Selina Concise Class 6 Math Chapter 23 Properties Of Angles And Lines Exercise 23B Solution

Selina Concise Class 6 Math Chapter 23 Properties Of Angles And Lines Exercise 23B Solution

EXERCISE 23B

(a) (i) Adjacent angles

(ii) Alternate exterior angles

(iii) Interior alternate angles

(iv) Corresponding angles

(v) Allied angles

(b) (i)  Alternate Interior angles

(ii) Corresponding angles

(iii) Alternate exterior angles

(iv) Corresponding angles

(v) Allied angles

(C) (i) Corresponding angles

(ii) Alternate exterior

(iii) Alternate interior angles

(iv) Alternate interior angles

(v) Alternate exterior angles

(vi) Vertically opposite angles

(2) Each figure given below shows a pair of parallel lines cut by a transversal. For each case, find a and b, giving reasons.

(i) a + 140o = 180o       (linear pair)

⇒ a = 180 – 140 = 40o

But, b = a            (alternate angles)

⇒ b = 40o

(ii) b + 60o = 180o   (Linear pair)

⇒ b = 180 – 60 = 120o

But, a = 60o          (corresponding angles)

(iii) a = 110o (vertically opposite angles)

But b = 180o – a (co-interior angles)

⇒ b = 180 – 110 = 70o

(iv) a = 60o    (alternate interior angles)

And, b = 180o – a   (Co-interior angles)

⇒ b = 180 – 60 = 120o

(v) a = 72o   (Alternate interior angles)

b = a = 72o   (Vertically opposite angles)

(vi) b = 100o (Corresponding angles)

a = 180o – b      (Linear pair angles)

⇒ a = 180o – 100o

⇒ a = 80o

(vii) a = 180o – 130o  (Co-interior angles)

⇒ a = 50o

b = 130o      (Vertically opposite angles)

(viii) b = 62o (Corresponding angles)

a = 180o – b (linear pair of angles)

⇒ a = 180 – 62 = 118o

(ix) a = 180o – 90o   (Linear pair of angles)

⇒ a = 90o

b = 90o   (Corresponding angles)

(3) If ∠1 = 120o, find the measure of: ∠2, ∠3, ∠4, ∠5, ∠6 and ∠8.

Give reasons.

Solution: ∠1 + ∠2 = 180o (straight angle)

∴ 120o + ∠2 = 180o

⇒ ∠2 = 180o – 120o = 60o

But ∠1 = ∠3 (vertically opposite angles)

∴ ∠3 = ∠1 = 120o

Similarly, ∠4 = ∠2 (vertically opposite)

∴ ∠4 = 60o

∠5 = ∠1 (Corresponding angles)

∴ ∠5 = 120o

Similarly ∠6 = ∠2 (corresponding angles)

∴ ∠6 = 60o

∠7 = ∠5 (vertically opposite angles)

∴ ∠7 = 120o

And ∠8 = ∠6 (vertically opposite angles)

∴ ∠8 = 60o

(4) In the figure given alongside, find the measure of the angles denoted by x, y, z, p, q and r.

Solution: x = 180o – 100o (linear pair of angles)

⇒ x = 80o

y = x      (alternate exterior angles)

⇒ y = 80o

z = 100o (Corresponding angles)

p = x (vertically opposite angles)

⇒ p = 80o

q = 100o (Vertically opposite angles)

r = q (Corresponding angles)

⇒ r = 100o

(5) Using the figure given alongside, fill in the blanks:

∠x = 60o (corresponding angles)

∠z = ∠x (corresponding angles)

⇒ ∠z = 60o

∠p = ∠z (vertically opposite angles)

⇒ ∠p = 60o

∠q = 180 – ∠p (Linear pair of angles)

= 180o – 60o = 120o

∠s = ∠r (vertically opposite angles)

⇒ ∠s = 120o

(6) In the figure given alongside, find the angles shown by x, y, z and w. give reasons.

Solution: ∠x = 115o (Vertically opposite of angles)

∠y = 70o (vertically opposite angles)

∠z = 70o (alternate interior angles)

∠w = 115o (alternate interior angles)

(7) Find a, b, c and d in the figure given below:

Ans: ∠a = 130o (vertically opposite angles)

∠b = 150o (vertically opposite angles)

∠c = 150o (alternate interior angles)

∠d = 130o (alternate interior angles)

(8) Find x, y and z in the figure given below:

Ans: ∠x = 180 – 75 = 105o (co-interior angles)

∠y = 180 – ∠x (co-interior angles)

⇒ ∠y = 180 – 105 = 75o

z = 75o (corresponding angles)

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