CBSE Class 10 Maths Revision Notes for Constructions
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 To divide a line segment internally in a given ratio m: n, where both m and n are positive integers, we follow the steps given below:
Step 1: Draw a line segment AB of given length by using a ruler.
Step 2: Draw any ray AX making an acute angle with AB.
Step 3: Along AX mark off (m + n) points A_{1}, A2,………A_{m1}, Am+1,………,Am+n, such that AA_{1} = A_{1}A_{2} = A_{m+n1} A_{m+n}.
Step 4: Join BA_{m+n}
Step 5: Through the point Am, draw a line parallel to A_{m+n}B by making an angle equal to AA_{m+n}B at A_{m}, intersecting AB at point P.
The point P so obtained is the required point which divides AB internally in the ratio m: n.
In ΔABA_{m+n, }we observe that A_{m}Pis parallel to A_{m+n}B. 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 A_{1}, A_{2}, A_{3} (m = 3) on AX and B_{1}, B_{2} (n = 2) on BY such that
AA_{1} = A_{1}A_{2} = A_{2}A_{3} = BB_{1} = B_{1}B_{2}.
4. Join A3B2. Let it intersect AB at a point C
Then BC : CA = 2:3
Justification
Here ΔBB_{2}C ∼ΔAA_{3}C …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 A_{1}, A_{2}, A_{3},………, An such that
AA_{1} = A_{1}A_{2} = ……… = A_{n1} A_{n}Step 5: Join A_{n}B
Step 6: Draw A_{m}B’ parallel to A_{n}B 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_{m}B'A_{n}B 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' ∼ ΔABCSteps 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 A_{1}, A_{2}, A_{3},………Am of AX such that AA_{1} = A_{1}A_{2} = ………= A_{m1}A_{m}.
Step 5: Join A_{n}B to B and draw a line through Am parallel to A_{n}B, 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 CircleConstruction 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 semicircle. We know that angle in a semicircle 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.
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