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NCERT Exemplar · Class 9 Mathematics Constructions

22 questions · 22 still being checked

EXERCISE 11.3 1–8 (part 3 of 4)

  1. Exercise 1

    Draw an angle of 110\displaystyle 110° with the help of a protractor and bisect it. Measure each angle.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    1. Draw ray \(\displaystyle BC \); at \(\displaystyle B \), mark \(\displaystyle \angle ABC = 110^\circ \) with a protractor and draw ray \(\displaystyle BA \). 2. Bisect \(\displaystyle \angle ABC \) (equal-radius arcs); draw ray \(\displaystyle BD \).NCERT_Solution_Class9_Maths_Exemplar_Ch11_Ex11-3_Q1\[\angle ABD = \angle DBC = \tfrac12 \times 110^\circ = 55^\circ \]Answer: each half \(\displaystyle = 55^\circ\).

  2. Exercise 2

    Draw a line segment AB of 4\displaystyle 4 cm in length. Draw a line perpendicular to AB through A and B, respectively. Are these lines parallel?

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    Yes.
    1. Draw \(\displaystyle AB = 4 \) cm. 2. At \(\displaystyle A \), construct \(\displaystyle \angle 90^\circ \) to \(\displaystyle AB \) (equal-radius arcs); draw ray \(\displaystyle AP \). 3. At \(\displaystyle B \), construct \(\displaystyle \angle 90^\circ \) to \(\displaystyle AB \) the same way, on the same side; draw ray \(\displaystyle BQ \).NCERT_Solution_Class9_Maths_Exemplar_Ch11_Ex11-3_Q2\[AP \perp AB,\ BQ \perp AB \ \Rightarrow\ AP \parallel BQ \]Answer: Yes, the two perpendiculars are parallel.

  3. Exercise 3

    Draw an angle of 80\displaystyle 80° with the help of a protractor. Then construct angles of (i) 40\displaystyle 40° (ii) 160\displaystyle 160° and (iii) 120\displaystyle 120°.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    1. Draw ray \(\displaystyle OA \); mark \(\displaystyle \angle AOB = 80^\circ \) with a protractor and draw ray \(\displaystyle OB \). 2. Bisect \(\displaystyle \angle AOB \) (equal-radius arcs) to get ray \(\displaystyle OC \). 3. Copy \(\displaystyle \angle AOB \) adjacent to \(\displaystyle OB \), away from \(\displaystyle OA \), to get ray \(\displaystyle OD \). 4. Copy \(\displaystyle \angle BOC \) adjacent to \(\displaystyle OB \), between \(\displaystyle OB \) and \(\displaystyle OD \), to get ray \(\displaystyle OE \).NCERT_Solution_Class9_Maths_Exemplar_Ch11_Ex11-3_Q3\[\text{(i) } \angle AOC = \tfrac12 \times 80^\circ = 40^\circ \] \[\text{(ii) } \angle AOD = 80^\circ + 80^\circ = 160^\circ \] \[\text{(iii) } \angle AOE = 80^\circ + 40^\circ = 120^\circ \]Answer: (i) \(\displaystyle 40^\circ\) (ii) \(\displaystyle 160^\circ\) (iii) \(\displaystyle 120^\circ\).

  4. Exercise 4

    Construct a triangle whose sides are 3.6\displaystyle 3.6 cm, 3.0\displaystyle 3.0 cm and 4.8\displaystyle 4.8 cm. Bisect the smallest angle and measure each part.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    1. Draw \(\displaystyle BC = 3.0 \) cm. 2. With centre \(\displaystyle B \), radius \(\displaystyle 3.6 \) cm, and centre \(\displaystyle C \), radius \(\displaystyle 4.8 \) cm, draw arcs meeting at \(\displaystyle A \); join \(\displaystyle AB, AC \). 3. \(\displaystyle \angle A \) is smallest (opposite the smallest side \(\displaystyle BC \)); bisect it (equal-radius arcs), meeting \(\displaystyle BC \) at \(\displaystyle D \).NCERT_Solution_Class9_Maths_Exemplar_Ch11_Ex11-3_Q4\[\cos A = \frac{CA^2+AB^2-BC^2}{2\cdot CA\cdot AB} = \frac{4.8^2+3.6^2-3.0^2}{2(4.8)(3.6)} = \frac{27}{34.56} = 0.78125 \] \[\angle A = 38.6^\circ,\qquad \angle BAD = \angle DAC = 19.3^\circ \]Answer: each bisected part \(\displaystyle \approx 19.3^\circ\).

