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NCERT Exemplar · Class 10 Mathematics Constructions

21 questions · 21 still being checked

EXERCISE 10.1 1–6 (part 1 of 4)

  1. Choose the correct answer from the given four options:

    Exercise 1

    To divide a line segment AB in the ratio 5\displaystyle 5:7\displaystyle 7, first a ray AX is drawn so that ∠BAX\displaystyle \angle \mathrm{BAX} is an acute angle and then at equal distances points are marked on the ray AX such that the minimum number of these points is
    (A)
    8\displaystyle 8 (B) 10\displaystyle 10
    (C)
    11\displaystyle 11
    (D)
    12\displaystyle 12

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    NCERT’s answer
    (D)
    (D) \(\displaystyle 12\)Equal points marked \(\displaystyle = m+n\), the ratio's two terms. \[m+n = 5+7 = 12 \]

  2. Exercise 2

    To divide a line segment AB in the ratio 4\displaystyle 4:7\displaystyle 7, a ray AX is drawn first such that ∠BAX\displaystyle \angle \mathrm{BAX} is an acute angle and then points A1, A2, A3,…\displaystyle \mathrm{A}_1, \mathrm{~A}_2, \mathrm{~A}_3, \ldots. are located at equal distances on the ray AX and the point B is joined to
    (A)
    A12\displaystyle \mathrm{A}_{12}
    (B)
    A11\displaystyle \mathrm{A}_{11}
    (C)
    A10\displaystyle \mathrm{A}_{10}
    (D)
    A9\displaystyle \mathrm{A}_9

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    NCERT’s answer
    (B)
    (B) \(\displaystyle A_{11}\)B joins the \(\displaystyle (m+n)\)-th point on AX. \[A_{m+n} = A_{4+7} = A_{11} \]

  3. Exercise 3

    To divide a line segment AB in the ratio 5\displaystyle 5 : 6\displaystyle 6, draw a ray AX such that ∠BAX\displaystyle \angle \mathrm{BAX} is an acute angle, then draw a ray BY parallel to AX and the points A1, A2, A3,…\displaystyle \mathrm{A}_1, \mathrm{~A}_2, \mathrm{~A}_3, \ldots and B1, B2, B3,…\displaystyle \mathrm{B}_1, \mathrm{~B}_2, \mathrm{~B}_3, \ldots are located at equal distances on ray AX and BY, respectively. Then the points joined are
    (A)
    A5\displaystyle \mathrm{A}_5 and B6\displaystyle \mathrm{B}_6
    (B)
    A6\displaystyle \mathrm{A}_6 and B5\displaystyle \mathrm{B}_5
    (C)
    A4\displaystyle \mathrm{A}_4 and B5\displaystyle \mathrm{B}_5
    (D)
    A5\displaystyle \mathrm{A}_5 and B4\displaystyle \mathrm{B}_4

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    NCERT’s answer
    (A)
    (A) \(\displaystyle A_5\) and \(\displaystyle B_6\)Join \(\displaystyle A_5B_6\); it meets AB at P. \[\frac{AP}{PB} = \frac{AA_5}{BB_6} = \frac{5}{6} \quad (\triangle AA_5P \sim \triangle BB_6P,\ AX \parallel BY) \]

  4. Exercise 4

    To construct a triangle similar to a given ΔABC\displaystyle \Delta \mathrm{ABC} with its sides 37\displaystyle \frac{3}{7} of the corresponding sides of ΔABC\displaystyle \Delta \mathrm{ABC}, first draw a ray BX such that ∠CBX\displaystyle \angle \mathrm{CBX} is an acute angle and X lies on the opposite side of A with respect to BC. Then locate points B1, B2, B3,…\displaystyle \mathrm{B}_1, \mathrm{~B}_2, \mathrm{~B}_3, \ldots on BX at equal distances and next step is to join
    (A)
    B10\displaystyle \mathrm{B}_{10} to C
    (B)
    B3\displaystyle \mathrm{B}_3 to C
    (C)
    B7\displaystyle \mathrm{B}_7 to C
    (D)
    B4\displaystyle \mathrm{B}_4 to C

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    NCERT’s answer
    (C)
    (C) \(\displaystyle B_7\) to CJoin \(\displaystyle B_7C\); draw \(\displaystyle B_3C' \parallel B_7C\), meeting BC at \(\displaystyle C'\). \[\frac{BC'}{BC} = \frac{BB_3}{BB_7} = \frac{3}{7} \quad (B_3C' \parallel B_7C) \]

  5. Exercise 5

    To construct a triangle similar to a given ΔABC\displaystyle \Delta \mathrm{ABC} with its sides 85\displaystyle \frac{8}{5} of the corresponding sides of ΔABC\displaystyle \Delta \mathrm{ABC} draw a ray BX such that ∠CBX\displaystyle \angle \mathrm{CBX} is an acute angle and X is on the opposite side of A with respect to BC. The minimum number of points to be located at equal distances on ray BX is
    (A)
    5\displaystyle 5 (B) 8\displaystyle 8 (C) 13\displaystyle 13

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    NCERT’s answer
    (B)
    (B) \(\displaystyle 8\)Points needed \(\displaystyle = \max(p,q)\) for scale \(\displaystyle \frac{p}{q}\). \[\max(8,5) = 8 \]

  6. Exercise 6

    To draw a pair of tangents to a circle which are inclined to each other at an angle of 60\displaystyle 60°, it is required to draw tangents at end points of those two radii of the circle, the angle between them should be
    (A)
    135\displaystyle 135°
    (B)
    90\displaystyle 90°
    (C)
    60\displaystyle 60°
    (D)
    120\displaystyle 120°

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    NCERT’s answer
    (D)
    (D) \(\displaystyle 120^\circ\)For tangents PA, PB with \(\displaystyle \angle APB = 60^\circ\), the radii OA, OB meet them at right angles.NCERT_Solution_Class10_Maths_Exemplar_Ch10_Ex10-1_Q6\[\angle OAP = \angle OBP = 90^\circ \quad \text{(radius} \perp \text{tangent)} \] \[\angle AOB = 180^\circ - \angle APB = 180^\circ - 60^\circ = 120^\circ \quad \text{(angle sum of OAPB)} \]