Selina Concise Class 7 Math Chapter 14 Lines And Angles Exercise 14A Solution
EXERCISE 14A
(1) State, true or false:
(i) False
(ii) True
(iii) True
(iv) True
(v) Write the number of end points in
(a) 2
(b) 1
(c) 0
(vii) Infinite
(viii) (a) 1
(b) 3
(ix) No
(x) No
(2) In which of the following figures, are ∠AOB and ∠AOC adjacent angles?
Give, in each case, reason for your answer.
(i) ∠AOB and ∠AOC not adjacent angles; because OB is their common arm and they are not on opposite sides.
(ii) ∠AOB and ∠AOC not adjacent angles; because OC is their common arm and they are not on opposite sides.
(iii) ∠AOB and ∠AOC are adjacent angles, because OA is their common arm and they are on opposite sides.
(iv) ∠AOB and ∠AOC not adjacent angles; because OB is their common arm and they are not on opposite sides.
(3) In the given figure, AC is a straight line. Find:
(i) ∠AOB + ∠AOC = 180° (linear pairs)
Then, x + 25° + 3x + 15° = 180°
⇒ 4x + 40° = 180°
⇒ 4x = 180° – 40° = 140°
⇒ x = 140°/4 = 35°
(ii) ∠AOB = x + 25° = (35 + 25)° = 60°
(iii) ∠BOC = 3x + 15°
= (3 × 35)° + 15°
= 105° + 15° = 120°
(4) Find y in the given figure.
Solution: We know, ∠AOB + ∠BOD + ∠DOC = 180°
⇒ y + 150 – x + x = 180°
⇒ y = (180 – 150)°
⇒ y = 30°
(5) In the given figure, find ∠PQR.
Solution: We know, ∠SQP + ∠PQR = 180°
⇒ x° + 70° + 20° – x° + ∠PQR = 180°
⇒ ∠PQR + 90° = 180°
⇒ ∠PQR = 180° – 90° = 90°
(6) In the given figure, if p° = q° = r°, find each.
Solution: Let, p° = q° = r° = x°
Then, x° + x° + x° = 180° (linear pair)
⇒ 3x° = 180°
⇒ x° = 180°/3
⇒ x° = 60°
Hence, p° = q° = r° = 60°
(7) In the given figure, if x = 2y, find x and y.
Solution: We know, x + y = 180° (linear pair)
⇒ 2y + y = 180°
⇒ 3y = 180°
⇒ y = 180°/3
⇒ y = 60°
Hence, x = (2 × 60) = 120°
(8) In the adjoining figure, if b° = a° + c°, find b.
Solution: a° + b° + c° = 180° (linear pair)
⇒ a° + c° + b° = 180°
⇒ b° + b° = 180°
⇒ 2b° = 180°
⇒ b° = 180°/2
⇒ b° = 90°
(9) In the given figure, AB is perpendicular to BC at B.
(i) x + 20° + 2x + 1° + 7x – 11° = 90°
⇒ 10x + 21° – 11° = 90°
⇒ 10x + 10° = 90°
⇒ 10x = 90° – 10°
⇒ x = 80°/10
⇒ x = 8
(ii) Complement of angle x = 90° – 8 = 82°
(10) Write the complement of:
(i) 90° – 25° = 65°
(ii) 90° = 90° – 90° = 0
(iii) 90° – a°
(iv) 90° – (x + 5) °
= 90° – x – 5°
= 80° – x
(v) 90° – (30 – a) °
= 90° – 30° + a°
= 60° + a°
(vi) ½ of a right angle
= 90° – (90/2)°
= 90° – 45° = 45°
(vii) 1/3 of 180°
= 90° – (180/3)°
= 90° – 60° = 30°
(viii) 90° – 21° 17’
= 68° 43’
(11) Write the supplement of:
(i) 180° – 100° = 80°
(ii) 180° – 0° = 180°
(iii) 180° – x°
(iv) 180° – (x + 35)°
= 180° – x° – 35°
= 145° – x°
(v) 180° – (90 + a + b)°
= 180° – 90° – a° – b°
= 90° – (a + b) °
(vi) 180° – (110 – x – 2y)°
= 180° – 110° + x + 2y
= 70° + (x + 2y) °
(vii) 180° – (90/5) °
= 180° – 18°
= 162°
(viii) 180° – 80° 49’ 25’’
= 99° 10’ 35”
(12) Are the following pairs of angles complementary?
(i) 10° + 80° = 90°, so it is complementary.
(ii) 37° 28’ + 52° 33’ = 90° 01’ ≠ 90°, so it is not a complementary.
(iii) x + 16° + 74° – x
= 90°, so it is complementary.
= 90°, so it is complementary.
(13) Are the following pairs of angles supplementary?
(i) 139° + 39° = 178° ≠ 180°, so it is not supplementary.
(ii) 26° 59’ + 153° 1,
= 180°, so it is supplementary.
= 75° ≠ 180°, so it is not supplementary.
(iv) 2x° + 65° + 115° – 2x°
= 180°, so it is supplementary.
(14) 3x + 18° + 2x + 25° = 180°
⇒ 5x + 43° = 180°
⇒ 5x = 180° – 43°
⇒ x = 137°/5
⇒ x = 27° 4’
(15) Let the angles are x and 5x.
Then, x + 5x = 90°
⇒ 6x = 90°
⇒ x = 90°/6
⇒ x = 15°
Hence, the two angles are 15° and (5×15°)= 75°.
(16) Let the two angles are 2x and 7x.
Then, 2x + 7x = 180°
⇒ 9x = 180°
⇒ x = 180°/9
⇒ x = 20°
Hence, the two angles are (2 × 20°) = 40° and (7 × 20°) = 140°
(17) Let the angles are 2x, 3x and 7x.
Then, 2x + 3x + 7x = 180°
⇒ 12x = 180°
⇒ x = 180°/12
⇒ x = 15°
Hence, the angles are (2 × 15°) = 30°
(3 × 15°) = 45°
(7 × 15°) = 105°
(20) 4x + 3x + 2x+ x = 180° (linear pair)
⇒ 10x = 180°
⇒ x = 180°/10
⇒ x = 18°
Its supplement = 180° – 18° = 162°
(21) K – 15° + 30° + K + 150° + 90° = 360°
⇒ 2K + 15° + 150° + 90° = 360°
⇒ 2K = 360° – 255°
⇒ 2K = 105°
⇒ K = 105°/2 = 52° 5’
(ii) 42° + 3K + 2K + K = 360°
⇒ 6K = 360° – 42°
⇒ 6K = 318°
⇒ K = 318°/6 = 53°