RD Chapter 8 Lines and Angles Ex 8.4 Solutions
Question - 1 : - In figure, AB, CD and ∠1 and ∠2 are in the ratio 3 : 2.Determine all angles from 1 to 8.
Answer - 1 : -
Let ∠1 = 3xand ∠2 = 2x
From figure: ∠1 and ∠2 are linear pair of angles
Therefore, ∠1 + ∠2 = 180
3x + 2x = 180
5x = 180
x = 180 / 5
=> x = 36
So, ∠1 = 3x= 1080 and ∠2 = 2x= 720
As we know, vertically opposite angles are equal.
Pairs of vertically opposite angles are:
(∠1 = ∠3); (∠2 = ∠4) ; (∠5, ∠7) and (∠6 , ∠8)
∠1 = ∠3 = 108°
∠2 = ∠4 = 72°
∠5 = ∠7
∠6 = ∠8
We also know, if a transversal intersects any parallel lines,then the corresponding angles are equal
∠1 = ∠5 = ∠7 = 108°
∠2 = ∠6 = ∠8 = 72°
Answer: ∠1 =108°, ∠2 =72°, ∠3 =108°, ∠4 =72°, ∠5 =108°, ∠6 =72°, ∠7 =108° and ∠8 = 72°
Question - 2 : - In figure, I, m andn are parallel lines intersected by transversal p at X, Y and Z respectively.Find ∠1, ∠2 and ∠3.
Answer - 2 : -
From figure:
∠Y =120° [Vertical opposite angles]
∠3 + ∠Y = 180° [Linear pair angles theorem]
=> ∠3= 180– 120
=> ∠3= 60°
Line l is parallel to line m,
∠1 = ∠3 [ Corresponding angles]
∠1 = 60°
Also, line m is parallel to line n,
∠2 = ∠Y [Alternate interior angles are equal]
∠2 =120°
Answer: ∠1 =60°, ∠2 =120° and ∠3 =60°.
Question - 3 : - In figure, AB || CD|| EF and GH || KL. Find ∠HKL.
Answer - 3 : -
Extend LK to meet line GF at point P.
From figure, CD || GF, so, alternate angles are equal.
∠CHG =∠HGP = 60°
∠HGP =∠KPF = 60° [Corresponding angles of parallellines are equal]
Hence, ∠KPG=180 – 60 = 120°
=> ∠GPK = ∠AKL= 120° [Corresponding angles of parallellines are equal]
∠AKH = ∠KHD = 25° [alternate angles of parallellines]
Therefore, ∠HKL = ∠AKH + ∠AKL = 25 + 120 = 145°
Question - 4 : - In figure, showthat AB || EF.
Answer - 4 : -
Produce EF to intersect AC at point N.
From figure, ∠BAC =57° and
∠ACD =22°+35° = 57°
Alternative angles of parallel lines are equal
=> BA || EF …..(1)
Sum of Co-interior angles of parallel lines is 180°
EF || CD
∠DCE + ∠CEF = 35 + 145 = 180° …(2)
From (1) and (2)
AB || EF
[Since, Lines parallel tothe same line are parallel to each other]
Hence Proved.
Question - 5 : - In figure, if AB ||CD and CD || EF, find ∠ACE.
Answer - 5 : -
Given: CD || EF
∠ FEC + ∠ECD = 180°
[Sum of co-interior anglesis supplementary to each other]
=> ∠ECD =180° – 130° = 50°
Also, BA || CD
=> ∠BAC = ∠ACD = 70°
[Alternative angles ofparallel lines are equal]
But, ∠ACE + ∠ECD =70°
=> ∠ACE =70° — 50° = 20°
Question - 6 : - In figure, PQ || ABand PR || BC. If ∠QPR = 102°, determine ∠ABC. Give reasons.
Answer - 6 : -
Extend line AB to meet line PR at point G.
Given: PQ || AB,
∠QPR = ∠BGR =102°
[Corresponding angles ofparallel lines are equal]
And PR || BC,
∠RGB+ ∠CBG =180°
[Corresponding angles aresupplementary]
∠CBG =180° – 102° = 78°
Since, ∠CBG = ∠ABC
=>∠ABC =78°
Question - 7 : - In figure, state which lines are parallel and why?
Answer - 7 : -
We know, If a transversal intersects two lines such that a pairof alternate interior angles are equal, then the two lines are parallel
From figure:
=> ∠EDC = ∠DCA = 100°
Lines DE and AC are intersected by a transversal DC such thatthe pair of alternate angles are equal.
So, DE || AC
Question - 8 : - In figure, if l||m,n || p and ∠1 = 85°, find ∠2.
Answer - 8 : -
Given: ∠1 = 85°
As we know, when a line cuts the parallel lines, the pair ofalternate interior angles are equal.
=> ∠1 = ∠3 = 85°
Again, co-interior angles are supplementary, so
∠2 + ∠3 = 180°
∠2 + 55°=180°
∠2 =180° – 85°
∠2 = 95°
Question - 9 : - If two straightlines are perpendicular to the same line, prove that they are parallel to eachother.
Answer - 9 : -
Let lines l and m are perpendicular to n, then
∠1= ∠2=90°
Since, lines l and m cut by a transversal line n and thecorresponding angles are equal, which shows that, line l is parallel to line m.
Question - 10 : - Prove that if thetwo arms of an angle are perpendicular to the two arms of another angle, thenthe angles are either equal or supplementary.
Answer - 10 : -
Let the angles be ∠ACB and ∠ABD
Let AC perpendicular to AB, and CD is perpendicular to BD.
To Prove : ∠ACD = ∠ABD OR ∠ACD + ∠ABD=180°
Proof :
In a quadrilateral,
∠A+ ∠C+ ∠D+ ∠B =360°
[ Sum of angles ofquadrilateral is 360° ]
=> 180° + ∠C + ∠B = 360°
=> ∠C + ∠B = 360° –180°
Therefore, ∠ACD + ∠ABD = 180°
And ∠ABD = ∠ACD = 90°
Hence, angles are equal as well as supplementary.