Ex 11.1, 4
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1 1/2 times the corresponding sides of the isosceles triangle.
Given base and altitude of an isosceles triangle.
We make the triangle by drawing its perpendicular bisector, as in isosceles triangle, altitude and perpendicular bisector are the same
 Draw a line segment AB of 8 cm. Draw arcs of same radius on both sides of the line segment while taking point A and B as its centre. Let these arcs intersect each other at O and O'. Join OO'. Let OO' intersect AB at D.

Taking D as centre, draw an arc of 4 cm radius which cuts the extended line segment OO' at point C.
An isosceles ΔABC is formed, having CD (altitude) 4 cm and AB (base) 8 cm.  Draw a ray AX making an acute angle with line segment AB on the opposite side of vertex C.
 Locate 3 points (as 3 is greater between 3 and 2) A_1, A_2, and A_3 on AX such that 〖AA〗_1 = A_1 A_2 = A_2 A_3.
 Join 〖BA〗_2 and draw a line through A_3 parallel to 〖BA〗_2 to intersect extended line segment AB at point B'.
 Draw a line through B' parallel to BC intersecting the extended line segment AC at C'.
ΔAB'C' is the required triangle.
Justification
Here,
(AB^′)/AB=(AA_3)/(AA_2 ) = 3/2
Also, A’C’ is parallel to AC
So, the will make the same angle with line BC
∴ ∠ AB’C’ = ∠ ABC
Now,
In Δ AB’C’ and Δ ABC
∠ A = ∠ A
∠ AB’C’ = ∠ ABC
Δ A’BC’ ∼ Δ ABC
Since corresponding sides of
similar triangles are in the same ratio
(AB^′)/AB=(B^′ C^′)/BC=(AC^′)/AC
So, (AB^′)/AB=(B^′ C^′)/BC=(AC^′)/AC =3/2
This justifies the construction.