Question

draw a triangle ABC with side BC=6cm,Ab=5cm and and angle ABC =60 . then construct a triangle whose side are 3/4 of the corresponding sides of the triangle ABC

Answer 1

Draw a line segment of 6cm.then from point b draw an angle of 60 degree and then from the same point b draw an arc of 5 cm.join the point where the arc cuts the line coming from b to c. Now from b make an acute angle downwards and then cut the line into four parts. From the fourth point join it to c then at the 3rd point make fourths line's parallel line. Then where the coming from 3ed line cuts line bc make parallel line of that line made at the point where line coming from 4th point. Now ur required triangle is ready.

Answer 2

Construction Procedure:

1. Draw a line segment AB which measures 4 cm, i.e., AB = 4 cm.

2. Take the point A as centre, and draw an arc of radius 5 cm.

3. Similarly, take the point B as its centre, and draw an arc of radius 6 cm.

4. The arcs drawn will intersect each other at point C.

5. Now, we obtained AC = 5 cm and BC = 6 cm and therefore ΔABC is the required triangle.

6. Draw a ray AX which makes an acute angle with the line segment AB on the opposite side of vertex C.

7. Locate 3 points such as A1, A2, A3 (as 3 is greater between 2 and 3) on line AX such that it becomes AA1= A1A2 = A2A3.

8. Join the point BA3 and draw a line through A2which is parallel to the line BA3 that intersect AB at point B’.

9. Through the point B’, draw a line parallel to the line BC that intersect the line AC at C’.

10. Therefore, ΔAB’C’ is the required triangle.

hope this helps u :)