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NCERT Exemplar · Class 9 Mathematics Constructions

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EXERCISE 11.2 1–6 (part 2 of 4)

  1. Write True or False in each of the following. Give reasons for your answer:

    Exercise 1

    An angle of 52.5\displaystyle 52.5° can be constructed.

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    NCERT’s answer
    True. As $\displaystyle 52.5^{\circ}=\frac{210^{\circ}}{4}$ and $\displaystyle 210^{\circ}=180^{\circ}+30^{\circ}$ which can be constructed.
    True. \[60^\circ - 45^\circ = 15^\circ \] \[\frac{15^\circ}{2} = 7.5^\circ \] \[7 \times 7.5^\circ = 52.5^\circ \] Both \(\displaystyle 60^\circ\) and \(\displaystyle 45^\circ\) are ruler-compass angles, so \(\displaystyle 7.5^\circ\) and every integer multiple of it are constructible.

  2. Exercise 2

    An angle of 42.5\displaystyle 42.5° can be constructed.

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    NCERT’s answer
    False. As $\displaystyle 42.5^{\circ}=\frac{1}{2} \times 85^{\circ}$ and $\displaystyle 85$° cannot be constructed.
    False. \[\frac{42.5^\circ}{15^\circ} = \frac{17}{6} \] Every angle built from \(\displaystyle 60^\circ\) and \(\displaystyle 90^\circ\) by bisection is \(\displaystyle 15^\circ\) times a power-of-two fraction; \(\displaystyle \tfrac{17}{6}\) has the factor \(\displaystyle 3\) in its denominator, so \(\displaystyle 42.5^\circ\) is not constructible.

  3. Exercise 3

    A triangle ABC can be constructed in which AB=5 cm,∠ A=45∘\displaystyle \mathrm{AB}=5 \mathrm{~cm}, \angle \mathrm{~A}=45^{\circ} and BC+\displaystyle \mathrm{BC}+ AC=5 cm\displaystyle \mathrm{AC}=5 \mathrm{~cm}.

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    NCERT’s answer
    False. As $\displaystyle \mathrm{BC}+\mathrm{AC}$ must be greater than AB which is not so.
    False. \[BC + AC = 5\text{ cm} \] \[AB = 5\text{ cm} \] \[BC + AC > AB \quad \text{(triangle inequality, required)} \] The sum equals the base instead of exceeding it, so A, B, C would be collinear — no triangle, hence no construction.

  4. Exercise 4

    A triangle ABC can be constructed in which BC=6 cm,∠C=30∘\displaystyle \mathrm{BC}=6 \mathrm{~cm}, \angle \mathrm{C}=30^{\circ} and AC - AB=4 cm\displaystyle \mathrm{AB}=4 \mathrm{~cm}.

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    NCERT’s answer
    True. As AC - AB < BC, i.e., $\displaystyle \mathrm{AC}<\mathrm{AB}+\mathrm{BC}$.
    True. NCERT_Solution_Class9_Maths_Exemplar_Ch11_Ex11-2_Q4 \[AC - AB = 4\text{ cm} \] \[BC = 6\text{ cm} \] \[AC - AB < BC \] The difference of the two sides is less than the base, satisfying the triangle inequality, so the difference construction closes at a valid A.

  5. Exercise 5

    A triangle ABC can be constructed in which ∠B=105∘,∠C=90∘\displaystyle \angle \mathrm{B}=105^{\circ}, \angle \mathrm{C}=90^{\circ} and AB+BC+\displaystyle \mathrm{AB}+\mathrm{BC}+ AC=10 cm\displaystyle \mathrm{AC}=10 \mathrm{~cm}.

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    NCERT’s answer
    False. As $\displaystyle \angle \mathrm{B}+\angle \mathrm{C}=105^{\circ}+90^{\circ}=195^{\circ}>180^{\circ}$.
    False. \[\angle B + \angle C = 105^\circ + 90^\circ = 195^\circ \] \[\angle A + \angle B + \angle C = 180^\circ \] Two angles alone exceed \(\displaystyle 180^\circ\), forcing \(\displaystyle \angle A\) negative; no such triangle exists.

  6. Exercise 6

    A triangle ABC can be constructed in which ∠B=60∘,∠C=45∘\displaystyle \angle \mathrm{B}=60^{\circ}, \angle \mathrm{C}=45^{\circ} and AB+BC+AC=\displaystyle \mathrm{AB}+\mathrm{BC}+\mathrm{AC}= 12\displaystyle 12 cm.

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    NCERT’s answer
    True. As $\displaystyle \angle \mathrm{B}+\angle \mathrm{C}=60^{\circ}+45^{\circ}=105^{\circ}<180^{\circ}$.
    True. NCERT_Solution_Class9_Maths_Exemplar_Ch11_Ex11-2_Q6 \[\angle B + \angle C = 60^\circ + 45^\circ = 105^\circ \] \[\angle A = 180^\circ - 105^\circ = 75^\circ \] All three angles are positive, so a triangle with this perimeter exists.