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.

#### Solution 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, A_{1}, A_{2}, A_{3}, A_{4} A_{5}, A_{6}, A_{7} (as 7 is greater between 5and 7), on line AX such that AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5} = A_{5}A_{6} = A_{6}A_{7}.

**Step 4**

Join BA_{5} and draw a line through A_{7} parallel to BA_{5} 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 ΔAA_{5}B and ΔAA_{7}B',

∠A_{5}AB = ∠A_{7}AB' (Common)

∠AA_{5}B = ∠AA_{7}B' (Corresponding angles)

∴ ΔAA_{5}B ∼ ΔAA_{7}B' (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.

#### Solution 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 A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6} and A_{7} such that AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5} = A_{5}A_{6} = A_{6}A_{7}

Step: VII-Join A_{5}B.

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 A_{3}B' on *AX *from point A_{7} which is A_{7}B' || A_{5}B 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*.*