Q. 55.0( 2 Votes )
Construct a ΔABC whose perimeter is 15 cm and sides are in the ratio 3 : 4 : 5.
The sides of the triangle are in the ratio 3:4:5
Now, 3+4+5 = 12
Step 1: Construct PQ = 15cm
Step 2: We have to divide PQ into 12 equal parts and consider the 1st three, the next four and the last five seperately for construction
A line inclined with any arbitary angle with the line PQ is drawn with the help of scale and pencil.
12 equal parts are taken with the help of compass and after joining the end points of both the lines, parallel lines are drawn with the help of pencil and set squares.
The line PQ is thus equally divided and points B and C are named.
Step 3: Arcs with B as centre and PB as radius and C as centre and CD as radius are intersected at A.
A,B and A,C is joined to yield the required triangle.
∆ABC is the required triangle with AB:BC:AC = 3:4:5
PQ = 15cm
AB+BC+AC = 15cm
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