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NCERT Exemplar · Class 9 Mathematics Constructions

22 questions · 22 still being checked

EXERCISE 11.1 1–3 (part 1 of 4)

  1. Exercise 1

    With the help of a ruler and a compass it is not possible to construct an angle of : (A) 37.5\displaystyle 37.5° (B) 40\displaystyle 40° (C) 22.5\displaystyle 22.5° (D) 67.5\displaystyle 67.5°

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 40^\circ\). \[60^\circ \to 30^\circ \to 15^\circ, \qquad \frac{60^\circ+15^\circ}{2} = 37.5^\circ \] \[90^\circ \to 45^\circ \to 22.5^\circ \] \[\frac{90^\circ+45^\circ}{2} = 67.5^\circ \] Each of (A), (C), (D) is a bisection or sum of the ruler-compass angles \(\displaystyle 60^\circ\) and \(\displaystyle 90^\circ\); \(\displaystyle 40^\circ\) needs trisecting \(\displaystyle 60^\circ\), which straightedge and compass cannot do.

  2. Exercise 2

    The construction of a triangle ABC, given that BC=6 cm,∠ B=45∘\displaystyle \mathrm{BC}=6 \mathrm{~cm}, \angle \mathrm{~B}=45^{\circ} is not possible when difference of AB and AC is equal to: (A) 6.9\displaystyle 6.9 cm (B) 5.2\displaystyle 5.2 cm (C) 5.0\displaystyle 5.0 cm (D) 4.0\displaystyle 4.0 cm

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    NCERT’s answer
    (A)
    (A) \(\displaystyle 6.9\) cm. NCERT_Solution_Class9_Maths_Exemplar_Ch11_Ex11-1_Q2 \[AC + BC > AB \quad \text{(triangle inequality)} \] \[\Rightarrow AB - AC < BC = 6\text{ cm} \] Since \(\displaystyle 6.9 > 6\), so no such triangle exists.

  3. Exercise 3

    The construction of a triangle ABC, given that BC=3 cm,∠C=60∘\displaystyle \mathrm{BC}=3 \mathrm{~cm}, \angle \mathrm{C}=60^{\circ} is possible when difference of AB and AC is equal to : (A) 3.2\displaystyle 3.2 cm (B) 3.1\displaystyle 3.1 cm (C) 3\displaystyle 3 cm (D) 2.8\displaystyle 2.8 cm

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    NCERT’s answer
    (D)
    (D) \(\displaystyle 2.8\) cm. NCERT_Solution_Class9_Maths_Exemplar_Ch11_Ex11-1_Q3 \[AC + BC > AB \quad \text{(triangle inequality)} \] \[\Rightarrow AB - AC < BC = 3\text{ cm} \] Only \(\displaystyle 2.8 < 3\); \(\displaystyle 3.0\), \(\displaystyle 3.1\), \(\displaystyle 3.2\) each meet or exceed \(\displaystyle BC\), so no such triangle closes at A.