Question

HOLA ,



$\huge \fbox \red{❥ Question}$


Draw a triangle ABC in which AB = 5 cm, BC = 6 cm and ∠ABC = 60°. Then construct a triangle whose sides are 57 times the corresponding sides of ∆ABC​

Answer 1

ടσlᥙtiσɳ:

The ΔA BC whose sides are 4/3

of the corresponding sides of ΔABC can be drawn as follows:

↛ Step 1: Draw a ΔABC with side BC=6cm,AB=5cm,∠ABC=60∘

↛ Step 2: Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

↛ Step 3: Locate 4 points, B¹ ,B² ,B ³,B4 on line segment BX.

↛ Step 4: Join B4 C and draw a line through B³ , parallel to B4 C intersecting BC at C

↛ Step 5: Draw a line through C parallel to AC intersecting AB at A

➢ The triangle A BC is the required triangle.