# 1. Draw a line segment of length 7 cm. Find a point P on it which divides it in theratio 3:5.2. Draw a right triangle ABC in which BC = 12 cm, AB = 5 cm and ∠B = 90°.Construct a triangle similar to it and of scale factor 23. Is the new trianglealso a right triangle?3. Draw a triangle ABC in which BC = 6 cm, CA = 5 cm and AB = 4 cm.Construct a triangle similar to it and of scale factor 53.4. Construct a tangent to a circle of radius 4 cm from a point which is at adistance of 6 cm from its centre

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Tangents on the given circle can be drawn as follows.

Step 1

Draw a circle of 4 cm radius with centre as O on the given plane.

Step 2

Draw a circle of 6 cm radius taking O as its centre. Locate a point P on this circle and join OP.

Step 3

Bisect OP. Let M be the mid-point of PO.

Step 4

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.

Step 5

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.

In ΔPQO,

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 PQ = 4.47 cm

Justification

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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