Division of a Line Segment in a Given Ratio - Constructions, CBSE, Class 10, Mathematics PDF Download

DIVISION OF A LINE SEGMENT

Let us divide the given line segment in the given ratio say 5 : 8. This can be done in the following two ways:
(i) Use of Basic Proportionality Theorem.
(ii) Constructing a triangle similar to a given triangle.

Construction-1: Draw a segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.

 
Steps of Construction :
Step 1 : Draw any ray AX making an angle of 30° with AB.
Step 2 : Locate 13 points : A₁, A₂, A₃, A₄, A₅, A₆, A₇, A₈, A₉, A₁₀, A₁₁, A₁₂ and A₁₃ So that: AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄ A₅ =......= A₁₁ A₁₂ = A₁₂ A₁₃
Step 3 : Join B with A₁₃.
Step 4 : Through the point A₅, draw a line A₅C ║ A₁₃B such that ∠AA₅C = corr. ∠AA₁₃B intersecting AB at a point C. Then AC : CB = 5 : 8.
Let us see how this method gives us the required division.
Since A₅C is parallel to A₁₃ B.

This given that C divides AB in the ratio 5 : 8.
By measurement, we find, AC = 2.9 cm, CB = 4.7 cm.


Alternative Solution

Step 1 : Draw a line segment AB = 7.6 cm and to be divided in the ratio 5 : 8.
Step 2 : Draw any ray AX making an angle of 30° with AB.
Step 3 : Draw a ray BY parallel to AX by making ∠ABY equal to ∠BAX. i.e. ∠ABY = corr. ∠BAX.

Step 4 : Locate the points A₁, A₂, A₃, A₃, A₄, A₅ on AX and B₁, B₂, B₃, B₄, B₅, B₆, B₇, and B₈ on BY such that:

AA₁ = A₁A₂ = .............. = A₄A₅ = BB₁ = B₁B₂ = .................... = B₆B₇ = B₇B₈·
Step 5 : Join A₅B₈. Let it intersect AB at a point C. Then AC : CB = 5 : 8.
Here ΔAA₅C is similar to ΔBB₈C

Construction-2 : Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle.

Sol. First of all we are to construct a triangle ABC with given sides, AB = 6 cm, BC = 7 cm, CA = 5 cm.

Given a triangle ABC, we are required to construct a triangle whose sides are 7/5 of the corresponding sides of ΔABC

Steps of Construction :
Step 1 : Draw any ray BX making an angle of 30° with the base BC of ΔABC on the opposite side of the vertex A.

Step 2 : Locate seven points B₁, B₂, B₃, B₄, B₅, B₆ and B₇ on BX so that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇·
[Note that the number of points should be greater of m and n in the scale factor m/n.]

Step 3 : Join B₅ (the fifth point) to C and draw a line through B₇ parallel to B₅C, intersecting the extended line segment BC at C'.
Step 4 : Draw a line through C' parallel to CA intersecting the extended line segment BA at A'.
Then, A'BC' is the required triangle.

For justification of the construction.

Construction-3 : Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ÐABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.

Sol. Given a triangle ABC, we are required to construct another triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.

Steps of Construction :
Step 1 : Draw a line segment BC = 6 cm.
Step 2 : At B construct ∠CBY = 60° and cut off AB = 5 cm, join AB and AC. ABC is the required Δ.
Step 3 : Draw any ray BX making an acute angle say 30° with BC on the opposite side of the vertex A, ∠CBX = 30° downwards.
Step 4 : Locate four (the greater of 3 and 4 in 3/4) points B₁, B₂, B₃ and B₄ on BX, so that BB₁ = B₁B₂ = B₂B₃ = B₃B₄.
Step 5 : Join B₄C and draw a line through B₃ (the 3rd point) parallel to B₄C to intersect BC at C'.
Step 6 : Draw a line through C' parallel to the line CA to intersect BA at A'.

Then A'BC' is the required triangle whose each side is 3/4 times the corresponding sides of the ΔABC. Let us now see how this construction gives the required triangle.
For justification of the construction.