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  1. Asked: In:

    In Fig. 6.28, find the values of x and y and then show that AB || CD.

    deva989

    50° + x = 180° [∵ Linear Pair] ⇒ x = 180° - 50° = 130° and y = 130° [∵ Vertically Opposite Angles] Hence, x = y = 130° Since, alternate angles are equal. Hence, AB||CD.

    50° + x = 180° [∵ Linear Pair]
    ⇒ x = 180° – 50° = 130° and
    y = 130° [∵ Vertically Opposite Angles]
    Hence, x = y = 130°
    Since, alternate angles are equal. Hence, AB||CD.

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  2. Asked: In:

    In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.

    deva989

    Given: y:z = 3:7 Let, y = 3k, therefore z = 7k y = ∠1 = 3k [∵ Vertically Opposite Angles] Given: CD||EF, Therefore, ∠1 + z = 180° [∵ Sum of co-interior angles] ⇒ 3k + 7k = 180° ⇒ 10k = 180° ⇒ k = 180°/10 = 18° Hence, z = 7k = 7 × 18° = 126° Given that : AB||CD and CD||EF, therefore AB||EF ⇒ x = z =Read more

    Given: y:z = 3:7
    Let, y = 3k, therefore z = 7k
    y = ∠1 = 3k [∵ Vertically Opposite Angles]
    Given: CD||EF,
    Therefore,
    ∠1 + z = 180° [∵ Sum of co-interior angles]
    ⇒ 3k + 7k = 180°
    ⇒ 10k = 180°
    ⇒ k = 180°/10 = 18°
    Hence, z = 7k = 7 × 18° = 126°
    Given that : AB||CD and CD||EF, therefore AB||EF
    ⇒ x = z = 126° [∵ Alternate Angles]

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  3. Asked: In:

    In Fig. 6.30, if AB || CD, EF ⊥ CD and ∠ GED = 126°, find ∠ AGE, ∠ GEF and ∠ FGE.

    deva989

    Given that: AB||CD, Therefore, ∠AGE = ∠GED [∵ Alternater Angles] ⇒ ∠AGE = 126° From the figure, ∠GED = ∠GEF + ∠FED ⇒ 126° = ∠GEF + 90° ⇒ ∠GEF = 126° - 90° = 36° Given that: AB||CD, Therefore, ∠FGE + 126° = 180° [∵ Sum of co-interior angles] ⇒ ∠FGE = 180° - 126° = 54°

    Given that: AB||CD,
    Therefore,
    ∠AGE = ∠GED [∵ Alternater Angles]
    ⇒ ∠AGE = 126°
    From the figure,
    ∠GED = ∠GEF + ∠FED
    ⇒ 126° = ∠GEF + 90°
    ⇒ ∠GEF = 126° – 90° = 36°
    Given that: AB||CD,
    Therefore,
    ∠FGE + 126° = 180° [∵ Sum of co-interior angles]
    ⇒ ∠FGE = 180° – 126° = 54°

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  4. Asked: In:

    In Fig. 6.31, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠ QRS.

    deva989

    Contruction: Produce PQ, so that it intersect ST at M. Given that: PQ||ST, therefore ∠1 = ∠2 [∵ Correspomding Angles] ⇒ ∠2 = 130° ∠2 + ∠3 = 180° [∵ Linear pair] ⇒ 130° + ∠3 = 180° ⇒ ∠3 = 180° - 130° = 50° ∠5 + ∠4 = 180° [∵ Linear pair] ⇒ 110° + ∠4 = 180° ⇒ ∠4 = 180° - 110° = 70° In triangle QMR, ∠3Read more

    Contruction: Produce PQ, so that it intersect ST at M.
    Given that: PQ||ST, therefore
    ∠1 = ∠2 [∵ Correspomding Angles]
    ⇒ ∠2 = 130°
    ∠2 + ∠3 = 180° [∵ Linear pair]
    ⇒ 130° + ∠3 = 180°
    ⇒ ∠3 = 180° – 130° = 50°
    ∠5 + ∠4 = 180° [∵ Linear pair]
    ⇒ 110° + ∠4 = 180°
    ⇒ ∠4 = 180° – 110° = 70°
    In triangle QMR,
    ∠3 + ∠4 + ∠R = 180°
    ⇒ 50° + 70° + ∠R = 180°
    ⇒ 120° + ∠R = 180°
    ⇒ ∠R = 180° – 120° = 60°

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  6. Asked: In:

    In Fig. 6.39, sides QP and RQ of ∆ PQR are produced to points S and T respectively. If ∠ SPR = 135° and ∠ PQT = 110°, find ∠ PRQ.

    deva989

    ∠PQT + ∠PQR = 180° ⇒ 110° + ∠PQR = 180° ⇒ ∠PQT = 180° - 110° = 70° ∠SPR + ∠QPR = 180° ⇒ 135° + ∠QPR = 180° ⇒ ∠QPR = 180° - 135° = 45° In △PQR, ∠QPR + ∠PQR + ∠R = 180° ⇒ 70° + 45° + ∠R = 180° ⇒ 115° + ∠R = 180° ⇒ ∠R = 180° - 115° = 65°

    ∠PQT + ∠PQR = 180°
    ⇒ 110° + ∠PQR = 180°
    ⇒ ∠PQT = 180° – 110° = 70°
    ∠SPR + ∠QPR = 180°
    ⇒ 135° + ∠QPR = 180°
    ⇒ ∠QPR = 180° – 135° = 45°
    In △PQR,
    ∠QPR + ∠PQR + ∠R = 180°
    ⇒ 70° + 45° + ∠R = 180°
    ⇒ 115° + ∠R = 180°
    ⇒ ∠R = 180° – 115° = 65°

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