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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.
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