Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.
Answers
Explanation:
Construction Procedure:
1. Draw a ΔABC with base side BC = 6 cm, and AB = 5 cm and ∠ABC = 60°.
2. Draw a ray BX which makes an acute angle with BC on the opposite side of vertex A.
3. Locate 4 points (as 4 is greater in 3 and 4), such as B1, B2, B3, B4, on line segment BX.
4. Join the points B4C and also draw a line through B3, parallel to B4C intersecting the line segment BC at C’.
5. Draw a line through C’ parallel to the line AC which intersects the line AB at A’.
6. Therefore, ΔA’BC’ is the required triangle.
[ Refer to the attachment ]
Justification:
The construction of the given problem can be justified by proving that
Since the scale factor is 3/4 , we need to prove
↪A’B = (3/4)AB
↪BC’ = (3/4)BC
↪A’C’= (3/4)AC
From the construction, we get A’C’ || AC
In ΔA’BC’ and ΔABC,
∴ ∠ A’C’B = ∠ACB (Corresponding angles)
↪∠B = ∠B (common)
∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)
Since the corresponding sides of the similar triangle are in the same ratio, it becomes
↪Therefore, A’B/AB = BC’/BC= A’C’/AC
↪So, it becomes A’B/AB = BC’/BC= A’C’/AC = 3/4
Hence, justified!!
