CBSE Class 10 Maths Revision Notes for Constructions

In ΔABA_m+n, we observe that A_mPis parallel to A_m+nB. Therefore, by Basic Proportionality theorem, we have:


Hence, P divides AB in the ratio m: n.

Alternative method to divide a line segment internally in a given ratio m: n

Example
Find the point C such that it divides BA in ratio 2:3



Steps of Construction:
1. Draw any ray XA making an acute angle with BA.
2. Draw a ray YB parallel to XA by making ∠YBA equal to ∠XAB.
3. Locate the points A₁, A₂, A₃ (m = 3) on AX and B₁, B₂ (n = 2) on BY such that
AA₁ = A₁A₂ = A₂A₃ = BB₁ = B₁B₂.
4. Join A3B2. Let it intersect AB at a point C
Then BC : CA = 2:3

Justification
Here ΔBB₂C ∼ΔAA₃C …AA test

The ratio of the sides of the triangle to be constructed with the corresponding sides of the given triangle is known as a scale factor. The scale factor may be less or greater than 1.

If the scale factor is less than 1, then the new figure will be smaller in comparison to the given figure.

If the scale factor is greater than 1, then the new figure will be bigger in comparison to the given figure.


Construction of Triangle Similar to given Triangle

Consider a triangle ABC. Let us construct a triangle similar to ΔABC such that each of its sides is of the corresponding sides of ΔABC.
Steps of constructions when m < n: 
Step 1: Construct the given triangle ABC by using the given data.
Step 2: Take any one of the three side of the given triangle as base. Let AB be the base of the given triangle.
Step 3: At one end, say A, of base AB. Construct an acute angle  ∠BAX below the base AB.
Step 4: Along AX mark off n points A₁, A₂, A₃,………, An such that
       AA₁ = A₁A₂ = ……… = A_n-1 A_n
Step 5: Join A_nB
Step 6: Draw A_mB’ parallel to A_nB which meets AB at B’.
Step 7: From B' draw B'C'||BC meeting AC at C'.

Triangle AB'C' is the required triangle each of whose sides is  of the corresponding side of ΔABC.

Justification

Since A_mB'||A_nB Therefore


In triangles ABC and AB’C’, we have

and ∠BAC = ∠B'AC'
∠ABC = ∠AB'C'
So, by AA similarity criterion, we have

ΔAB'C' ∼ ΔABC

Steps of construction when m > n:

Step 1: Construct the given triangle by using the given data.
Step 2: Take any one of the three sides of the given triangle and consider it as the base. Let AB be the base of the given triangle.
Step 3: At one end, say A, of base AB. Construct an acute angle ∠BAX below base AB i.e., on the opposite side of the vertex C.
Step 4: Along AX mark off m (large of m and n) points A₁, A₂, A₃,………Am of AX such that AA₁ = A₁A₂ = ………= A_m-1A_m.
Step 5: Join A_nB to B and draw a line through Am parallel to A_nB, intersecting the extended line segment AB at B'.
Step 6: Draw a line through B' parallel to BC intersecting the extended line segment AC at C'.
Step 7: ΔAB'C' so obtained is the required triangle.

Justification
Consider triangle ABC and AB’ C’. We have:
∠BAC = ∠B'AC'
∠ABC = ∠AB'C'
So, by AA similarity criterion,
ΔABC ∼ ΔAB'C'




The tangent to a circle is a line that intersects the circle at exactly one point.
Tangent to a circle is perpendicular to the radius through the point of contact.


Construction of Triangle to a Circle from a point outside the Circle

Construction of a tangent to a circle from a point outside the circle, when its centre is known

The steps of constructions are as follows:

Step 1: Join the centre O of the circle to the point P.

Step 2: Draw perpendicular bisector of OP intersecting OP at Q.

Step 3: With Q as centre and radius OQ, draw a circle. This circle has OP as its diameter.

Step 4: Let this circle intersect the first circle at two points T and T'. Join PT and P T' .

PT and P T' are the two tangents to the given circle from the point P.

Justification

Join OT and O T'
It can be seen that ∠PTO is an angle in the semi-circle. We know that angle in a semi-circle is a right angle.
∴ ∠PTO = 90°

⇒ OT ⊥ PT

Since OT is the radius of the circle, PT has to be a tangent of the circle. Similarly, PT' is a tangent of the circle.