 Geometry Construction solution of ncert exercise 11.1 part 1 Class Ten Mathematics

# Geometry Construction

## 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 = 60o. 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 60o 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 = 45o, ∠A = 105o. 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 45o at point B and an angle of 30o 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.