1. Draw a line segment of length 7 cm. Find a point P on it which divides it in the
Tangents on the given circle can be drawn as follows.
Draw a circle of 4 cm radius with centre as O on the given plane.
Draw a circle of 6 cm radius taking O as its centre. Locate a point P on this circle and join OP.
Bisect OP. Let M be the mid-point of PO.
Taking M as its centre and MO as its radius, draw a circle. Let it intersect the given circle at the points Q and R.
Join PQ and PR. PQ and PR are the required tangents.
It can be observed that PQ and PR are of length 4.47 cm each.
Since PQ is a tangent,
∠PQO = 90°
PO = 6 cm
QO = 4 cm
Applying Pythagoras theorem in ΔPQO, we obtain
PQ2 + QO2 = PQ2
PQ2 + (4)2 = (6)2
PQ2 + 16 = 36
PQ2 = 36 − 16
PQ2 = 20
PQ = 4.47 cm
The construction can be justified by proving that PQ and PR are the tangents to the circle (whose centre is O and radius is 4 cm). For this, let us join OQ and OR.
∠PQO is an angle in the semi-circle. We know that angle in a semi-circle is a right angle.
∴ ∠PQO = 90°
⇒ OQ ⊥ PQ
Since OQ is the radius of the circle, PQ has to be a tangent of the circle. Similarly, PR is a tangent of the circle.
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