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Asked by Pradeep Last Modified
Answer 
Abhay
Let us assume that is an isosceles triangle having CA and CB of equal lengths, base AB of 8 cm, and AD is the altitude of 4 cm.
A whose sides are
times of can be drawn as follows.
Step 1
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.
Step 2
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) as 4 cm and AB (base) as 8 cm.
Step 3
Draw a ray AX making an acute angle with line segment AB on the opposite side of vertex C.
Step 4
Locate 3 points (as 3 is greater between 3 and 2) , and
on AX such that
.
Step 5
Join and draw a line through
parallel to
to intersect extended line segment AB at point B'.
Step 6
Draw a line through B' parallel to BC intersecting the extended line segment AC at C'. is the required triangle.
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