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. Construct AB a line segment of length 5 cm.
2 Draw the arcs of radius 6 cm and 5 cm taking A and B as the centre, respectively.
3. As these arcs will intersect each other at point C, therefore as a result ΔABC is the required triangle with the sides measuring as 5 cm, 6 cm, and 7 cm respectively.
4. Draw a ray AX that intersects the line segment AB and makes an acute angle on the other side of vertex C.
5. As 7 is greater between 5 and 7, on line AX locate the 7 points such as A1, A2, A3, A4, A5, A6, A7 , in such a way that it becomes AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7
6. Join the points BA5 and draw a line from point A7 to point BA5 that is parallel to the line BA5 where it intersects the extended line segment AB at point B’.
7. Draw a line which is parallel to the line BC and it intersects to make a triangle from B’ the extended line segment AC to C’.
8. As a result, ΔAB’C’ is the required triangle.
Justification:
It is possible to justify the construction of the given problem by proving that
AB’ = (7/5)AB
B’C’ = (7/5)BC
AC’= (7/5)AC
We get B’C’ || BC from the above construction
Therefore, ∠AB’C’ = ∠ABC as these are corresponding angles
Now in ΔAB’C’ and ΔABC,
∠ABC = ∠AB’C as proved above
∠BAC = ∠B’AC’ as these are common angles
Therefore, from the AA similarity criterion ΔAB’C’ ∼ ΔABC
As a result, AB’/AB = B’C’/BC= AC’/AC …. (1)
Now in ΔAA7B’ and ΔAA5B,
∠A7AB’=∠A5AB as these are common angles
We get, from the corresponding angles that,
∠A A5B = ∠A A7B’
Using the AA similarity criterion, we get
ΔA A2B’ and A A3B
Therefore, AA5/AA7 = AB’/AB
As a result, AB /AB’ = 5/7 ……. (2)
From the eq.(1) and eq.(2), we get
AB’/AB = B’C’/BC = AC’/ AC = 7/5
Now the above can be rewritten as
AB’ = (7/5)AB
B’C’ = (7/5)BC
AC’= (7/5)AC
As a result, the above construction is justified.