Question

QⲉSⲧⲓⲟⲛ:


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

Answer:

The ΔA′BC′ whose sides are 43 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,B3,B4 on line segment BX.

Step 4: Join B4C and draw a line through B3, parallel to B4C 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.

Answer 2

Sⲟlⲧⲓⲟⲛ:

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 ,B³ ,B4 on line segment BX.

Step 4: Join B 4 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.