Question 26

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.

 

This is a construction Question from the Maths Sample Paper .

Every year, 1 construction question is asked. This time, we have to just construct, and not justify.

Other times, you might be asked to justify the construction as well. Which you can look at in the Constructions - Class 10 Chapter.

The solution to this question is below

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Question 26 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 Join 𝐴_5B 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

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Davneet Singh
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.