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NCERT Exemplar · Class 9 Mathematics Lines and Angles

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EXERCISE 6.1 1–8 (part 1 of 4)

  1. Write the correct answer in each of the following:

    Exercise 1

    In Fig. 6.1\displaystyle 6.1, if AB∥CD∥EF,PQ∥RS,∠RQD\displaystyle \mathrm{AB}\|\mathrm{CD}\| \mathrm{EF}, \mathrm{PQ} \| \mathrm{RS}, \angle \mathrm{RQD} =25∘\displaystyle =25^{\circ} and ∠CQP=60∘\displaystyle \angle \mathrm{CQP}=60^{\circ}, then ∠QRS\displaystyle \angle \mathrm{QRS} is equal to (A) 85\displaystyle 85° (B) 135\displaystyle 135° (C) 145\displaystyle 145° (D) 110\displaystyle 110° NCERT_Question_Class9_Maths_Exemplar_Ch6_Ex6-1_Q1

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 145°\) NCERT_Solution_Class9_Maths_Exemplar_Ch6_Ex6-1_Q1 \[\angle ARQ = \angle RQD = 25° \quad \text{(alternate interior, }AB\parallel CD\text{)} \] \[\angle BRS = \angle CQP = 60° \quad \text{(vertically opposite at }Q\text{, then corresponding, }PQ\parallel RS\text{)} \] \[\angle ARS = 180° - 60° = 120° \quad \text{(linear pair on }AB\text{)} \] \[\angle QRS = \angle ARQ + \angle ARS = 25° + 120° = 145° \]

  2. Exercise 2

    If one angle of a triangle is equal to the sum of the other two angles, then the triangle is (A) an isosceles triangle (B) an obtuse triangle (C) an equilateral triangle (D) a right triangle

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    NCERT’s answer
    (D)
    (D) a right triangle. With \(\displaystyle A=B+C\) in a triangle, \[A+B+C=180^\circ \] \[A+A=180^\circ \] \[A=90^\circ \] one angle is \(\displaystyle 90^\circ\), so the triangle is right-angled.

  3. Exercise 3

    An exterior angle of a triangle is 105\displaystyle 105° and its two interior opposite angles are equal. Each of these equal angles is (A) 3712∘\displaystyle 37 \frac{1}{2}^{\circ} (B) 5212∘\displaystyle 52 \frac{1}{2}^{\circ} (C) 7212∘\displaystyle 72 \frac{1}{2}^{\circ} (D) 75\displaystyle 75°

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    (B) \(\displaystyle 52\tfrac12^{\circ}\). Let each equal interior opposite angle be \(\displaystyle y\); the exterior angle equals their sum: \[105^\circ = y+y \] \[2y=105^\circ \] \[y=52\tfrac12^{\circ} \]NCERT prints: (A) — that is \(\displaystyle 37\tfrac12^{\circ}\), reached only by treating \(\displaystyle 105^\circ\) itself as an interior angle and bisecting its \(\displaystyle 75^\circ\) supplement, not by \(\displaystyle 2y=105^\circ\).

  4. Exercise 4

    The angles of a triangle are in the ratio 5\displaystyle 5 : 3\displaystyle 3 : 7. The triangle is (A) an acute angled triangle (B) an obtuse angled triangle (C) a right triangle (D) an isosceles triangle

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    NCERT’s answer
    (A)
    (A) an acute angled triangle. \[5x+3x+7x=180^\circ \] \[15x=180^\circ \] \[x=12^\circ \] the angles are \(\displaystyle 60^\circ, 36^\circ, 84^\circ\), each less than \(\displaystyle 90^\circ\).

  5. Exercise 5

    If one of the angles of a triangle is 130\displaystyle 130°, then the angle between the bisectors of the other two angles can be (A) 50\displaystyle 50° (B) 65\displaystyle 65° (C) 145\displaystyle 145° (D) 155\displaystyle 155°

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    NCERT’s answer
    (D)
    (D) \(\displaystyle 155^{\circ}\). NCERT_Solution_Class9_Maths_Exemplar_Ch6_Ex6-1_Q5 \[\angle BIC = 90^\circ + \tfrac{1}{2}\angle A \quad \text{(bisectors of } B, C \text{ meet at } I\text{)} \] \[\angle BIC = 90^\circ + 65^\circ = 155^\circ \]

  6. Exercise 6

    In Fig. 6.2\displaystyle 6.2, POQ is a line. The value of x\displaystyle x is (A) 20\displaystyle 20° (B) 25\displaystyle 25° (C) 30\displaystyle 30° (D) 35\displaystyle 35° NCERT_Question_Class9_Maths_Exemplar_Ch6_Ex6-1_Q6

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    NCERT’s answer
    (A)
    (A) \(\displaystyle 20^{\circ}\). \(\displaystyle POQ\) is a straight line, so the angles at \(\displaystyle O\) sum to \(\displaystyle 180^\circ\): NCERT_Solution_Class9_Maths_Exemplar_Ch6_Ex6-1_Q6 \[40^\circ + 4x + 3x = 180^\circ \] \[7x = 140^\circ \] \[x = 20^\circ \]

  7. Exercise 7

    In Fig. 6.3\displaystyle 6.3, if OPIIRS, ∠OPQ=110∘\displaystyle \angle \mathrm{OPQ}=110^{\circ} and ∠QRS=130∘\displaystyle \angle \mathrm{QRS}=130^{\circ}, then ∠PQR\displaystyle \angle \mathrm{PQR} is equal to (A) 40\displaystyle 40° (B) 50\displaystyle 50° (C) 60\displaystyle 60° (D) 70\displaystyle 70° NCERT_Question_Class9_Maths_Exemplar_Ch6_Ex6-1_Q7

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 60^{\circ}\). Draw \(\displaystyle QX\) through \(\displaystyle Q\) parallel to \(\displaystyle OP\) and \(\displaystyle RS\). NCERT_Solution_Class9_Maths_Exemplar_Ch6_Ex6-1_Q7 \[\angle PQX = \angle OPQ = 110^\circ \quad \text{(alternate angles, } QX\parallel OP\text{)} \] \[\angle XQR = 180^\circ - \angle QRS = 50^\circ \quad \text{(co-interior angles, } QX\parallel RS\text{)} \] \[\angle PQR = \angle PQX - \angle XQR = 60^\circ \]

  8. Exercise 8

    Angles of a triangle are in the ratio 2\displaystyle 2 : 4\displaystyle 4 : 3. The smallest angle of the triangle is (A) 60\displaystyle 60° (B) 40\displaystyle 40° (C) 80\displaystyle 80° (D) 20\displaystyle 20°

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 40^{\circ}\). \[2x+4x+3x=180^\circ \] \[9x=180^\circ \] \[x=20^\circ \] the angles are \(\displaystyle 40^\circ, 80^\circ, 60^\circ\); the smallest is \(\displaystyle 40^\circ\).