Question

Construct a triangle of sides 4 cm, 5cm and 6cm and then a triangle similar to it whose sides are ${\displaystyle \frac{2}{3}}$ of the corresponding sides of the first triangle. Give the justification of the construction.

 

Construct a triangle with sides 5 cm, 4 cm and 6 cm. Then construct another triangle whose sides are ${\displaystyle \frac{2}{3}}$ times the corresponding sides of first triangle.

 

 

Answer 1

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

${\displaystyle AB\prime =\frac{2}{3}AB,B\prime C\prime =\frac{2}{3}BC,AC\prime =\frac{2}{3}AC}$

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 (Proved above)

∴ ΔAB'C' ~ ΔABC (AA similarity criterion)

${\displaystyle \Rightarrow \frac{AB\prime }{AB}=\frac{B\prime C\prime }{BC}=\frac{AC\prime }{AC}....\left(1\right)}$

In ΔAA₂B' and ΔAA₃B,

∠A₂AB' = ∠A₃AB (Common)

∠AA₂B' = ∠AA₃B (Corresponding angles)

∴ ΔAA₂B' ∼ ΔAA₃B (AA similarity criterion)

${\displaystyle \Rightarrow \frac{AB\prime }{AB}=(}$

${\displaystyle \Rightarrow \frac{AB\prime }{AB}=\frac{2}{3} ....\left(2\right)}$

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

${\displaystyle \frac{AB\prime }{AB}=\frac{B\prime C\prime }{BC}=\frac{AC\prime }{AC}=\frac{2}{3}}$

${\displaystyle \Rightarrow AB\prime =\frac{2}{3}\left(AB\right),B\prime C\prime =\frac{2}{3}\left(BC\right),AC\prime =\frac{2}{3}\left(AC\right)}$

This justifies the construction.