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## 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.

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 Bto 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.

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.

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.

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.

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.

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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