Construct an Isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are \(1\frac{1}{2}\) times the corresponding sides of the Isosceles triangle.
Construction Procedure:
1. Draw a line segment BC with the measure of 8 cm.
2. Now draw the perpendicular bisector of the line segment BC and intersect at the point D
3. Take the point D as centre and draw an arc with the radius of 4 cm which intersect the perpendicular bisector at the point A
4. Now join the lines AB and AC and the triangle is the required triangle.
5. Draw a ray BX which makes an acute angle with the line BC on the side opposite to the vertex A.
6. Locate the 3 points B1, B2 and B3 on the ray BX such that BB1 = B1B2 = B2B3
7. Join the points B2C and draw a line from B3 which is parallel to the line B2C where it intersects the extended line segment BC at point C’.
8. Now, draw a line from C’ the extended line segment AC at A’ which is parallel to the line AC and it intersects to make a triangle.
9. Therefore, ΔA’BC’ is the required triangle.
The construction of the given problem can be justified by proving that
A’B = (3/2)AB
BC’ = (3/2)BC
A’C’= (3/2)AC
From the construction, we get A’C’ || AC
∴ ∠ A’C’B = ∠ACB (Corresponding angles)
In ΔA’BC’ and ΔABC,
∠B = ∠B (common)
∠A’BC’ = ∠ACB
∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)
Therefore, A’B/AB = BC’/BC= A’C’/AC
Since the corresponding sides of the similar triangle are in the same ratio, it becomes
A’B/AB = BC’/BC= A’C’/AC = 3/2