JAC Class 10 Maths Solutions Chapter 11 Constructions Ex 11.1

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
AA1 = A1A2 = A2A3 = A3A4 = A4A5 in Ax.
5. With the same radius cut off eight equal parts.
BB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7 = B7B8.
6. Join A5B8. Let it cut AB at C.
AC : CB = 5 : 8.

In ΔACA5 and CBB8
1. A$$\hat{C}$$A5 = B$$\hat{C}$$B8 (V.O.A.)
2. C$$\hat{A}$$A5 = A$$\hat{B}$$B8 (Alternate angles)
3. C$$\hat{A}$$5A = B$$\hat{B}$$8C (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:

1. Construct ΔABC given AB = 4 cm, BC = 5 cm, AC = cm.
2. At B, make an acute angle CBx.
3. Divide Bx into three equal parts with a convenient radius.
4. Join B3C.
5. From B2 draw a parallel to B3C.
6. Let it cut BC at C’.
7. 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.
∴ $$\frac{\mathrm{AC}}{\mathrm{A}^{\prime} \mathrm{C}^{\prime}}=\frac{\mathrm{AB}}{\mathrm{A}^{\prime} \mathrm{B}^{\prime}}=\frac{\mathrm{BC}}{\mathrm{BC} \mathrm{C}^{\prime}}$$ = $$\frac{3}{2}$$ ∴ $\frac{BC’}{BC}=\frac{2}{3}$

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:

1. Construct a triangle ABC given BC= 7 cm, AB = 5 cm, AC = 6 cm.
2. At B, below BC, make an acute angle CBx.
3. With a convenient radius cut off seven equal parts BB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7.
4. Join B5C.
5. From B1 draw a parallel to B5C to cut BC produced at C’.
6. 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:

1. Draw a line segment BC = 8 cm.
2. Draw its perpendicular bisector.
3. Cut off DA = 4 cm. (altitude given)
4. Join AB and AC. ABC is the required triangle.
5. At B, below BC, draw an acute angle CBx.
6. With a convenient radius cut off three equal parts BB1 = B1B2 = B2B3.
7. Join B2C.
8. At B3 draw a parallel to B2C to meet BC extended at C’.
9. At C’ draw a parallel to CA to meet BA produced at A’.
10. 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:

1. Construct ΔABC given BC = 6 cm, AB = 5 cm, A$$\hat{B}$$C = 60°.
2. At B, below BC, make an acute angle CBx.
3. In Bx cut off four equal parts BB1 = B1B2 = B2B3 = B3B4
4. Join B4C.
5. From B3 draw a parallel to B4C to meet BC at C’.
6. 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 BB1 = B1B2 = B2B3 = B3B4 with a convenient radius.
4. Join B3C.
5. At B4 draw a parallel to B3C 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’BB4 CB3 || C’B4

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:

1. Construct ΔABC given BC= 3 cm, $$\hat{B}$$ = 90°, BA = 4 cm.
2. At B, make an acute angle CBx.
3. Cut off five equal parts BB1 = B1B2 = B2B3 = B3B4 = B4B5 along Bx with a convenient radius.
4. Join B3C.
5. At B5 draw a parallel to B3C to meet BC produced at C’.
6. At C’ draw a parallel to CA to meet BA produced at A’.

A’BC’ is the required triangle similar to ΔABC.
BC’ : BC = BB5 = BB3
$$\frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{BB}_5}{\mathrm{BB}_3}$$ = $$\frac{5}{3}$$

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.
∴ $$\frac{A’C’}{AC}$$ = $$\frac{BC’}{BC}$$ = $$\frac{BA’}{BA}$$ = $$\frac{5}{3}$$