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Constructions : Exercise 11.2 (Mathematics NCERT Class 10th)


Q.1      In each of the following, give the justification of the construction :

1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Sol.         Step of Construction :

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                 1. Take a point O and draw a circle of radius 6 cm.
                 2. Mark a point P at a distance of 10 cm from the centre O.
                 3. Join OP and bisect it. Let M be its mid- point.
                 4. With M as centre and MP as radius, draw a circle to intersect the circle at Q and R.
                 5. Join PQ and PR. Then PQ and PR are the required tangents.
                  On measuring, we find PQ = PR = 8 cm.
                 Justification :
                 On joining OQ, we find that \angle PQO = {90^o}, as \angle PQO is the angle in the semi-circle.
                 Therefore, PQ \bot OQ
                 Since OQ is the radius of the given circle, so PQ has to be a tangent to the circle. Similarly, PR is also a tangent to the circle.


Q.2      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.
Sol.         Steps of Construction :-

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                 1. Take a point O and draw two concentric circles of radii 4 cm and 6 cm respectively.
                 2. Mark a point P on the circumference of the bigger circle.
                 3. Join OP and bisect it. Let M be its mid- point.
                 4. With M as centre and MP as radius, draw a circle to intersect the circle at Q and R.
                 5. Join PQ and PR. Then, PQ and PR are the required tangents.
                 On measuring, we find that PQ = PR = 4.8 cm (approx).
                 Justification : -
                 On joining OQ, we find that \angle PQO = {90^o}, as \angle PQO is the angle in the semi-circle.
                 Therefore, PQ \bot OQ
                 Since OQ is the radius of the given circle, so PQ has to be a tangent to the circle. Similarly, PR is also a tangent to the circle.


Q.3      Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Sol.        Steps of Construction :

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               1. Take a point O, draw a circle of radius 3 cm with this point as centre.
               2. Take two points P and Q on one of its extended diameter such that OP = OQ = 7 cm.
               3. Bisect OP and OQ. Let their respective mid- points be  {M_1}\,and\,{M_2}.
               4. With {M_1} as centre and {M_1}P as radius , draw a circle to intersect the circle at {T_1}\,and\,{T_2}
               5. Join P{T_1}\,and\,P{T_2}. Then P{T_1}\,and\,P{T_2} are the required tangents.
               Similarly, the tangents Q{T_3}\,and\,Q{T_4} can be obtained.
               Justification :
               On joining O{T_1} we find \angle P{T_1}O = {90^o}, as an angle in the semi- circle.
               Therefore, P{T_1} \bot \,O{T_1}
               Since O{T_1} is a radius of the given circle, so P{T_1} has to be a tangent to the circle.
               Similarly, P{T_2},\,Q{T_3}\,and\,Q{T_4} are also tangents to the circle.


Q.4      Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60º
Sol.         Steps of Construction :
               1. With O as centre of radius = 5 cm, draw a circle.
               2. Draw any diameter AOC.
               3. Draw a radius OL such that \angle COL = {60^o} (i.e., the given angle).

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                 4. At L draw LM \bot OL
                 5. At A, draw AN \bot OA
                 6. These two perpendiculars intersect each other at P. Then PA and PL are the required tangents.
                 Justification :
                 Since, OA is the radius , so PA has to be a tangent to the circle.
                 Similarly, PL is the tangent to the circle.
                 \angle APL = {360^o} - \angle OAP - \angle OLP - \angle AOL
                 = 360º – 90º – 90º – (180º – 60º)
                 = 360º – 360º + 60º = 60º
                 Thus tangents PA and PL are inclined to each other at an angle of 60º.


Q.5       Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Sol.         Steps of Construction :

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                1. Draw a line segment AB = 8 cm
                2. Taking A as centre , draw a circle of radius 4 cm and taking B as centre, draw a circle of radius 3 cm.
                3. Clearly, O is the centre of AB. With O as centre, draw a circle of radius OA or OB, intersecting circle with centre B at {T_1}\,and\,{T_2}, circle with centre A at {T_3}\,and\,{T_4}.
                4. Join A{T_1}\,,\,A{T_2},\,B{T_3}\,and\,B{T_4}. Then these are the required tangents.
                Justification :
                On joining B{T_1}, we find that, B{T_1}A = 90^\circ , as \angle B{T_1}A is the angle in the semi- circle.
                Therefore, A{T_1} \bot B{T_1}
                Since, B{T_1} is the radius of the given circle, so A{T_1} has to be a tangent to the circle.
                Similarly, A{T_2},B{T_3}\,and\,B{T_4} are the tangents.


Q.6      Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and \angle B = {90^o}. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Sol.        Steps of a Construction :

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               1. With the given data, draw a \Delta \,ABC, in which AB = 6 cm, BC = 8 cm and \angle B = {90^o}.
               2. Draw BD \bot AC.
               3. Draw perpendicular bisectors of BC and BD. They meet at a point O.
               4. Taking O as centre and OD as radius, draw a circle. This circle passes through B, C,D.
               5. Join OA.
               6. Draw perpendicular bisector of OA. This perpendicular bisector meets OA at M.
              7. Taking M as centre and MA as radius, draw a circle which intersects the circle drawn in step 4 at points P and Q.
               8. Join AP and AQ.
               These are the required tangents from A.


Q.7      Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
Sol.        Steps of Construction :

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             1. Draw a circle with the help of a bangle.
             2. Draw a secant ARS from an external point A. Produce RA to C such that AR = AC.
             3. With CS as diameter, draw a semi-circle.
             4. At the point A, draw AB \bot AS, cutting the semi- circle at B.
            5. With A as centre and AB as radius, draw an arc to intersect the given circle, in T and T'. Join AT and AT'. Then, AT and AT' are the required tangent lines.



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