Jharkhand Board JAC Class 10 Maths Solutions Chapter 11 Constructions Ex 11.1 Textbook Exercise Questions and Answers.

## JAC Board Class 10 Maths Solutions Chapter 11 Constructions Exercise 11.1

In each of the following, give the justification of the construction also:

Question 1.

Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.

Solution :

Steps of construction:

1. Draw the line segment AB = 7.6 cm.

2. At A, below AB, make an angle BAx = 30°.

3. At B, above AB, make an angle ABy = 30°.

B\(\hat{A}\)x = A\(\hat{B}\)y = 30°

These are alternate angles. ∴ Ax || By.

4. With a convenient radius cut off five equal parts

AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5} in Ax.

5. With the same radius cut off eight equal parts.

BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4} = B_{4}B_{5} = B_{5}B_{6} = B_{6}B_{7} = B_{7}B_{8}.

6. Join A_{5}B_{8}. Let it cut AB at C.

AC : CB = 5 : 8.

In ΔACA_{5} and CBB_{8}

1. A\(\hat{C}\)A_{5} = B\(\hat{C}\)B_{8} (V.O.A.)

2. C\(\hat{A}\)A_{5} = A\(\hat{B}\)B_{8} (Alternate angles)

3. C\(\hat{A}\)_{5}A = B\(\hat{B}\)_{8}C (Alternate angles)

Δs are equiangular.

∴\(\frac{\mathrm{AA}_5}{\mathrm{BB}_8}=\frac{\mathrm{CA}_5}{\mathrm{AB}_8}=\frac{\mathrm{CA}}{\mathrm{BC}}\)

∴ \(\frac{\mathrm{CA}}{\mathrm{BC}}=\frac{\mathrm{AA}_5}{\mathrm{BB}_8}=\frac{5}{8}\)

Question 2.

Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle.

Solution :

Steps of construction:

- Construct ΔABC given AB = 4 cm, BC = 5 cm, AC = cm.
- At B, make an acute angle CBx.
- Divide Bx into three equal parts with a convenient radius.
- Join B
_{3}C. - From B
_{2}draw a parallel to B_{3}C. - Let it cut BC at C’.
- At C’ make angle A’C’B = ACB.

Join A’C’.

∴ ΔABC ||| A’BC’.

In ΔABC and ΔA’BC’,

1.A\(\hat{B}\)C = A’\(\hat{B}\)C’ (Common angle)

2. A\(\hat{C}\)B = A’\(\hat{C’}\)B (Corresponding angles)

3. B\(\hat{A}\)C = B\(\hat{A’}\)C’ (Remaining angles)

Δs are equiangular.

∴ \(\frac{\mathrm{AC}}{\mathrm{A}^{\prime} \mathrm{C}^{\prime}}=\frac{\mathrm{AB}}{\mathrm{A}^{\prime} \mathrm{B}^{\prime}}=\frac{\mathrm{BC}}{\mathrm{BC} \mathrm{C}^{\prime}}\) = \(\frac{3}{2}\) ∴ [latex]\frac{BC’}{BC}=\frac{2}{3}[/latex]

Question 3.

Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \(\frac{7}{5}\) of the corresponding sides of the first triangle.

Solution :

Steps of construction:

- Construct a triangle ABC given BC= 7 cm, AB = 5 cm, AC = 6 cm.
- At B, below BC, make an acute angle CBx.
- With a convenient radius cut off seven equal parts BB
_{1}= B_{1}B_{2}= B_{2}B_{3}= B_{3}B_{4}= B_{4}B_{5}= B_{5}B_{6}= B_{6}B_{7}. - Join B
_{5}C. - From B
_{1}draw a parallel to B_{5}C to cut BC produced at C’. - At C’ draw a parallel to CA to meet BA produced at A’.

A’BC’ is the required triangle.

In ΔABC and ΔA’BC’

Question 4.

Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1\(\frac{1}{2}\) times the corresponding sides of the isosceles triangle.

Solution :

Steps of construction:

- Draw a line segment BC = 8 cm.
- Draw its perpendicular bisector.
- Cut off DA = 4 cm. (altitude given)
- Join AB and AC. ABC is the required triangle.
- At B, below BC, draw an acute angle CBx.
- With a convenient radius cut off three equal parts BB
_{1}= B_{1}B_{2}= B_{2}B_{3}. - Join B
_{2}C. - At B
_{3}draw a parallel to B_{2}C to meet BC extended at C’. - At C’ draw a parallel to CA to meet BA produced at A’.
- Join A’C’. A’BC’ is the required triangle similar to ΔABC.

