NCERT Solutions for Exercise 11.1 Class 10 Maths Chapter 11 - Constructions

Q2 Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.

Answer:

Steps of construction are:-

(i) Firstly draw a line segment AB of length 4 cm.

(ii) Now cut an arc of radius 5 cm from point A and an arc of 6 cm from point B.

(iii) Name the point of intersection of arcs to be point C.

(iv) Now join point AC and BC. Thus $\mathrm{\Delta }$ ABC is the required triangle.

(v) Draw a line AD which makes an acute angle with AB and is opposite of vertex C.

(vi) Cut three equal parts of line AD namely AA ₁ , AA ₂ , AA ₃ .

(vii) Now join A ₃ to B. Draw a line A ₂ B' parallel to A ₃ B.

(viii) And then draw a line B'C' parallel to BC.

Hence $\mathrm{\Delta }$ AB'C' is the required triangle.

Q3 Construct a triangle with sides 5 cm, 6 cm, and 7 cm and then another triangle whose sides are 7 / 5 of the corresponding sides of the first triangle.

Answer:

Steps of construction are:-

(i) Firstly draw a line segment AB of length 5 cm.

(ii) Now cut an arc of radius 6 cm from point A and an arc of 7 cm from point B.

(iii) Name the point of intersection of arcs to be point C.

(iv) Now join point AC and BC. Thus $\mathrm{\Delta }$ ABC is the required triangle.

(v) Draw a line AD which makes an acute angle with AB and is opposite of vertex C.

(vi) Cut seven equal parts of line AD namely AA ₁ , AA ₂ , AA ₃ , AA ₄ , AA ₅ , AA ₆ , AA ₇ ,.

(vii) Now join A ₅ to B. Draw a line A ₇ B' parallel to A ₅ B.

(viii) And then draw a line B'C' parallel to BC.

Hence $\mathrm{\Delta }$ AB'C' is the required triangle.

Q4 Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 3/ 2 times the corresponding sides of the isosceles triangle.

Answer:

Steps of construction:-

(i) Draw a line segment AB of length 8 cm.

(ii) Cut arcs taking point A and point B as the center. Draw the line to intersect on line segment AB. Mark the intersecting point as point D.

(iii) Cut arc of length 4 cm on the same line which will be the altitude of the triangle.

(iv) Name the point as C. Then $\mathrm{\Delta }$ ABC is the isosceles triangle.

(v) Draw a line AX which makes an acute angle with AB and is opposite of vertex C.

(vi) Cut seven equal parts of line AX namely AA ₁ , AA ₂ , AA ₃ .

(vii) Now join A ₂ to B. Draw a line A ₃ B' parallel to A ₂ B.

(viii) And then draw a line B'C' parallel to BC.

Hence $\mathrm{\Delta }$ AB'C' is the required triangle.

Q5 Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and $\mathrm{\angle }$ ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.

Answer:

Steps of construction:-

(i) Draw a line segment BC with a measurement of 6 cm.

(ii) Now construct angle 60 ^o from point B and draw AB = 5 cm.

(iii) Join point C with point A. Thus $\mathrm{\Delta }$ ABC is the required triangle.

(iv) Draw a line BX which makes an acute angle with BC and is opposite of vertex A.

(v) Cut four equal parts of line BX namely BB ₁ , BB ₂ , BB ₃ , BB _4.

(vi) Now join B ₄ to C. Draw a line B ₃ C' parallel to B ₄ C.

(vii) And then draw a line B'C' parallel to BC.

Hence $\mathrm{\Delta }$ AB'C' is the required triangle.

Q6 Draw a triangle ABC with side BC = 7 cm, $\mathrm{\angle }$ B = 45°, $\mathrm{\angle }$ A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of D ABC.

Answer:

Steps of construction:-

(i) Draw a line segment BC.

(ii) Now draw an angle $\mathrm{\angle }$ B = 45 ^o and $\mathrm{\angle }$ C = 30 ^o and draw rays in these directions.

(iii) Name the intersection of these lines as A.

(iv) Thus $△ABC$ is the required triangle.

(v) Draw a line BX which makes an acute angle with BC and is opposite of vertex A.

(vi) Cut four equal parts of line BX namely BB ₁ , BB ₂ , BB ₃ , BB _4.

(vii) Now join B ₃ to C. Draw a line B ₄ C' parallel to B ₃ C.

(viii) And then draw a line B'C' parallel to BC.

Hence $\mathrm{\Delta }$ AB'C' is the required triangle.

Q7 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 5/3 times the corresponding sides of the given triangle.

Answer:

Steps of construction:-

(i) Draw a line segment AB having a length of 4 cm.

(ii) Now, construct a right angle at point A and make a line of 3 cm.

(iii) Name this point C. Thus $\mathrm{\Delta }$ ABC is the required triangle.

(iv) Draw a line AX which makes an acute angle with AB and is opposite of vertex C.

(v) Cut four equal parts of line AX namely AA ₁ , AA ₂ , AA ₃ , AA ₄ , AA ₅ .

(vi) Now join A ₃ to B. Draw a line A ₅ B' parallel to A ₃ B.

(vii) And then draw a line B'C' parallel to BC.

Hence $\mathrm{\Delta }$ AB'C' is the required triangle.