Ex 11.1, 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.
- Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6 cm and 7 cm radius respectively. Let these arcs intersect each other at point C. ΔABC is the required triangle having length of sides as 5 cm, 6 cm, and 7 cm respectively.
- Draw a ray AX making acute angle with line AB on the opposite side of vertex C.
- Mark 7 points, A_1, A_2, A_3, A_4 A_5, A_6, A_7 (as 7 is greater between 5and 7), on line AX such that 〖AA〗_1 = A_1 A_2 = A_2 A_3= A_3 A_4= A_4 A_5= A_5 A_6= A_6 A_7.
- Join BA5 and draw a line through A7 parallel to BA5 to intersect extended line segment AB at point B'.
- Draw a line through B’ parallel to BC intersecting the extended line segment AC at C’.
ΔAB’C’ is the required triangle.
(AB^′)/AB=(AA_7)/(AA_5 ) = 7/5
Also, B’C’ is parallel to BC
So, the will make the same angle with line AB
∴ ∠ AB’C’ = ∠ ABC
In Δ AB’C’ and Δ ABC
∠ A = ∠ A
∠ AB’C’ = ∠ ABC
Δ AB’C’ ∼ Δ ABC
Since corresponding sides of
similar triangles are in the same ratio
So, (AB^′)/AB=(B^′ C^′)/BC=(AC^′)/AC=7/5
This justifies the construction.