NCERT Solutions
Class 10 Maths
Constructions

Ex.11.1 Q.7
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm.
the construct another triangle whose sides are times the corresponding sides of the given triangle.
Give the justification of the construction.
It is given that sides other than hypotenuse are of lengths 4 cm and 3 cm.
Clearly, these will be perpendicular to each other. The required triangle can be drawn as follows:
Step 1. Draw a line segment AB = 4 cm. Draw a ray SA making 900 with it.
Step 2. Draw an arc of 3 cm radius while taking A as its centre to intersect SA at C.
Join BC. So, ∆ABC is the required triangle.
Step 3. Draw a ray AX making an acute angle with AB, opposite to vertex C.
Step 4. Locate 5 points (as 5 is greater in 5 and 3), A1, A2, A3, A4, A5,
on line segment AX such that: AA1 = A1 A2 = A2 A3 = A3 A4 = A4 A5.
Step 5. Join A3 B. Draw a line through A5 parallel to A3 B intersecting extended line segment AB at B'.
Step 6. Through B', draw a line parallel to BC intersecting extended line segment AC at C'.
So, ∆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 ∆AA3B and ∆AA5B',
∠A3AB = ∠A5AB'
(Common)
∠AA3 B = ∠AA5 B'
(Corresponding angles)
So, ∆AA3B ∼ ∆AA5B'
(AA similarity criterion)
=> AB ÷ AB’ = AA3 ÷ AA5
=> 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.