  5. Exercise 5

    Construct a triangle ABC in which BC=5 cm,∠ B=60∘\displaystyle \mathrm{BC}=5 \mathrm{~cm}, \angle \mathrm{~B}=60^{\circ} and AC+AB=7.5 cm\displaystyle \mathrm{AC}+\mathrm{AB}=7.5 \mathrm{~cm}.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    1. Draw \(\displaystyle BC = 5 \) cm. 2. At \(\displaystyle B \), draw ray \(\displaystyle BX \) with \(\displaystyle \angle XBC = 60^\circ \); cut \(\displaystyle BD = 7.5 \) cm on it \(\displaystyle (=AC+AB)\). 3. Join \(\displaystyle DC \); draw its perpendicular bisector, meeting \(\displaystyle BD \) at \(\displaystyle A \). 4. Join \(\displaystyle AC \). \(\displaystyle \triangle ABC \) is the required triangle.NCERT_Solution_Class9_Maths_Exemplar_Ch11_Ex11-3_Q5\[AD = AC \quad (A \text{ on perp. bisector of } DC) \ \Rightarrow\ AB+AC = AB+AD = BD = 7.5 \text{ cm} \]Answer: \(\displaystyle \triangle ABC\) constructed; \(\displaystyle AB \approx 3.125\) cm, \(\displaystyle AC \approx 4.375\) cm.

  6. Exercise 6

    Construct a square of side 3\displaystyle 3 cm.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    1. Draw \(\displaystyle AB = 3 \) cm. 2. At \(\displaystyle A \) and \(\displaystyle B \), construct \(\displaystyle \angle 90^\circ \) to \(\displaystyle AB \), on the same side. 3. With centres \(\displaystyle A, B \) and radius \(\displaystyle 3 \) cm, mark \(\displaystyle D, C \) on the two perpendiculars. 4. Join \(\displaystyle DC \). \(\displaystyle ABCD \) is the required square.NCERT_Solution_Class9_Maths_Exemplar_Ch11_Ex11-3_Q6\[AB=BC=CD=DA=3\text{ cm},\quad \angle A=\angle B=\angle C=\angle D=90^\circ \]Answer: square \(\displaystyle ABCD\), side \(\displaystyle 3\) cm.

  7. Exercise 7

    Construct a rectangle whose adjacent sides are of lengths 5\displaystyle 5 cm and 3.5\displaystyle 3.5 cm.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    1. Draw \(\displaystyle AB = 5 \) cm. 2. At \(\displaystyle A \) and \(\displaystyle B \), construct \(\displaystyle \angle 90^\circ \) to \(\displaystyle AB \), on the same side. 3. With centres \(\displaystyle A, B \) and radius \(\displaystyle 3.5 \) cm, mark \(\displaystyle D, C \) on the two perpendiculars. 4. Join \(\displaystyle DC \). \(\displaystyle ABCD \) is the required rectangle.NCERT_Solution_Class9_Maths_Exemplar_Ch11_Ex11-3_Q7\[AB=DC=5\text{ cm},\quad BC=AD=3.5\text{ cm},\quad \angle A=\angle B=\angle C=\angle D=90^\circ \]Answer: rectangle \(\displaystyle ABCD\), \(\displaystyle 5\text{ cm}\times3.5\text{ cm}\).

  8. Exercise 8

    Construct a rhombus whose side is of length 3.4\displaystyle 3.4 cm and one of its angles is 45\displaystyle 45°.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    1. Draw \(\displaystyle AB = 3.4 \) cm. 2. At \(\displaystyle A \), construct \(\displaystyle \angle 90^\circ \) to \(\displaystyle AB \) and bisect it to get \(\displaystyle \angle DAB = 45^\circ \). 3. With centre \(\displaystyle A \), radius \(\displaystyle 3.4 \) cm, mark \(\displaystyle D \) on that ray. 4. With centres \(\displaystyle B, D \) and radius \(\displaystyle 3.4 \) cm, draw arcs meeting at \(\displaystyle C \); join \(\displaystyle BC, DC \).NCERT_Solution_Class9_Maths_Exemplar_Ch11_Ex11-3_Q8\[AB=BC=CD=DA=3.4\text{ cm},\quad \angle A = 45^\circ \]Answer: rhombus \(\displaystyle ABCD\), side \(\displaystyle 3.4\) cm, \(\displaystyle \angle A = 45^\circ\).