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NCERT Exemplar · Class 9 Mathematics Lines and Angles

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EXERCISE 6.3 1–10 (part 3 of 4)

  1. Exercise 1

    In Fig. 6.9\displaystyle 6.9, OD is the bisector of ∠AOC\displaystyle \angle \mathrm{AOC}, OE is the bisector of ∠BOC\displaystyle \angle \mathrm{BOC} and OD⊥OE\displaystyle \mathrm{OD} \perp \mathrm{OE}. Show that the points A, O and B are collinear. NCERT_Question_Class9_Maths_Exemplar_Ch6_Ex6-3_Q1

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    NCERT_Solution_Class9_Maths_Exemplar_Ch6_Ex6-3_Q1 \[\angle AOD = \angle DOC = x \quad \text{(OD bisects } \angle AOC\text{)} \] \[\angle COE = \angle EOB = y \quad \text{(OE bisects } \angle BOC\text{)} \] \[\angle DOE = \angle DOC + \angle COE \implies x + y = 90^\circ \quad (OD \perp OE) \] \[\angle AOD + \angle DOC + \angle COE + \angle EOB = 2x + 2y = 180^\circ \] A straight angle on ray OA and ray OB means they are opposite rays. Answer: A, O and B are collinear.

  2. Exercise 2

    In Fig. 6.10\displaystyle 6.10, ∠1=60∘\displaystyle \angle 1=60^{\circ} and ∠6=120∘\displaystyle \angle 6=120^{\circ}. Show that the lines m\displaystyle m and n\displaystyle n are parallel. NCERT_Question_Class9_Maths_Exemplar_Ch6_Ex6-3_Q2

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    NCERT_Solution_Class9_Maths_Exemplar_Ch6_Ex6-3_Q2 \[\angle 5 + \angle 6 = 180^\circ \quad \text{(linear pair)} \implies \angle 5 = 180^\circ - 120^\circ = 60^\circ \] \[\angle 1 = \angle 5 = 60^\circ \quad \text{(corresponding angles)} \] \[\Rightarrow m \parallel n \quad \text{(converse of corresponding angles axiom)} \] Answer: \(\displaystyle m \parallel n\).

  3. Exercise 3

    AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t\displaystyle t with parallel lines l\displaystyle l and m\displaystyle m (Fig. 6.11\displaystyle 6.11). Show that AP || BQ. NCERT_Question_Class9_Maths_Exemplar_Ch6_Ex6-3_Q3

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    NCERT_Solution_Class9_Maths_Exemplar_Ch6_Ex6-3_Q3 \[\angle lAB = \angle ABm \quad \text{(alternate interior angles, } l \parallel m, \text{ transversal } t\text{)} \] \[\angle PAB = \tfrac12 \angle lAB, \quad \angle QBA = \tfrac12 \angle ABm \quad \text{(} AP, BQ \text{ bisect these angles)} \] \[\angle PAB = \angle QBA \] \[\Rightarrow AP \parallel BQ \quad \text{(converse of alternate interior angles axiom, transversal } t\text{)} \] Answer: \(\displaystyle AP \parallel BQ\).

  4. Exercise 4

    If in Fig. 6.11\displaystyle 6.11, bisectors AP and BQ of the alternate interior angles are parallel, then show that l∥m\displaystyle l \| m. NCERT_Question_Class9_Maths_Exemplar_Ch6_Ex6-3_Q3

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    NCERT_Solution_Class9_Maths_Exemplar_Ch6_Ex6-3_Q4 \[AP \parallel BQ \implies \angle PAB = \angle QBA \quad \text{(alternate interior angles, transversal } t\text{)} \] \[\angle lAB = 2\angle PAB, \quad \angle ABm = 2\angle QBA \quad \text{(} AP, BQ \text{ bisect } \angle lAB, \angle ABm\text{)} \] \[\angle lAB = \angle ABm \] \[\Rightarrow l \parallel m \quad \text{(converse of alternate interior angles axiom, transversal } t\text{)} \] Answer: \(\displaystyle l \parallel m\).

  5. Exercise 5

    In Fig. 6.12\displaystyle 6.12, BA∥ED\displaystyle \mathrm{BA} \| \mathrm{ED} and BC∥EF\displaystyle \mathrm{BC} \| \mathrm{EF}. Show that ∠ABC=∠DEF\displaystyle \angle \mathrm{ABC}=\angle \mathrm{DEF} [Hint: Produce DE to intersect BC at P (say)]. NCERT_Question_Class9_Maths_Exemplar_Ch6_Ex6-3_Q5

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    NCERT_Solution_Class9_Maths_Exemplar_Ch6_Ex6-3_Q5 Produce DE to meet BC at P. \[\angle ABC = \angle DPC \quad \text{(corresponding angles, } AB \parallel ED, \text{ transversal } BC\text{)} \] \[\angle DPC = \angle DEF \quad \text{(corresponding angles, } BC \parallel EF, \text{ transversal } DP\text{)} \] \[\angle ABC = \angle DEF \] Answer: \(\displaystyle \angle ABC = \angle DEF\).

