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Question
Divide a line segment of length 14 cm internally in the ratio 2 : 5. Also, justify your construction.
Solution
Given that
Determine a point which divides a line segment of length 14cm internally in the ratio of 2:5.
We follow the following steps to construct the given
Step of construction
Step: I-First of all we draw a line segment AB = 14cm.
Step: II- We draw a ray AX making an acute angle ∠BAX = 60° with AB.
Step: III- Draw a ray BY parallel to AX by making an acute angle ∠ABY = ∠BAX.
Step IV- Mark of two points A1, A2 on AX and three points B1, B2, B3, B4, B5 on BY in such a way that AAA1 = A1A2 = B1B2 = B2B3 = B3B4 = B4B5.
Step: V- Joins A2B3 and this line intersects AB at a point P.
Thus, P is the point dividing AB internally in the ratio of 2:5
Justification:-
In ΔAA2P and ΔBB5P we have
∠A2AP = ∠PBB5 [∠ABY = ∠BAX]
And ∠APA2 = ∠BPB5 [Vertically opposite angle]
So, AA similarity criterion, we have
ΔAA2P ≈ ΔBB5P
`(A A_2)/(BB_5)=(AP)(BP)`
`(AP)/(BP)=2/5`
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