In ΔA’BC’ and ABC

1. A’\(\hat{B}\)C’ = A\(\hat{B}\)C (Common angle)

2. B\(\hat{A}\)‘C’ = B\(\hat{A}\)C (Corresponding angles)

3. A’\(\hat{C}\)‘B = A\(\hat{C}\)B

Δs are equiangular.

∴ \(\frac{\mathrm{A}^{\prime} \mathrm{C}^{\prime}}{\mathrm{AC}}=\frac{\mathrm{BC}}{\mathrm{BC}}=\frac{\mathrm{A}^{\prime} \mathrm{B}}{\mathrm{AB}}\)

\(\frac{BC’}{BC}\) = \(\frac{3}{2}\)

BC = 8 cm, BC’ = 8 × \(\frac{3}{2}\) = 12 cm.

BA = CA = 5.6 cm ∴ A’B = A’C’ = 5.6 × \(\frac{3}{2}\) = 8.4 cm.

Question 5.

Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are \(\frac{3}{4}\) of the corresponding sides of the triangle ABC.

Solution :

Steps of construction:

- Construct ΔABC given BC = 6 cm, AB = 5 cm, A\(\hat{B}\)C = 60°.
- At B, below BC, make an acute angle CBx.
- In Bx cut off four equal parts BB
_{1}= B_{1}B_{2}= B_{2}B_{3}= B_{3}B_{4} - Join B
_{4}C. - From B
_{3}draw a parallel to B_{4}C to meet BC at C’. - At C’ draw a parallel to CA to meet CA at A’.

A’BC’ is the required triangle similar to ΔABC.

In Δs A’BC’ and ABC

Question 6.

Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are \(\frac{4}{3}\) times the corresponding sides of ΔABC.

Solution :

\(\hat{A}\) + \(\hat{B}\) + \(\hat{C}\) = 180°

\(\hat{A}\) + \(\hat{B}\) = 150° (105° + 45°)

∴ \(\hat{C}\) = 30°

Steps of construction:

1. In ΔABC, BC = 7 cm.

\(\hat{B}\) = 45°, \(\hat{C}\) = 105°

∴ \(\hat{A}\) = 180° – 45° – 105°

= 180° – 150° = 30°.

Construct ΔABC given BC = 7 cm, \(\hat{B}\) = 45°, \(\hat{C}\) = 30°.

2. At B draw an acute angle CBx.

3. In Bx cut off four equal parts BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4} with a convenient radius.

4. Join B_{3}C.

5. At B_{4} draw a parallel to B_{3}C to meet BC produced at C’.

6. At C’ make angle of 30° equal to A\(\hat{C}\)B. Let it meet BA produced at A’.

A’BC’ is the required triangle similar to ΔABC.

In ΔC’BB_{4} CB_{3} || C’B_{4}

Question 7.

Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are \(\frac{5}{3}\) times the corresponding sides of the given triangle.

Solution :

BC’ = \(\frac{5}{3}\) × BC = \(\frac{5}{3}\) × 3 = 5 cm

BA’= \(\frac{5}{3}\) × 4 = \(\frac{20}{3}\) = 6.6 cm

A’C’ = \(\frac{5}{3}\) × 5 = \(\frac{25}{3}\) = 8.3 cm.

Steps of construction:

- Construct ΔABC given BC= 3 cm, \(\hat{B}\) = 90°, BA = 4 cm.
- At B, make an acute angle CBx.
- Cut off five equal parts BB
_{1}= B_{1}B_{2}= B_{2}B_{3}= B_{3}B_{4}= B_{4}B_{5}along Bx with a convenient radius. - Join B
_{3}C. - At B
_{5}draw a parallel to B_{3}C to meet BC produced at C’. - At C’ draw a parallel to CA to meet BA produced at A’.

A’BC’ is the required triangle similar to ΔABC.

BC’ : BC = BB_{5} = BB_{3}

\(\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{BB}_5}{\mathrm{BB}_3}\) = \(\frac{5}{3}\)

In Δs A’BC’ and ABC

1. A’\(\hat{B}\)C’ = A\(\hat{B}\)C = 90°

2. B\(\hat{A}\)‘C’ = B\(\hat{A}\)C (Corresponding angles)

3. B\(\hat{C}\)‘A’ = B\(\hat{C}\)A

Δs are equiangular.

∴ \(\frac{A’C’}{AC}\) = \(\frac{BC’}{BC}\) = \(\frac{BA’}{BA}\) = \(\frac{5}{3}\)