  6. Exercise 6

    In Fig. 6.13\displaystyle 6.13, BA∥ED\displaystyle \mathrm{BA} \| \mathrm{ED} and BC∥EF\displaystyle \mathrm{BC} \| \mathrm{EF}. Show that ∠ABC+∠DEF=180∘\displaystyle \angle \mathrm{ABC}+\angle \mathrm{DEF}=180^{\circ} NCERT_Question_Class9_Maths_Exemplar_Ch6_Ex6-3_Q6

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    NCERT_Solution_Class9_Maths_Exemplar_Ch6_Ex6-3_Q6 Produce ED to meet BC at P. \[\angle ABC = \angle EPC \quad \text{(corresponding angles, } AB \parallel EP, \text{ transversal } BC\text{)} \] \[\angle EPC + \angle DEF = 180^\circ \quad \text{(co-interior angles, } BC \parallel EF, \text{ transversal } DP\text{)} \] \[\angle ABC + \angle DEF = 180^\circ \] Answer: \(\displaystyle \angle ABC + \angle DEF = 180^\circ\).

  7. Exercise 7

    In Fig. 6.14\displaystyle 6.14, DE∥QR\displaystyle \mathrm{DE} \| \mathrm{QR} and AP and BP are bisectors of ∠EAB\displaystyle \angle \mathrm{EAB} and ∠RBA\displaystyle \angle \mathrm{RBA}, respectively. Find ∠APB\displaystyle \angle \mathrm{APB}. NCERT_Question_Class9_Maths_Exemplar_Ch6_Ex6-3_Q7

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    NCERT’s answer
    $\displaystyle 90$°
    NCERT_Solution_Class9_Maths_Exemplar_Ch6_Ex6-3_Q7 \[\angle EAB + \angle ABR = 180^\circ \quad \text{(co-interior angles, } DE \parallel QR, \text{ transversal } AB\text{)} \] \[\angle PAB = \tfrac12\angle EAB, \quad \angle PBA = \tfrac12\angle ABR \quad \text{(AP, BP bisect } \angle EAB, \angle RBA\text{)} \] \[\angle PAB + \angle PBA = \tfrac12(\angle EAB + \angle ABR) = 90^\circ \] \[\angle APB = 180^\circ - (\angle PAB + \angle PBA) = 90^\circ \quad \text{(angle sum of } \triangle APB\text{)} \] Answer: \(\displaystyle \angle APB = 90^\circ\).

  8. Exercise 8

    The angles of a triangle are in the ratio 2\displaystyle 2 : 3\displaystyle 3 : 4. Find the angles of the triangle.

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    NCERT’s answer
    $\displaystyle 40$°, $\displaystyle 60$, $\displaystyle 80$°
    \[\text{Let the angles be } 2x,\ 3x,\ 4x. \] \[2x + 3x + 4x = 180^\circ \quad \text{(angle sum property of a triangle)} \] \[9x = 180^\circ \ \Rightarrow\ x = 20^\circ \] \[2x = 40^\circ, \quad 3x = 60^\circ, \quad 4x = 80^\circ \] Answer: \(\displaystyle 40^\circ, 60^\circ, 80^\circ\).

  9. Exercise 9

    A triangle ABC is right angled at A . L is a point on BC such that AL ⟂ BC. Prove that ∠BAL=∠ACB\displaystyle \angle \mathrm{BAL}=\angle \mathrm{ACB}.

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    NCERT_Solution_Class9_Maths_Exemplar_Ch6_Ex6-3_Q9\[\angle BAC = 90^\circ \quad \text{(given)} \] \[\angle ABC + \angle ACB = 90^\circ \quad \text{(angle sum property, }\triangle ABC\text{)} \] \[\angle ALB = 90^\circ \quad \text{(}AL \perp BC\text{, given)} \] \[\angle BAL + \angle ABL = 90^\circ \quad \text{(angle sum property, }\triangle ABL\text{)} \] \[\angle ABL = \angle ABC \quad \text{(}L\text{ lies on }BC\text{)} \] \[\angle BAL + \angle ABC = \angle ACB + \angle ABC \quad \text{(both equal } 90^\circ\text{)} \] \[\angle BAL = \angle ACB \]Answer: \(\displaystyle \angle BAL = \angle ACB\).

  10. Exercise 10

    Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.

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    NCERT_Solution_Class9_Maths_Exemplar_Ch6_Ex6-3_Q10 \[\angle BPE = 90^\circ \quad \text{(} EF \perp AB \text{, given)} \] \[\angle DQE = \angle BPE = 90^\circ \quad \text{(corresponding angles, } AB \parallel CD\text{)} \] \[\angle DRG = 90^\circ \quad \text{(} GH \perp CD \text{, given)} \] \[\angle DQE = \angle DRG \] \[\Rightarrow EF \parallel GH \quad \text{(converse of corresponding angles axiom, transversal } CD\text{)} \] Answer: \(\displaystyle EF \parallel GH\).