Construction of Triangle

NCERT Exercise 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: Draw a line segment AB = 7.6 cm.

Draw a ray AX which makes an acute angle with AB.

Locate 13 points (5 + 8) A₁, A₂, A₃, …….A₁₃ on AX so that;

AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇ = A₇A₈ = A₈A₉ = A₉A₁₀ = A₁₀A₁₁ = A₁₁A₁₂ = A₁₂A₁₃

Join A₁₃ to B.

Through the point A₅, draw a line A₅C || A₁₃B; which intersects AB at point C.

We get; AC : CB = 5 : 8

AC = 2.92 cm and CB = 4.68 cm

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

Solution: Making the triangle.

Draw a line segment AB = 4 cm
Draw an arc at 5 cm from point A.
Draw an arc at 6 cm from point B so that it intersects the previous arc.
Joint the point of intersection from A and B.
This gives the required triangle ABC.

Dividing the base in 2 : 3 ratio:

Draw a ray AX at an acute angle from AB.
Plot three points on AX so that; AA₁ = A₁A₂ = A₂A₃
Join A₃ to B.
Draw a line from point A₂ so that this line is parallel to A₃B and intersects AB at point B’.
Draw a line from point B’ parallel to BC so that this line intersects AC at point C’.

Triangle AB’C’ is the required triangle.

Question 3: 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.

Solution: Making the triangle

Draw a line segment AB = 5 cm
Draw an arc at 6 cm from point A.
Draw an arc at 7 cm from point B so that it intersects the previous arc.
Joint the point of intersection from A and B.
This gives the required triangle ABC.

Dividing the base:

Draw a ray AX at and acute angle from AB.
Plot seven points on AX; so that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇.
Join A₅ to B.
Draw a line from point A₇ parallel to A₅B so that it joins AB’ (AB extended to AB’).
Draw a line B’C’ || BC.

Triangle AB’C’ is the required triangle.

Question 4: Construct and 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.

Solution: Making the triangle:

Draw a line segment AB = 8 cm.
Draw two intersecting arcs at 4 cm distance from points A and B; on either side of AB.
Join these arcs to get perpendicular bisector CD of AB. (Because altitude is the perpendicular bisector of base of isosceles triangle).
Join A and B to C to get the triangle ABC.

Draw a ray DX at an acute angle from point D.
Plot 3 points on DX so that DD₁ = D₁D₂ = D₂D₃.
Join D₂ to point B.
Draw a line from D₃ parallel to D₂B so that it meets the extension of AB at B’.
Now, draw B’C’ || BC.
Draw A’C’ || AC.

Triangle A’B’C’ is the required triangle.

Question 5: Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60^o. Then construct a triangle whose sides are ¾ of the corresponding sides of the triangle ABC.

Solution: Making the triangle:

Draw a line segment AB = 5 cm.
Make a 60^o angle from point B and draw BC = 6 cm.
Join A and C to get the triangle ABC.

Dividing the base:

Draw a ray at an acute angle from BA.
Plot 4 points on BA so that BB₁ = B₁B₂ = B₂B₃ = B₃3B₄.
Join B₄ to point A.
Draw a line from B₃ parallel B₄A so that it meets AB at point A’.
Draw A’C’ || AC.

Triangle A’C’B is the required triangle.

Question 6: Draw a triangle ABC with side BC = 7 cm, ∠B = 45^o, ∠A = 105^o. Then construct a triangle whose sides are 4/3 times the corresponding sides of triangle ABC.

Solution: Making the triangle:

Draw a line segment BC = 7 cm.
Make an angle of 45^o at point B and an angle of 30^o at point C (because 45 + 30 + 105 = 180).
Join the lines from points B and C at point A.
This gives the required triangle ABC.

Dividing the Base:

Draw a ray BX at an acute angle from point B.
Plot 4 points on BX so that BB₁ = B₁B₂ = B₂B₃ = B₃B₄.
Join B₃ to C.
Now, draw a line from B₄ parallel to B₃C so that it meets line BC at C’.
Draw A’C’ || AC.

Triangle A’BC’ is the required triangle.

Question 7: Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 m and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.

Solution: Making the triangle:

Draw a line segment AB = 3 cm.

Make a right angle at point A and draw AC = 4 cm from this point.

Join points A and B to get the right triangle ABC.

Dividing the base:

Draw a ray AX at an acute angle from AB.
Plot 5 points on AX so that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅.
Join A₃ to point B.
Draw a line from A₅ parallel to A₃B so that it meets AB at point B’.
Draw C’B’ || CB.

Triangle A’B’C’ is the required triangle.