NCERT Solutions
Class 10 Maths
Constructions

Ex.11.1 Q.4
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm
and then another triangle whose side are 1 times the corresponding sides of the isosceles triangle.
Give the justification of the construction.
Let us assume that ∆ABC is an isosceles triangle having CA and CB of equal lengths, base AB of 8 cm, and AD is the altitude of 4 cm.
The ∆AB'C' whose sides are times of ∆ABC 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 ∆ABC 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) A1, A2, and A3 on AX such that:
AA1 = A1A2 = A2A3.
Step 5. Join BA2 and draw a line through A3 parallel to BA2 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'.
Now, ∆AB'C' is the required triangle.
Justification:
The construction can be justified by proving that
AB’ = AB, B’C’ =
BC, AC’ =
AC
In ∆ABC and ∆AB'C',
∠ABC = ∠AB'C'
(Corresponding angles)
∠BAC = ∠B'AC'
(Common)
So, ∆ABC ∼ ∆AB'C'
(AA similarity criterion)
=> AB ÷ AB’ = BC ÷ B’C’= AC ÷ AC’ …...... (1)
In ∆AA2B and ∆AA3B',
∠A2 AB = ∠A3 AB'
(Common)
∠AA2B = ∠AA3B'
(Corresponding angles)
So, ∆AA2B ∼ ∆AA3B'
(AA similarity criterion)
=> AB ÷ AB’ = AA2 ÷ AA3
=> AB ÷ AB’ = ………... (2)
On comparing equations (1) and (2), we obtain
AB ÷ AB’ = BC ÷ B’C’ = AC ÷ AC’ =
AB’ = AB, B’C’ =
BC, AC’ =
AC
This justifies the construction.