# 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) A_{1}, A_{2}, A_{3}, …….A_{13} on AX so that;

AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5} = A_{5}A_{6} = A_{6}A_{7} = A_{7}A_{8} = A_{8}A_{9} = A_{9}A_{10} = A_{10}A_{11} = A_{11}A_{12} = A_{12}A_{13}

Join A_{13} to B.

Through the point A_{5}, draw a line A_{5}C || A_{13}B; 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; AA
_{1}= A_{1}A_{2}= A_{2}A_{3} - Join A
_{3}to B. - Draw a line from point A
_{2}so that this line is parallel to A_{3}B 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 AA
_{1}= A_{1}A_{2}= A_{2}A_{3}= A_{3}A_{4}= A_{4}A_{5}= A_{5}A_{6}= A_{6}A_{7}. - Join A
_{5}to B. - Draw a line from point A
_{7}parallel to A_{5}B 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 DD
_{1}= D_{1}D_{2}= D_{2}D_{3}. - Join D
_{2}to point B. - Draw a line from D
_{3}parallel to D_{2}B 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 BB
_{1}= B_{1}B_{2}= B_{2}B_{3}= B_{3}3B_{4}. - Join B
_{4}to point A. - Draw a line from B
_{3}parallel B_{4}A 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 BB
_{1}= B_{1}B_{2}= B_{2}B_{3}= B_{3}B_{4}. - Join B
_{3}to C. - Now, draw a line from B
_{4}parallel to B_{3}C 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 AA
_{1}= A_{1}A_{2}= A_{2}A_{3}= A_{3}A_{4}= A_{4}A_{5}. - Join A
_{3}to point B. - Draw a line from A
_{5}parallel to A_{3}B so that it meets AB at point B’. - Draw C’B’ || CB.

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