# Gujarat Board Solutions Class 10 Maths Chapter 11 Constructions Ex 11.1

Gujarat Board Solutions Class 10 Maths Chapter 11 Constructions Ex 11.1

## Gujarat Board Textbook Solutions Class 10 Maths Chapter 11 Constructions Ex 11.1

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 AB = 7.6 cm

2. At A, draw an acute angle ∠BAX below BA.

3. On AX mark 13 (5 + 8) arcs A_{1}, A_{2}, …, A_{13}.

4. Join B to A_{13}.

5. From A_{5}, draw A_{5}C || A_{13}B.

AC : CO = 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 of the corresponding sides of the first triangle.

Solution:

Steps of Construction:

1. Draw ΔABC with AC = 6cm, AB = 5 cm, BC = 4 cm.

2. At A_{1} draw an acute angle∠CAX below the base AC.

3. Mark 3 arcs on AX such that AA_{1} = A_{1}A_{2} = A_{2}A_{3}.

4. Join A_{3}C.

5. Draw A_{2}C’ || A_{3}C

6. From C’ draw B’C’ || BC

Question 3.

Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 5/8 of the corresponding sides of the first triangle.

Solution:

1. Construct a ΔABC with AB = 7 cm, AC = 5 cm, BC = 6 cm

2. At A draw an acute ∠BAX below AB.

3. On AX, mark 7 arcs A_{1}, A_{2}, …, A_{7} such that

AA_{1} = A_{1}A_{2} =…=A_{6}A_{7}.

4. Join A_{5}B.

5. From A_{7} draw A_{7}B’ || A_{5}B meeting AB at B’ (extent AB to B’)

6. From B’ draw B’C’ || BC meeting AC at C’ [extend AC to C’]

AB’C’ is the required triangle each of whose sides is 5/8 of the corresponding sides of ΔABC.

Question 4.

Construct an isosceles triangle whose base is 8 cm and altitude is 4 cm and then another triangle whose sides are 1 times the corresponding sides of the isosceles triangle. (CBSE 2017)

Solution:

Steps of Construction:

1. Draw a line segment BC = 8 cm.

2. Draw its perpendicular bisector AD (4 cm).

3. Joining AB and AC, we get isosceles triangle ABC

4. Construct an acute ∠CBX.

5. Along BX cut 3 arcs B_{1}, B_{2} and B_{3} such that

BB_{1} = B_{1}B_{2} = B_{2}B_{3}.

6. Join C to B_{2} and draw a line B_{3}C’ || B_{2}C [extend BC to C’]

7. From C’ draw C’A’ || CA [extend BA to A’]

∴ ΔA’BC’ is a required triangle.

Question 5.

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

Solution:

Steps of Construction:

1. Draw a line segment BC = 6 cm.

2. At B draw ∠CBY = 60° on which take AB = 5 cm.

3. Join AC. ΔABC is required triangle.

4. From B, draw an acute ∠CBX downwards.

5. Mark four arcs B_{1}, B_{2}, B_{3}, B_{4} on BX such that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4}.

6. Join B_{4}C and from B_{3} draw B_{3}C’ B_{4}C.

7. From C’, draw A’C’ || AC.

When ΔA’BC’ is the required triangle whose sides are 3/4 of the corresponding sides of ΔABC.

Question 6.

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

Solution:

Steps of Construction:

1. Draw a ABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°

[∠C = 180° – 45° – 105° = 180° – 150° = 30°]

2. At B, draw an acute ∠CBX below BC.

3. On BX, mark the arcs B_{1}, B_{2}, B_{3}, B_{4} such

that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{1}B_{4}.

4. Join B_{3}C.

5. Draw B_{4}C’|| B_{3}C [Extend BC to C’]

6. Draw C’A’ || CA [Extend BA to A’]

∴ ΔA’BC’ is the required triangle.

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 times the corresponding sides of the given triangle. (CBSE 2010)

Solution:

1. Draw a line segment AB = 4 cm.

2. By making right angle at B. Draw BC = 3 cm.

3. Join AC. ΔABC is the required right-angled triangle.

4. At A, Draw acute angle ∠BAX downwards.

5. On AX, mark 5 arcs A_{1}, A_{2}, A_{3}, A_{4}, A_{5} such

that AA_{1} = A_{1}A_{2} = … = A_{4}A_{5}

6. Join A_{3}B.

7. Draw A_{5}B’|| A_{3}B [Extend AB to B’]

8. Draw B_{1}C’ || BC [Extend AC to C’]

Hence ΔAB’C’ is the required triangle

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