Question 2:

Construct a triangle of sides 4 cm, 5cm and 6cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle. Give the justification of the construction.

Answer:

Step 1

Draw a line segment AB = 4 cm. Taking point A as centre, draw an arc of 5 cm radius. Similarly, taking point B as its centre, draw an arc of 6 cm radius. These arcs will intersect each other at point C. Now, AC = 5 cm and BC = 6 cm and ΔABC is the required triangle.

Step 2

Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.

Step 3

Locate 3 points A₁, A₂, A₃ (as 3 is greater between 2 and 3) on line AX such that AA₁ = A₁A₂ = A₂A₃.

Step 4

Join BA₃ and draw a line through A₂ parallel to BA₃ to intersect AB at point B’.

Step 5

Draw a line through B’ parallel to the line BC to intersect AC at C’.

ΔAB’C’ is the required triangle.

 

Justification

The construction can be justified by proving that

By construction, we have B’C’ || BC

∴ ∠AB’C’ = ∠ABC (Corresponding angles)

In ΔAB’C’ and ΔABC,

∠AB’C’ = ∠ABC (Proved above)

∠B’AC’ = ∠BAC (Common)

∴ ΔAB’C’ ∼ ΔABC (AA similarity criterion)

 ⇒ AB’/AB = B’C’/BC = AC’/AC ……….  (1)

In ΔAA₂B’ and ΔAA₃B,

∠A₂AB’ = ∠A₃AB (Common)

∠AA₂B’ = ∠AA₃B (Corresponding angles)

∴ ΔAA₂B’ ∼ ΔAA₃B (AA similarity criterion)

⇒ AB’/AB = AA₂/AA₃

⇒ AB’/AB = 2/3  ………………..  (2)

From equations (1) and (2), we obtain

AB’/AB = B’C’/BC = AC’/AC = 2/3

⇒ AB’ = 2/3 AB, B’C’ = 2/3 BC , AC’ = 2/3 AC

This justifies the construction.

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