Draw a ∆ABC with sides 6 cm, 8 cm and 9 cm and then construct a triangle similar to ∆ABC whose sides are 3/5 of the corresponding sides of ∆ABC.
Let’s first construct Δ ABC with sides 6 cm, 8 cm, 9 cm
Steps to draw Δ ABC
Draw base AB of side 6 cm
With A as center, and 8 cm as radius, draw an arc
With B as center, and 9 cm as radius, draw an arc
3. Let C be the point where the two arcs intersect. Join AC & BC
Thus, Δ ABC is the required triangle
Now, let’s make a similar triangle with Scale factor = 3/5
Steps of construction
Draw any ray AX making an acute angle with AB
on the side opposite to the vertex C.
Mark 5 (the greater of 3 and 5 in 3/5 ) points
𝐴_1, 𝐴_2, 𝐴_3, 𝐴_4, 𝐴_5 on AX so that 〖𝐴𝐴〗_1=𝐴_1 𝐴_2=𝐴_2 𝐴_3=𝐴_3 𝐴_4=𝐴_4 𝐴_5
and draw a line through 𝐴_3 (the 3rd point, 3 being smaller of 3 and 5 in 3/5) parallel to 𝐴_5 𝐵,
to intersect AB at B′.
4. Draw a line through B′ parallel to the line BC to intersect AC at C′.
Thus, Δ AB’C′ is the required triangle
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.