Tag: NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1

NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 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 a line segment AB = 7.6 cm.
2. Draw an acute angle BAX on base AB. Mark the ray as AX.
3. Locate 13 points A₁, A₂, A₃, …… , A₁₃ on the ray AX so that AA₁ = A₁A₂ = ……… = A₁₂A₁₃
4. Join A₁₃ with B and at A₅ draw a line ∥ to BA₁₃, i.e. A₅C. The line intersects AB at C.
5. On measure AC = 2.9 cm and BC = 4.7 cm.

Justification:
In ∆AA₅C and ∆AA₁₃B,

∴ AC : BC = 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 a ΔABC with AB = 4 cm, BC = 6 cm and AC = 5 cm.
2. Draw an acute angle CBX on the base BC at point B. Mark the ray as BX.
3. Mark the ray BX with B₁, B₂, B₃ such that
BB₁ = B₁B₂ = B₂B₃
4. Join B₃ to C.
5. Draw B₂C’ ∥ B3C, where C’ is a point on BC.
6. Draw C’A’ ∥ AC, where A’ is a point on BA.
7. ΔA’BC’ is the required triangle.

Justification: In ∆A’BC and ∆ABC,

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. Draw a ΔABC with AB = 5 cm, BC = 7 cm and AC = 6 cm.
2. Draw an acute angle CBX below BC at point B.
3. Mark the ray BX as B₁, B₂, B₃, B₄, B₅, B₆ and B₇ such that BB₁= B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇.
4. Join B₅ to C.
5. Draw B₇C’ parallel to B₅C, where C’ is a point on extended line BC.
6. Draw A’C’ ∥ AC, where A’ is a point on extended line BA.
A’BC’ is the required triangle.

Justification: In ∆ABC and ∆A’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 base AB = 8 cm.
2. Draw perpendicular bisector of AB. Mark CD = 4 cm, on ⊥ bisector where D is mid-point on AB.
3. Draw an acute angle BAX, below AB at point A.
4. Mark the ray AX with A₁, A₂, A₃ such that AA₁ =A₁A₂ = A₂A₃
5. Join A₂ to B. Draw A₃B’ ∥ A₂ B, where B’ is a point on extended line AB.
6. At B’, draw B’C’ 11 BC, where C’ is a point on extended line AC.
7. ∆AB’C’ is the required triangle.

Justification: In ∆ABC and ∆A’BC’,

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. Draw a line segment BC = 6 cm and at point B draw an ∠ABC = 60°.
2. Cut AB = 5 cm. Join AC. We obtain a ΔABC.
3. Draw a ray BX making an acute angle with BC on the side opposite to the vertex A.
4. Locate 4 points A₁, A₂, A₃ and A₄ on the ray BX so that BA₁ = A₁A₂ = A₂A₃ = A₃A₄.
5. Join A₄ to C.
6. At A₃, draw A₃C’ ∥ A₄C, where C’ is a point on the line segment BC.
7. At C’, draw C’A’ ∥ CA, where A’ is a point on the line segment BA.
∴ ∆A’BC’ is the required triangle.

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 ∆ABC.
Solution:
In ∆ABC, ∠A + ∠B + ∠C = 180°
⇒ 105° + 45° + ∠C = 180°
⇒ 150° + ∠C = 180°
⇒ ∠C = 30°
Steps of Construction:
1. Draw a line segment BC = 7 cm. At point B, draw an ∠B = 45° and at point C, draw an ∠C = 30° and get ΔABC.
2. Draw an acute ∠CBX on the base BC at point B. Mark the ray BX with B₁, B₂, B₃, B₄, such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄
3. Join B₃ to C.
4. Draw B₄C’ ∥ B₃C, where C’ is point on extended line segment BC.
5. At C’, draw C’A’ ∥ AC, where A’ is a point on extended line segment BA.
6. ∆A’BC’ is the required triangle.

Justification: In ∆ABC and ∆A’BC’,

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 \(\frac { 5 }{ 3F }\) times the corresponding sides of the given triangle.
Solution:
Steps of Construction:
1. Draw a right angled ∆ABC with AB = 4 cm, AC = 3 cm and ∠A = 90°.
2. Make an acute angle BAX on the base AB at point A.
3. Mark the ray AX with A₁, A₂, A₃, A₄, A₅ such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅.
4. Join A₃B. At A₅, draw A₅B’ ∥ A₃B, where B’ is a point on extended line segment AB.
5. At B’, draw B’C’ ∥ BC, where C’ is a point on extended line segment AC.
6. ∆AB’C’ is the required triangle.

Justification:
In ∆ABC and ∆AB’C’,

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