Question

Draw a right triangle in which the size and length 4 centimetre and 8 centimetre then construct another triangle whose sides are 5 by 3 times the corresponding sides of a given triangle

Answer 1

here is your answer OK dude

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), A1, A2, A3, A4, A5, on line segment AX such that AA1 = A1A2 = A2A3 = A3A4 = A4A5.

Step 5

Join A3B. Draw a line through A5 parallel to A3B 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 ΔAA3B and ΔAA5B',

∠A3AB = ∠A5AB' (Common)

∠AA3B = ∠AA5B' (Corresponding angles)

∴ ΔAA3B ∼ ΔAA5B' (AA similarity criterion)



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



⇒

This justifies the construction.

Answer 2

here it is ...

hope it helps ...