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 90° with it.
Step 2
Draw an arc of 3 cm radius while taking A as its centre to intersect SA at C. Join BC. ΔABC 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 = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅
Step 5
Join A₃B. Draw a line through A₅ parallel to AB intersecting extended line segment AB at B’.
Step 6
Through B’, Draw a line parallel to BC intersecting extended line segment AC at C’.
ΔAB’C’ is the required triangle.
Justification
The construction can be justified by proving that AB’ = 5/3 AB, B’C’ = 5/3 BC, AC’ = 5/3 AC
In ΔABC and ΔAB’C’,
∠ABC = ∠AB’C’ (corresponding angles)
∠BAC = ∠B’AC’ (Common)
∴ ΔABC ∼ ΔAB’C’ (AA similarity criterion)
⇒ AB/AB’ = BC / B’C’ = AC/AC’ …(1)
In ΔAA₃B and ΔAA₅B’,
∠A₃AB = ∠A₅AB’ (common)
∠AA₃B = ∠AA₅B’ (corresponding angles)
∴ ΔAA₃B ∼ ΔAA₅B’ (AA similarity criterion)
⇒ AB/AB, = AA₃/AA₅ ⇒ AB/AB’ = 3/5 …(2)
On comparing equations (1) and (2), we obtain
AB’/AB = B’C’/BC = AC’/AC = 5/3
⇒ AB’ = 5/3 AB, B’C’ = 5/3 BC, AC’ = 5/3 AC
This justifies the construction.