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Justification:
The construction of the given problem can be justified by proving that
AC/CB = 5/ 8
By construction, we have A5C || A13B. From the Basic proportionality theorem for the triangle AA13B, we get
AC/CB =AA
5
/A
5
A
13
….. (1)
From the figure constructed, it is observed that AA5 and A5A13 contain 5 and 8 equal divisions of line segments, respectively.
Therefore, it becomes
AA
5
/A
5
A
13
=5/8… (2)
Compare the equations (1) and (2), we obtain
AC/CB = 5/ 8
Hence, justified.
Justification
The construction of the given problem can be justified by proving that
AB’ = (2/3)AB
B’C’ = (2/3)BC
AC’= (2/3)AC
From the construction, we get B’C’ || BC
∴ ∠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,
∠A
2
AB’ =∠A
3
AB (Common)
From the corresponding angles, we get
∠AA
2
B’ =∠AA
3
B
Therefore, from the AA similarity criterion, we obtain
ΔAA
2
B’ and AA
3
B
So, AB’/AB = AA
2
/AA
3
Therefore, AB’/AB = 2/3 ……. (2)
From equations (1) and (2), we get
AB’/AB=B’C’/BC = AC’/ AC = 2/3
This can be written as
AB’ = (2/3)AB
B’C’ = (2/3)BC
AC’= (2/3)AC
Hence, justified.
Justification
The construction of the given problem can be justified by proving that
AB’ = (7/5)AB
B’C’ = (7/5)BC
AC’= (7/5)AC
From the construction, we get B’C’ || BC
∴ ∠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 ΔAA
7
B’ and ΔAA
5
B,
∠A
7
AB’=∠A
5
AB (Common)
From the corresponding angles, we get
∠A A
7
B’=∠A A
5
B
Therefore, from the AA similarity criterion, we obtain
ΔA A
2
B’ and A A
3
B
So, AB’/AB = AA
5
/AA
7
Therefore, AB /AB’ = 5/7 ……. (2)
From equations (1) and (2), we get
AB’/AB = B’C’/BC = AC’/ AC = 7/5
This can be written as
AB’ = (7/5)AB
B’C’ = (7/5)BC
AC’= (7/5)AC
Hence, justified.
Construction Procedure:
1. Draw a line segment BC with a measure of 8 cm.
2. Now, draw the perpendicular bisector of the line segment BC and intersect at point D.
3. Take the point D as the centre and draw an arc with a radius of 4 cm, which intersects the perpendicular bisector at the point A.
4. Now, join the lines AB and AC, and the triangle is the required triangle.
5. Draw a ray BX which makes an acute angle with the line BC on the side opposite to the vertex A.
6. Locate the 3 points B
1
, B
2
and B
3
on the ray BX such that BB
1
= B
1
B
2
= B
2
B
3
7. Join the points B
2
C and draw a line from B
3,
which is parallel to the line B
2
C where it intersects the extended line segment BC at point C’.
8. Now, draw a line from C’ to the extended line segment AC at A’, which is parallel to the line AC, and it intersects to make a triangle.
9. Therefore, ΔA’BC’ is the required triangle.
Justification
The construction of the given problem can be justified by proving that
A’B = (3/2)AB
BC’ = (3/2)BC
A’C’= (3/2)AC
From the construction, we get A’C’ || AC
∴ ∠ A’C’B = ∠ACB (Corresponding angles)
In ΔA’BC’ and ΔABC,
∠B = ∠B (Common)
∠A’BC’ = ∠ACB
∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)
Therefore, A’B/AB = BC’/BC= A’C’/AC
Since the corresponding sides of the similar triangle are in the same ratio, it becomes
A’B/AB = BC’/BC= A’C’/AC = 3/2
Hence, justified.
Justification
The construction of the given problem can be justified by proving that
Since the scale factor is 4/3, we need to prove
A’B = (4/3)AB
BC’ = (4/3)BC
A’C’= (4/3)AC
From the construction, we get A’C’ || AC
In ΔA’BC’ and ΔABC,
∴ ∠A’C’B = ∠ACB (Corresponding angles)
∠B = ∠B (Common)
∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)
Since the corresponding sides of the similar triangle are in the same ratio, it becomes
Therefore, A’B/AB = BC’/BC= A’C’/AC
So, it becomes A’B/AB = BC’/BC= A’C’/AC = 4/3
Hence, justified.
Justification
The construction of the given problem can be justified by proving that
Since the scale factor is 5/3, we need to prove
A’B = (5/3)AB
BC’ = (5/3)BC
A’C’= (5/3)AC
From the construction, we get A’C’ || AC
In ΔA’BC’ and ΔABC,
∴ ∠ A’C’B = ∠ACB (Corresponding angles)
∠B = ∠B (Common)
∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)
Since the corresponding sides of the similar triangle are in the same ratio, it becomes
Therefore, A’B/AB = BC’/BC= A’C’/AC
So, it becomes A’B/AB = BC’/BC= A’C’/AC = 5/3
Hence, justified.
