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Asked by Ashish Last Modified
Answer 
Abhay
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.
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