Question

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.

Answer 1

Let us assume that ΔABC is an isosceles triangle having CA and CB of equal lengths, base AB of 8 cm, and AD is the altitude of 4 cm.

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.

A ΔAB'C' whose sides aretimes of ΔABC can be drawn as follows.

step 3

Draw a line segment AB' = 1.5 * 8= 12cm along AB

step 4

Draw a line through B' and parallel to BC

step 5

Extend AC to meet the above line at C'

$△A{B}^{\prime }{C}^{\prime }$ is the required triangle