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 times the corresponding sides of the given triangle. Give the justification of the construction.

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

In ΔABC and ΔAB'C',

∠ABC = ∠AB'C' (Corresponding angles)

∠BAC = ∠B'AC' (Common)

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

… (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)

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

⇒

This justifies the construction.

**
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