Question

Divide a line segment of length 9 cm internally in the ratio 4 : 3. Also, give justification of the construction.

Answer 1

Given that

Determine a point which divides a line segment of length 9 cm internally in the ratio of 4:3.

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment AB = 9cm.

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, A₃, A₄ on AX and three points  B₁, B₂, B₃ on BY in such a way that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = BB₁ = B₁B₂ = B₂B₃.

Step: V- Joins A₄B₃  and this line intersects AB at a point P.

Thus, P is the point dividing AB internally in the ratio of 4:3

Justification:

In ΔAA₄P and ΔBB₃P we have

∠A₄AP = ∠PBB₃          [∠ABY = ∠BAX]

And ∠APA₄ = ∠BPB₃    [Vertically opposite angle]

So, AA similarity criterion, we have

ΔAA₄P ≈ ΔBB₃P

`(A A_4)/(BB_3)=(AP)/(BP)`

`(AP)/(BP)=4/3`