Question

Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are `7/5` of the corresponding sides of the first triangle. Give the justification of the construction.

Answer 1

Step 1

Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6 cm and 7 cm radius respectively. Let these arcs intersect each other at point C. ΔABC is the required triangle having length of sides as 5 cm, 6 cm, and 7 cm respectively.

Step 2

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

Step 3

Locate 7 points, A₁, A₂, A₃, A₄ A₅, A₆, A₇ (as 7 is greater between 5and 7), on line AX such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇.

Step 4

Join BA₅ and draw a line through A₇ parallel to BA₅ to intersect extended line segment AB at point B'.

Step 5

Draw a line through B' parallel to BC intersecting the extended line segment AC at C'. ΔAB'C' is the required triangle.

Justification

The construction can be justified by proving that

AB' = 7/5AB, B'C' = 7/5(BC), AC'=7/5 (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 ΔAA₅B and ΔAA₇B',

∠A₅AB = ∠A₇AB' (Common)

∠AA₅B = ∠AA₇B' (Corresponding angles)

∴ ΔAA₅B ∼ ΔAA₇B' (AA similarity criterion)

`=> (AB)/(AB') = (`

`=>(AB)/(AB')=5/7 ....(2)`

On comparing equations (1) and (2), we obtain

`(AB)/(AB')=(BC)/(B'C') = (AC)/(AC') = 5/7`

`=>AB' = 7/5 AB, B'C' =  7/5 BC, AC' = 7/5 AC`

This justifies the construction.

Answer 2

Given that

Construct a triangle of sides AB = 5cm, BC = 6cm and AC = 7cm and then a triangle similar to it whose sides are 7/5^th of the corresponding sides of ΔABC .

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment AB = 5cm.

Step: II- With A as centre and radius AC = 7cm, draw an arc.

Step: III- With B as centre and radius = BC = 6cm, draw an arc, intersecting the arc drawn in step II at C.

Step: IV- Joins AC and BC to obtain ΔABC.

Step: V- Below AB, makes an acute angle ∠BAX = 60°.

Step: VI- Along AX, mark off seven points A₁, A₂, A₃, A₄, A₅, A₆ and A₇ such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆A₇

Step: VII-Join A₅B.

Step: VIII- Since we have to construct a triangle each of whose sides is 7/5^th of the corresponding sides of ΔABC.

So, we draw a line A₃B' on AX from point A₇ which is A₇B' || A₅B and meeting AB at B'.

Step: IX- From B' point draw B'C' || BC, and meeting ACat C'

Thus, ΔAB'C' is the required triangle, each of whose sides is 7/5^th of the corresponding sides of ΔABC.