CBSE Class 10 Maths Revision Notes for Constructions

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Constructions

  1. 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 to angleAAm+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:
    begin mathsize 12px style fraction numerator A A subscript m over denominator A subscript m A subscript m plus n end subscript end fraction equals fraction numerator A P over denominator P B end fraction
rightwards double arrow fraction numerator A P over denominator P B end fraction equals m over n open square brackets because fraction numerator A A subscript m over denominator A subscript m A subscript m plus n end subscript end fraction equals m over n comma space b y space c o n s t r u c t i o n close square brackets
rightwards double arrow A P colon P B equals m colon n end style

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

  2. 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

    begin mathsize 12px style BB subscript 2 over AA subscript 3 equals BC over AC... open parentheses straight c. straight p. straight s. straight t. close parentheses
2 over 3 equals BC over AC end style

     

  3. 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.

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

  5. 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 begin mathsize 12px style open parentheses straight m over straight n close parentheses to the power of th end styleof 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 is begin mathsize 12px style open parentheses straight m over straight n close parentheses to the power of th end style of the corresponding side of ΔABC.

    Justification

    Since AmB'||AnB Therefore
    begin mathsize 12px style fraction numerator A B italic apostrophe over denominator B italic apostrophe B end fraction italic equals fraction numerator A A subscript m over denominator A subscript m A subscript n end fraction
space space space space space space space space space space space space space space space space space space space space space space open square brackets by space basic space proportionality space theorem close square brackets
rightwards double arrow fraction numerator A B italic apostrophe over denominator B italic apostrophe B end fraction equals fraction numerator m over denominator n minus m end fraction
rightwards double arrow fraction numerator B italic apostrophe B over denominator A B apostrophe end fraction equals fraction numerator n minus m over denominator m end fraction
N o w space fraction numerator A B over denominator A B apostrophe end fraction equals fraction numerator A B apostrophe plus B apostrophe B over denominator A B apostrophe end fraction
rightwards double arrow space fraction numerator A B over denominator A B apostrophe end fraction equals 1 plus fraction numerator B apostrophe B over denominator A B apostrophe end fraction equals 1 plus fraction numerator n minus m over denominator m end fraction equals n over m
rightwards double arrow fraction numerator A B apostrophe over denominator A B end fraction equals m over n end style

    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
    begin mathsize 12px style rightwards double arrow fraction numerator A B apostrophe over denominator A B end fraction equals fraction numerator B apostrophe C apostrophe over denominator B C end fraction equals fraction numerator A C apostrophe over denominator A C end fraction
rightwards double arrow fraction numerator A B apostrophe over denominator A B end fraction equals fraction numerator B apostrophe C apostrophe over denominator B C end fraction equals fraction numerator A C apostrophe over denominator A C end fraction equals m over n end style


    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 = ………= 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'

    begin mathsize 12px style rightwards double arrow fraction numerator A B over denominator A B apostrophe end fraction equals fraction numerator B C over denominator B apostrophe C apostrophe end fraction equals fraction numerator A C over denominator A C apostrophe end fraction
I n space increment A space A subscript m space B apostrophe comma space A subscript n space B parallel to A subscript m space B apostrophe
therefore space fraction numerator A B over denominator B B apostrophe end fraction equals fraction numerator A A subscript n over denominator A subscript n A subscript m end fraction
rightwards double arrow fraction numerator B B apostrophe over denominator A B end fraction equals fraction numerator A subscript n A subscript m over denominator A A subscript n end fraction
rightwards double arrow fraction numerator B B apostrophe over denominator A B end fraction equals fraction numerator m minus n over denominator n end fraction
rightwards double arrow fraction numerator A B apostrophe minus A B over denominator A B end fraction equals fraction numerator m minus n over denominator n end fraction
rightwards double arrow negative 1 equals fraction numerator m minus n over denominator n end fraction
rightwards double arrow fraction numerator A B apostrophe over denominator A B end fraction equals m over n
From space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space have
fraction numerator AB apostrophe over denominator AB end fraction equals fraction numerator straight B apostrophe straight C apostrophe over denominator BC end fraction equals fraction numerator AC apostrophe over denominator AC end fraction equals straight m over straight n end style



    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°

    OTPT

    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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