Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. the construct another triangle whose sides are `5/3` times the corresponding sides of the given triangle. Give the justification of the construction.

#### Solution

It is given that sides other than hypotenuse are of lengths 4 cm and 3 cm. Clearly, these will be perpendicular to each other.

The required triangle can be drawn as follows.

**Step 1**

Draw a line segment AB = 4 cm. Draw a ray SA making 90° with it.

**Step 2**

Draw an arc of 3 cm radius while taking A as its centre to intersect SA at C. Join BC. ΔABC is the required triangle.

**Step 3**

Draw a ray AX making an acute angle with AB, opposite to vertex C.

**Step 4**

Locate 5 points (as 5 is greater in 5 and 3), A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, on line segment AX such that AA_{1} = A_{1}A_{2}= A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5}.

**Step 5**

Join A_{3}B. Draw a line through A_{5} parallel to A_{3}B intersecting extended line segment AB at B'.

**Step 6**

Through B', draw a line parallel to BC intersecting extended line segment AC at C'. ΔAB'C' is the required triangle.

**Justification**

The construction can be justified by proving that

`AB' =5/3 AB, B'C' = 5/3 BC, AC' = 5/3 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_{3}B and ΔAA_{5}B',

∠A_{3}AB = ∠A_{5}AB' (Common)

∠AA_{3}B = ∠AA_{5}B' (Corresponding angles)

∴ ΔAA_{3}B ∼ ΔAA_{5}B' (AA similarity criterion)

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

`=>(AB)/(AB') = 3/5 ....2`

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

`(AB)/(AB') = (BC)/(B'C') = (AC)/(AC') = 3/5`

`=> AB' =5/3 AB, B'C' = 5/3 BC, AC' = 5/3 AC`

This justifies the construction.