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Questions
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.
Construct a triangle with sides 5 cm, 4 cm and 6 cm. Then construct another triangle whose sides are `2/3` times the corresponding sides of first triangle.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are `2/3` of the corresponding sides of it.
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Solution
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
`AB' = 2/3AB, B'C' = 2/3BC, AC' = 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)
`=> (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) = (`
`=> (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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