Construction, Class 10, Math Detailed Chapter Notes PDF Download

INTRODUCTION

In class IX, we have discussed a number of constructions with the help of ruler and compass e.g. bisecting a line segment, bisecting an angle, perpendicular bisector of line segment, some more constructions of triangles etc. with their justifications. In this chapter we will discuss more constructions by using the knowledge of the earlier construction.

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 PA₁₃B 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.

Therefore  (Basic Proportionality Therom)
By construction, 
Therefore 

This given that C divides AB in the ratio 5 : 8.

By measurement, we find, AC = 2.9 cm, CB = 4.7 cm.

By Calcuation

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

Then

Since by consteuction, 

By measurement : AC = 2.9 cm, BC = 4.7 cm.

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 ΔABC = 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.

CONSTRUCTION OF TANGENTS TO A CIRC LE

(a) If a point lies inside a circle, we can not draw any tangent to the circle i.e., No tangent is possible in this case.

If a point lies on the circle, then there is only one tangent to the circle at this point. The tangent to a circle at any point is perpendicular to the radius passing through the point of contact.

Two tangents are drawn from an external point to circle, they are equal in length.

Construction 4 : Draw a circle of radius 5 cm. From a point 8 cm away from its centre, construct pair of tangents to the circle measure their lengths.
Sol.

Steps of Construction :
Step-1 : Draw a circle with radius 5 cm whose centre is O.
Step-2 : Take a point P at a distance 8 cm from the centre O such that OP = 8cm.
Step-3 : Bisect the line segment OP at the point C such that OC =CP =4 cm.
Step-4 : Taking C as centre and OC as arc, draw a dotted circle to intersect the given circle at the points T and T'.
Step-5 : Join PT and PT'
PT and PT' are the required pair of tangents to the circle.
By measurement we obtain PT = PT' = 6.2 cm 

Construction 5. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.

Steps of Construction:
Step 1 : Draw two concentric circles with centre O and radii 4 cm and 6 cm such that OP = 6 cm, OQ = 4 cm.
Step 2 : Join OP and bisect it at M. i.e. M is the mid-point of OP i.e. OM = PM = 3 cm.
Step 3 : Taking M as centre with OM as radius draw a circle intersecting the smaller circle in two points namely T and S.
Step 4 : Join PT and PS.
PT and PS are the required tangents from a point P to the smaller circle, whose radius is 4 cm. By measurement: PT = 4.5 cm.
Verification. OTP is right D at T

Verification. OTP is right Δ at T
OP² = OT² + PT²
6² = 4² + PT² ⇒ PT² = 36 - 16 = 20