SolveIt is under development
SolveItNCERT · CBSE · NEET

NCERT Exemplar · Class 10 Mathematics Constructions

21 questions · 21 still being checked

EXERCISE 10.3 1–4 (part 3 of 4)

  1. Exercise 1

    Draw a line segment of length 7\displaystyle 7 cm. Find a point P on it which divides it in the ratio 3\displaystyle 3:5.

    Not cross-checked

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

    1. Draw \(\displaystyle AB = 7\text{ cm}\). 2. Draw ray \(\displaystyle AX\) making an acute angle with \(\displaystyle AB\). 3. Mark points \(\displaystyle A_1,A_2,\ldots,A_8\) on \(\displaystyle AX\) with \(\displaystyle AA_1=A_1A_2=\cdots=A_7A_8\). 4. Join \(\displaystyle A_8B\). 5. Through \(\displaystyle A_3\), draw \(\displaystyle A_3P \parallel A_8B\), meeting \(\displaystyle AB\) at \(\displaystyle P\). NCERT_Solution_Class10_Maths_Exemplar_Ch10_Ex10-3_Q1 \[\frac{AP}{PB} = \frac{AA_3}{A_3A_8} = \frac{3}{5} \]

  2. Exercise 2

    Draw a right triangle ABC in which BC=12 cm,AB=5 cm\displaystyle \mathrm{BC}=12 \mathrm{~cm}, \mathrm{AB}=5 \mathrm{~cm} and ∠B=90∘\displaystyle \angle \mathrm{B}=90^{\circ}. Construct a triangle similar to it and of scale factor 23\displaystyle \frac{2}{3}. Is the new triangle also a right triangle?

    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 \(\displaystyle BC = 12\text{ cm}\). 2. At \(\displaystyle B\), draw \(\displaystyle BX \perp BC\); with centre \(\displaystyle B\), radius \(\displaystyle 5\text{ cm}\), cut \(\displaystyle BX\) at \(\displaystyle A\). Join \(\displaystyle AC\). 3. Draw ray \(\displaystyle BY\) making an acute angle with \(\displaystyle BC\), on the side opposite \(\displaystyle A\). 4. Mark \(\displaystyle B_1,B_2,B_3\) on \(\displaystyle BY\) with \(\displaystyle BB_1=B_1B_2=B_2B_3\). 5. Join \(\displaystyle B_3C\); through \(\displaystyle B_2\), draw \(\displaystyle B_2C' \parallel B_3C\), meeting \(\displaystyle BC\) at \(\displaystyle C'\). 6. Through \(\displaystyle C'\), draw \(\displaystyle C'A' \parallel CA\), meeting \(\displaystyle AB\) at \(\displaystyle A'\). NCERT_Solution_Class10_Maths_Exemplar_Ch10_Ex10-3_Q2 \[\angle A'BC' = \angle ABC = 90^\circ \quad \text{(}A',C'\text{ lie on rays }BA,BC\text{)} \]

  3. Exercise 3

    Draw a triangle ABC in which BC=6 cm,CA=5 cm\displaystyle \mathrm{BC}=6 \mathrm{~cm}, \mathrm{CA}=5 \mathrm{~cm} and AB=4 cm\displaystyle \mathrm{AB}=4 \mathrm{~cm}. Construct a triangle similar to it and of scale factor 53\displaystyle \frac{5}{3}.

    Not cross-checked

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

    1. Draw \(\displaystyle BC = 6\text{ cm}\). 2. With centres \(\displaystyle B,C\) and radii \(\displaystyle 4\text{ cm},5\text{ cm}\), draw arcs meeting at \(\displaystyle A\). Join \(\displaystyle AB,AC\). 3. Draw ray \(\displaystyle BX\) making an acute angle with \(\displaystyle BC\), on the side opposite \(\displaystyle A\). 4. Mark \(\displaystyle B_1,\ldots,B_5\) on \(\displaystyle BX\) with \(\displaystyle BB_1=B_1B_2=\cdots=B_4B_5\). 5. Join \(\displaystyle B_3C\); through \(\displaystyle B_5\), draw \(\displaystyle B_5C' \parallel B_3C\), meeting \(\displaystyle BC\) produced at \(\displaystyle C'\). 6. Through \(\displaystyle C'\), draw \(\displaystyle C'A' \parallel CA\), meeting \(\displaystyle BA\) produced at \(\displaystyle A'\). NCERT_Solution_Class10_Maths_Exemplar_Ch10_Ex10-3_Q3 \[\triangle A'BC' \sim \triangle ABC, \quad \frac{BA'}{BA}=\frac{BC'}{BC}=\frac{5}{3} \]

  4. Exercise 4

    Construct a tangent to a circle of radius 4\displaystyle 4 cm from a point which is at a distance of 6\displaystyle 6 cm from its centre.

    Not cross-checked

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

    1. Draw a circle with centre \(\displaystyle O\), radius \(\displaystyle 4\text{ cm}\). 2. Mark \(\displaystyle P\) with \(\displaystyle OP = 6\text{ cm}\); join \(\displaystyle OP\). 3. Draw the perpendicular bisector of \(\displaystyle OP\), meeting it at \(\displaystyle M\). 4. With centre \(\displaystyle M\), radius \(\displaystyle MO\), draw a circle cutting the given circle at \(\displaystyle T,T'\). 5. Join \(\displaystyle PT\) and \(\displaystyle PT'\). NCERT_Solution_Class10_Maths_Exemplar_Ch10_Ex10-3_Q4 \[\angle OTP = 90^\circ \quad \text{(angle in a semicircle on } OP\text{)} \] \[PT = \sqrt{OP^2-OT^2} = \sqrt{36-16} = \sqrt{20} = 2\sqrt5\text{ cm} \]