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For each of the following, give the justification of the constructions made:
Question 1. Draw a line segment of length 7.6 cm and divide it in the ratio of 5:8. Measure the two parts.
Answer.
Procedures for construction:
A line segment with a length of 7.6 cm is divided in the ratio of 5:8.
1. Create a line segment AB with a length of 7.6 cm.
2. Create an acute angle between ray AX and line segment AB by drawing a ray AX.
3. Locate the points on the ray AX, i.e., 13 (= 5+8) points, such as A1, A2, A3, A4........ A13, so that AA1 = A1A2 = A2A3 and so on.
4. Create a connection between the line segment and the ray, BA13.
5. Parallel to BA13, draw a line through point A5 that produces an angle equal to AA13B.
6. The location A5, where the line AB intersects at point C.
7. C is the point that divides the 7.6cmlong line segment AB in the appropriate 5:8 ratio.
8. Now, determine the lengths of the AC and CB lines. It comes to 2.9 cm and 4.7 cm in length.
Justification:
The given problem's structure can be justified by demonstrating that
AC/CB = 5/8
A5C  A13B is our construction. We obtain the following from the fundamental proportionality theorem for the triangle AA13B:
AC/CB = AA5/A5A13.... (1)
AA5 and A5A13 contain five and eight equal divisions of line segments, respectively, as seen in the figure.
As a result, it becomes
AA5/A5A13=5/8… (2)
By comparing equations (1) and (2), we obtain at
AC/CB = 5/8
Hence proved.
Question 2. 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.
Procedure for Construction:
1. Draw a 4centimetre line segment AB, i.e. AB = 4 centimetres.
2. Draw a 5 cm radius arc with point A at the centre.
3. Draw an arc with a radius of 6 cm with point B as the centre.
4. At point C, the sketched arcs will cross.
5. With AC = 5 cm and BC = 6 cm, the needed triangle is ΔABC.
6. On the opposite side of vertex C, draw a ray AX that forms an acute angle with the line segment AB.
7. Draw a line AX with three points A1, A2, and A3 (since 3 is higher than 2 and 3) so that AA1= A1A2 = A2A3.
8. Draw a line through A2 parallel to BA3 and intersecting AB at point B'.
9. Draw a line parallel to the line BC from point B' to point C', intersecting the line AC.
10. Therefore, ΔAB’C’ is the required triangle.
Justification:
It is possible to justify the problem's construction by demonstrating that
(2/3)AB = AB'
(2/3)BC = B'C'
AC' equals (2/3)AC
We get B'C'  BC from the construction.
∴ ∠AB’C’ = ∠ABC (Corresponding angles)
In ΔAB’C’ and ΔABC,
∠ABC = ∠AB’C (Proved above)
∠BAC = ∠B’AC’ (Common)
∴ ΔAB’C’ ∼ ΔABC (From AA similarity criterion)
Therefore, AB’/AB = B’C’/BC= AC’/AC …. (1)
In ΔAAB’ and ΔAAB,
∠A2AB’ =∠A3AB (Common)
We get, from the corresponding angles,
∠AA2B’ =∠AA3B
As a result of applying the AA similarity criterion, we obtain:
ΔAA2B’ and AA3B
So,
AB’/AB = AA2/AA3
AB'/AB = 2/3 as a result..... (2)
We get (1) and (2) from the equations.
AB’/AB=B’
C’/BC = AC’/ AC = 2/3
This can be expressed as
(2/3)AB = AB'
(2/3)BC = B'C'
AC' = (2/3)AC
Thus the result is justified.
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.
Answer.
Construction Procedures:
1. Draw a line segment with the length AB = 5 cm.
2. Using A and B as the centres, draw arcs with radii of 6 and 7 cm, respectively.
3. Because these arcs will cross at point C, ΔABC is the requisite triangle with sides of 5 cm, 6 cm, and 7 cm, respectively.
4. Draw a ray AX on the opposite side of vertex C that forms an acute angle with the line segment AB.
5. On the line AX, locate the seven points A1, A2, A3, A4, A5, A6, and A7 (since 7 is greater than 5 and 7), such that AA1 = A1A2 = A2A3 = A3A4 = A3A5 = A4A6 = A6A7
6. Connect the points BA5 and create a parallel line from A7 to BA5 that intersects the extended line segment AB at point B'.
7. Now, draw a line from B' to C' that is parallel to the extended line segment AC, intersecting the line BC to form a triangle.
8. So, the needed triangle is ΔAB’C’.
Justification:
The given problem's structure can be justified by demonstrating that
AB' = (7/5)AB
B’C’ = (7/5)BC
AC’= (7/5)AC
We get B'C'  BC from the construction.
∴ ∠AB’C’ = ∠ABC (Corresponding angles)
In ΔAB’C’ and ΔABC,
∠ABC = ∠AB’C (Proved above)
∠BAC = ∠B’AC’ (Common)
∴ ΔAB’C’ ∼ ΔABC (From AA similarity criterion)
Therefore, AB’/AB = B’C’/BC= AC’/AC …. (1)
In ΔAA7B’ and ΔAA5B,
∠A7AB’=∠A5AB (Common)
We obtain the following from the corresponding angles:
∠A A7B’=∠A A5B
As a result of the AA similarity criterion, we arrive at
ΔA A2B’ and A A3B
Thus, AB'/AB = AA5/AA7.
As a result, AB /AB' = 5/7... (2)
We get from equations (1) and (2)
AB'/AB = B'C'/BC = AC'/AC = 7/5
This can also be expressed as
AB' = (7/5)AB
B’C’ = (7/5)BC
AC’= (7/5)AC
Hence justified.
Question 4. Construct an isosceles triangle whose base is 8 cm and altitude is 4cm, then another triangle whose sides are 112 times the corresponding sides of the isosceles triangle.
Answer.
Procedure for Construction:
1. Draw a line segment BC with an 8 cm length.
2. Now, create a perpendicular bisector of the line segment BC that intersects at point D.
3. Using D as the centre, draw an arc with a radius of 4 cm that intersects the perpendicular bisector at A.
4. Now, connect the lines AB and AC to form the needed triangle.
5. On the side opposite the vertex A, draw a ray BX that forms an acute angle with the line BC.
6. On the ray BX, locate the three points B1, B2, and B3 such that BB1 = B1B2 = B2B3.
7. Connect the points B2C and construct a line parallel to the line B2C from B3 to point C' where it intersects the extended line segment BC.
8. Now, draw a line from C' to A' so it is parallel to the extended line segment AC and intersects the line AC to form a triangle.
9. As a result, the needed triangle is ΔA’BC’'.
Justification:
The given problem's structure can be justified by demonstrating that
A'B = (3/2)AB
BC' = (3/2)BC
A’C’= (3/2)AC
We get A'C'  AC from the construction.
∴ ∠ A’C’B = ∠ACB (Corresponding angles)
In ΔA’BC’ and ΔABC,
∠B = ∠B (common)
∠A’BC’ = ∠ACB
∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)
As a result, A'B/AB = BC'/BC = A'C'/AC.
Because the similar triangle's corresponding sides have the same ratio, it becomes
A'B/AB = BC'/BC = A'C'/AC equals 3/2
As a result, justified.
Points to remember

A point that moves in the same direction as two other points has a locus that is normal to the line linking them.

A bisector divides a line segment in half, whereas perpendicular or normal refers to straight angles.

Using a pair of compasses and a ruler, you can make a variety of forms.

Examine the situation and offer a clear solution.