+∠SRN=∠QRN ]\ `
` \=>∠MRQ+110°=180°\ `
` \=>∠MRQ=180°-110°\ `
` \=>∠MRQ=70°\ `
Then,
` \=>∠MRS=130°\ `
` \=>∠MRQ+∠QRS=130°\ `` \[ ∴∠MQR+∠QRS=∠MRS ]\ `
` \=>70°+∠QRS=130°\ `
` \=>∠QRS=130°-70°\ `
` \=>∠QRS=60°\ `
So,
The value of ` \∠QRS=60°\ `.
5. In Fig. 6.32, if AB ` \||\ ` CD, ∠APQ=50° and ∠PRD=127°, find ` \x\ ` and ` \y\ `.
solution :
Given,
AB ` \||\ ` CD, ` \∠APQ=50°\ ` and ` \∠PRD=127°\ `
Find,
` \x=?\ ` and ` \y=?\ `
In the Fig.
AB ` \||\ ` CD and transversal line PQ.
` \∠APQ=∠PQR\ `[ Alternate interior angles ]
` \=>∠PQR=50°\ `` \[ ∴∠APQ=50°(Given) ]\ `
` \=>x=50°\ `
Similarly,
AB ` \||\ ` CD and transversal line PR.
` \∠PRD=∠APR\ `[ Alternate interior angles ]
` \=>∠APR=127°\ `` \[ ∴∠PRD=127°(Given) ]\ `
` \=>∠APR=∠APQ+∠QPR\ `
` \=>127°=50°+∠QPR\ `` \[ ∴∠APQ=50°(Given) ]\ `
` \=>∠QPR=127°-50°\ `
` \=>∠QPR=77°\ `
` \=>y=77°\ `
So,
The values of ` \x=50°\ ` and ` \y=77°\ `.
6. In Fig. 6.33, PQ and RS and two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. prove that AB ∥ CD.
solution :
Given,
PQ ` \||\ ` RS
Find,
AB ` \||\ ` CD
Let,
Draw two lines BD and CN such BO` \⊥\ `PQ and CN` \⊥\ `RS.
` \∠1=∠2\ `----- (i)[ Angle of incident ` \=\ ` Angle of reflection ]
` \∠3=∠4\ `----- (ii)[ Angle of incident ` \=\ ` Angle of reflection ]
BO ` \||\ ` CN[ ∵BO and CN are perpendicular on parallel mirrors PQ and RS ]
` \∠2=∠3\ `----- (iii)[ Alternate interior angles ]
From (i), (ii) and (iii),
We get,
` \∠1=∠4\ `----- (iv)
Adding (ii) and (iv),
We get,
` \∠1+∠2=∠3+∠4\ `
` \=>∠ABC=∠BCD\ `[ Alternate interior angles ]
So,
We say,
AB ` \||\ ` CD
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