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Brilliant’s Composite Mathematics Class 8 Solutions Chapter 10 Parallel Lines
Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Brilliant’s Composite Mathematics Class 8 Math Book, Chapter 10, Parallel Lines. Here students can easily find step by step solutions of all the problems for Parallel Lines, Exercise 10.1, 10.2, 10.3 and 10.4 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.
Parallel Lines Solution Exercise 10.1
Question no – (1)
Solution :
Required figure,
We extend the line BP till E (a point on CD).
Now, ∠DEP = ∠PBA [Alternate angles]
In △DEP, we have,
∠EDP = 55°, ∠PED = 45°,
then ∠DPE = 180° – (45° + 55°)
= 80°
∴ ∠DPB = 180°- 80°
= 100°
Question no – (5)
Solution :
Required figure,
Since AB ∥ CE and ∠DCE = 45°
Therefore ∠ABC = 45°
∴ ∠ACB = 180° – (65° + 45°)
= 180° – 110°
= 70°
Question no – (7)
Solution :
Required figure,
(i) Since, in AA’B’B, the opposite sides are parallel to each other,
Therefore, this is a parallelogram.
(ii) Since the points B, C, B’, C’ lie on a single line,
Therefore, BB’C’C is not a parallelogram.
Question no – (9)
Solution :
Required figure,
Since ∠1 : ∠2 = 3:6,
And ∠1 + ∠2 = 180°,
We have = ∠1 = 3/(3 + 6) × 180°
= 3/9× 180° = 60°
∠2 = 180° – 60° = 120°
Now, ∠1 = ∠3 (Opposite angles)
∠1 = ∠5 (Corresponding angles)
∠3 = ∠1 (Corresponding angles)
∴ ∠1 = ∠3 = ∠5 = ∠7 = 60°
Similarly, ∠2 = ∠4 = ∠6 = ∠8 = 120°
Parallel Lines Exercise 10.2 Solution
Question no – (1)
Solution :
Required figure,
We have DE ∥ AB
and DE = 1/2 AB [Mid – point theorem]
Similarly, DF = 1/2 AC, EF = 1/2 BC
∴ ∠DE = DF = EF
∴ △DEF is an equilateral triangle.
Question no – (2)
Solution :
∠AED = ∠ECB = 45°
∴ ∠ADE = 180° – (65° + 45°)
= 180° – 110°
= 70°
∴ ∠EDB = 180° – 70°
= 110°
Question no – (3)
Solution :
(i) EF ∥ BC
= True, [Mid Point theorem]
(ii) ED ∥ AB
= True, [Mid Point theorem]
(iii) FE = BD
= True [FE = 1/2 BC = BD]
(iv) EF = 1/2 BC
= True, [Mid Point theorem]
(v) BDEF is a parallelogram.
= True [BD ∥ FE, DE ∥ BF]
Question no – (4)
Solution :
Since, D, E, F are midpoints of BC, CA, AB
DE = 1/2 AB, EF = 1/2 BC, DF = 1/2 CA.
∴ Perimeter of △DEF = DE + EF + DF
= 1/2 AB + 1/2 BC + 1/2 CA
= 1/2 (AB + BC + CA)
= 1/2 (Perimeter of △ ABC)
Question no – (6)
Solution :
Comparing, △BLN and △CMN, We have
∠BLN = ∠CMN [both are 90°]
∠LNB = ∠MNC [Opposite angles]
∠LBN = ∠MCN [∠LBN = 180° – {90° + ∠LNB}
= 90° – ∠LNB
= 90° – ∠MNC = ∠MCN
∴ Since all the angles are equal we have
△BLN ≅ △CMN.
Therefore, LN = NM
Parallel Lines Exercise 10.3 Solution
Question no – (1)
Solution :
Required Figure,
Since, AD ⊥ l, CE ⊥ l, BF ⊥ l,
We have, AD ∥ CE ∥ BF.
Now, DE/EF = AC/CB
[Since AD ∥ CE ∥ BF and AB, l intersect them]
Or, DE/EF = l ….[Since AC = CB]
Or, DE = EF
= E is the midpoint of DF.
Question no – (4)
Solution :
Since, ∠Def = ∠GHE, we have GH || ED.
So, EH/HF = DG/GF
Or, 5/x = 3/2
Or, x = 2 × 5/3
= 3 1/3 cm
Question no – (5)
Solution :
Since, EF || BC and FG || AD
We have, AE/EB = AF/FC = DG/GC
Or, 3/2 = DG/3
Or, DG = 9/2
∴ DG = 4.5 cm
Question no – (6)
Solution :
Since, M and N divide AB into three equal parts
We have AM : MN : NB = 1 : 1 : 1,
AM : MB = 1 : 2,
AN : NB = 2 : 1
Since MP || BC and NQ || BC,
We have AM/MN = AP/PQ
Or, 1/1 = AP/PQ
∴ AP = PQ ….. (i)
Again MN/NB + PQ/QC
Or, 1/1 = PQ/QC
Or, PQ = QC ….. (ii)
From (i) and (ii), we have AP = PQ = QC
∴ P < Q divide AC into three equal parts.
Question no – (7)
Solution :
Required Figure,
Since, DA ⊥ AB, OM ⊥ AB, CA ⊥ AB,
We have, DA || OM || CB.
∴ AO/OC = DO/OB
Or, 2.4/3.6 = DO/3
Or, 2/3 = DO/3
Or, DO = 2 cm
Question no – (9)
Solution :
Required Figure,
Since, AB || CD || EF, and AC : CE = 2 : 3,
AC/CE = BD/DF = 2/3 ….. (i)
Now, BP || DQ || FR
∴ BD/DF = PQ/QR = 2/3 …… (ii)
Question no – (10)
Solution :
(i) PM/MQ = 0.2/(1.3 – 0.27) = 0.2/1.1 = 2/11
PN/NR = 0.3/(2.6 – 0.3) = 0.3/2.3 = 3/23
(ii) PM/QM = 4/4.5 = 8/9, PN/NR = 4/4.5 = 8/9 = PM/QM
∴ Since PM/QM = PN/NR,
∴ MN || QR
Question no – (11)
Solution :
(i) MQ/PM = NR/PN
Or, 4/6 = NR/8
Or, NR = 4 × 8/6 = 16/3 cm
(ii) PM/MQ = PN/NR
Or, PM/MQ = (12 -4)/4
Or, PM/MQ = 8/4 = 2/1
Now, PM : MQ = 2 : 1 and PM + MQ = 12 cm
∴ Clearly, PM = 8 cm
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