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Draw an isosceles  \Delta ABC in which BC = 5.5 cm and altitude AL = 3 cm .

Then construct  another triangle whose  sides are 3/4 of the corresponding sides of \Delta ABC

 

 

 

 
 
 
 
 

Answers (1)

1) Draw a line BC = 5.5 cm 

2) Draw \perp bisector of BC 

3 ) mark a 3 cm arc on the bisector and call the point of introduction A 

4) Join AB and AC 

5) \Delta ABC is formed 

6) make a ray BX making an acute angle with line BC 

7) take scale factor 3/4 < 1 

mark 4 points on BX (B _ 1 , B _2 , B _3 , B_4)

such that  BB _ 1 =B_1 B _2 = B_2 B _3 = B_3B_4

9) Join B_4 C

10 ) Make B_3 C '|| B_ 4 C  such that it intersects BC at C' 

11 ) Make A'C'  || AC such that it intersect AB at A ' 

12) \Delta A'B'C' is formed which is 

\frac{3}{4} \Delta ABC

 

Posted by

Ravindra Pindel

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