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.
Construction Procedure:
1. Draw a line segment AB =5 cm.
2. Take A and B as centre, and draw the arcs of radius 6 cm and 5 cm respectively.
3. These arcs will intersect each other at point C and therefore ΔABC is the required triangle with the length of sides as 5 cm, 6 cm, and 7 cm respectively.
4. Draw a ray AX which makes an acute angle with the line segment AB on the opposite side of vertex C.
5. Locate the 7 points such as A1, A2, A3, A4, A5, A6, A7 (as 7 is greater between 5 and 7), on line AX such that it becomes AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7
6. Join the points BA5 and draw a line from A7 to BA5 which is parallel to the line BA5 where it intersects the extended line segment AB at point B’.
7. Now, draw a line from B’ the extended line segment AC at C’ which is parallel to the line BC and it intersects to make a triangle.
8. Therefore, ΔAB’C’ is the required triangle.
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 ΔAA7B’ and ΔAA5B,
∠A7AB’=∠A5AB (Common)
From the corresponding angles, we get,
∠A A7B’=∠A A5B
Therefore, from the AA similarity criterion, we obtain
ΔA A2B’ and A A3B
So, AB’/AB = AA5/AA7
Therefore, AB /AB’ = 5/7 ……. (2)
From the 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