A line segment of length 7.6 cm can be divided in the ratio of 5:8 as follows.
Step 1 Draw line segment AB of 7.6 cm and draw a ray AX making an acute angle with line segment AB.
Step 2 Locate 13 (= 5 + 8) points, A1, A2, A3, A4 …….. A13, on AX such that AA1 = A1A2 = A2A3 and so on.
Step 3 Join BA13.
Step 4 Through the point A5, draw a line parallel to BA13 (by making an angle equal to ∠AA13B) at A5intersecting AB at point C.
C is the point dividing line segment AB of 7.6 cm in the required ratio of 5:8.
The lengths of AC and CB can be measured. It comes out to 2.9 cm and 4.7 cm respectively.
Justification
The construction can be justified by proving that
`(AC)/(CB) = 5/8`
By construction, we have A5C || A13B. By applying Basic proportionality theorem for the triangle AA13B, we obtain
`(AC)/(CB) =(`
From the figure, it can be observed that AA5 and A5A13 contain 5 and 8 equal divisions of line segments respectively
`:. (`
On comparing equations (1) and (2), we obtain
`(AC)/(CB) = 5/8`
This justifies the construction
Draw a line segment of length 7 cm. Find a point P on it which divides it in the ratio 3:5.
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