Q.1 Â Â Â In each of the following, give the justification of the construction :
Â Â Â Â Â Â Â Â Â 1. Take a point O and draw a circle of radius 6 cm.
Â Â Â Â Â Â Â Â Â 2. Mark a point P at a distance of 10 cm from the centre O.
Â Â Â Â Â Â Â Â Â 3. Join OP and bisect it. Let M be its mid- point.
Â Â Â Â Â Â Â Â Â 4. With M as centre and MP as radius, draw a circle to intersect the circle at Q and R.
Â Â Â Â Â Â Â Â Â 5. Join PQ and PR. Then PQ and PR are the required tangents.
Â Â Â Â Â Â Â Â Â On measuring, we find PQ = PR = 8 cm.
Â Â Â Â Â Â Â Â Â Justification :
Â Â Â Â Â Â Â Â Â On joining OQ, we find that , as is the angle in the semi-circle.
Â Â Â Â Â Â Â Â Â Therefore,
Â Â Â Â Â Â Â Â Â 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 Â Â Â 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.Â Â Â Â Â Steps of Construction :-
Â Â Â Â Â Â Â Â Â 1. Take a point O and draw two concentric circles of radii 4 cm and 6 cm respectively.
Â Â Â Â Â Â Â Â Â 2. Mark a point P on the circumference of the bigger circle.
Â Â Â Â Â Â Â Â Â 3. Join OP and bisect it. Let M be its mid- point.
Â Â Â Â Â Â Â Â Â 4. With M as centre and MP as radius, draw a circle to intersect the circle at Q and R.
Â Â Â Â Â Â Â Â Â 5. Join PQ and PR. Then, PQ and PR are the required tangents.
Â Â Â Â Â Â Â Â Â On measuring, we find that PQ = PR = 4.8 cm (approx).
Â Â Â Â Â Â Â Â Â Justification : -
Â Â Â Â Â Â Â Â Â On joining OQ, we find that , as is the angle in the semi-circle.
Â Â Â Â Â Â Â Â Â Therefore,
Â Â Â Â Â Â Â Â Â 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 Â Â Â 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.Â Â Â Â Â Steps of Construction :
Â Â Â Â Â Â Â Â 1. Take a point O, draw a circle of radius 3 cm with this point as centre.
Â Â Â Â Â Â Â Â 2. Take two points P and Q on one of its extended diameter such that OP = OQ = 7 cm.
Â Â Â Â Â Â Â Â 3. Bisect OP and OQ. Let their respective mid- points beÂ .
Â Â Â Â Â Â Â Â 4. With as centre and as radius , draw a circle to intersect the circle at
Â Â Â Â Â Â Â Â 5. Join . Then are the required tangents.
Â Â Â Â Â Â Â Â Similarly, the tangents can be obtained.
Â Â Â Â Â Â Â Â Justification :
Â Â Â Â Â Â Â Â On joining we find , as an angle in the semi- circle.
Â Â Â Â Â Â Â Â Therefore,
Â Â Â Â Â Â Â Â Since is a radius of the given circle, so has to be a tangent to the circle.
Â Â Â Â Â Â Â Â Similarly, are also tangents to the circle.
Q.4 Â Â Â Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60Âº
Sol. Â Â Â Â Steps of Construction :
Â Â Â Â Â Â Â Â 1. With O as centre of radius = 5 cm, draw a circle.
Â Â Â Â Â Â Â Â 2. Draw any diameter AOC.
Â Â Â Â Â Â Â Â 3. Draw a radius OL such that (i.e., the given angle).
Â Â Â Â Â Â Â Â Â 4. At L draw
Â Â Â Â Â Â Â Â Â 5. At A, draw
Â Â Â Â Â Â Â Â Â 6. These two perpendiculars intersect each other at P. Then PA and PL are the required tangents.
Â Â Â Â Â Â Â Â Â Justification :
Â Â Â Â Â Â Â Â Â Since, OA is the radius , so PA has to be a tangent to the circle.
Â Â Â Â Â Â Â Â Â Similarly, PL is the tangent to the circle.
Â Â Â Â Â Â Â Â Â
Â Â Â Â Â Â Â Â Â = 360Âº â€“ 90Âº â€“ 90Âº â€“ (180Âº â€“ 60Âº)
Â Â Â Â Â Â Â Â Â = 360Âº â€“ 360Âº + 60Âº = 60Âº
Â Â Â Â Â Â Â Â Â Thus tangents PA and PL are inclined to each other at an angle of 60Âº.
Q.5 Â Â Â 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.Â Â Â Â Â Steps of Construction :
Â Â Â Â Â Â Â Â 1. Draw a line segment AB = 8 cm
Â Â Â Â Â Â Â Â 2. Taking A as centre , draw a circle of radius 4 cm and taking B as centre, draw a circle of radius 3 cm.
Â Â Â Â Â Â Â Â 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 , circle with centre A at .
Â Â Â Â Â Â Â Â 4. Join . Then these are the required tangents.
Â Â Â Â Â Â Â Â Justification :
Â Â Â Â Â Â Â Â On joining , we find that, , as is the angle in the semi- circle.
Â Â Â Â Â Â Â Â Therefore,
Â Â Â Â Â Â Â Â Since, is the radius of the given circle, so Â has to be a tangent to the circle.
Â Â Â Â Â Â Â Â Similarly, are the tangents.
Â Â Â Â Â Â Â Â 1. With the given data, draw a ABC, in which AB = 6 cm, BC = 8 cm and .
Â Â Â Â Â Â Â Â 2. Draw .
Â Â Â Â Â Â Â Â 3. Draw perpendicular bisectors of BC and BD. They meet at a point O.
Â Â Â Â Â Â Â Â 4. Taking O as centre and OD as radius, draw a circle. This circle passes through B, C,D.
Â Â Â Â Â Â Â Â 5. Join OA.
Â Â Â Â Â Â Â Â 6. Draw perpendicular bisector of OA. This perpendicular bisector meets OA at M.
Â Â Â Â Â Â Â 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.
Â Â Â Â Â Â Â Â 8. Join AP and AQ.
Â Â Â Â Â Â Â Â These are the required tangents from A.
Q.7 Â Â Â 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. Â Â Â Â Steps of Construction :
Â Â Â Â Â Â Â 1. Draw a circle with the help of a bangle.
Â Â Â Â Â Â Â 2. Draw a secant ARS from an external point A. Produce RA to C such that AR = AC.
Â Â Â Â Â Â Â 3. With CS as diameter, draw a semi-circle.
Â Â Â Â Â Â Â 4. At the point A, draw , cutting the semi- circle at B.
Â Â Â Â Â Â 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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