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 A1, A2,………Am-1, Am+1,………,Am+n, such that AA1 = A1A2 = Am+n-1 Am+n.
Step 4: Join BAm+n
Step 5: Through the point Am, draw a line parallel to Am+nB by making an angle equal toAAm+nB at Am, intersecting AB at point P.
The point P so obtained is the required point which divides AB internally in the ratio m: n.
In ΔABAm+n, we observe that AmPis parallel to Am+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.
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If the scale factor is less than 1, then the new figure will be smaller in comparison to the given figure.
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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 isof 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 = ……… = An-1 An
Step 5: Join AnB
Step 6: Draw AmB’ parallel to AnB 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 isof the corresponding side of ΔABC.
Justification
Since AmB'||AnB 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 A1, A2, A3,………Am of AX such that AA1 = A1A2 = ………= Am-1Am.
Step 5: Join AnB to B and draw a line through Am parallel to AnB, 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 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.
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