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**1) To divide a line segment in a given ratio.**

*Construction:*

Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.**Steps of Construction:**

1: Draw a line segment AB = 7.6 cm

2: Draw a ray AC making any acute angle with AB, as shown in the figure.

3: On ray AC, starting from A, mark 5 + 8 = 13 equal line segments:

4: JoinÂ

5: FromÂ , drawÂ , meeting AB at P.

6: Thus, P divides AB in the ratio 5:8.

On measuring the two parts, we find AP = 2.9 cm and PB = 4.7 cm (approx).

**Justification :
**InÂ ,Â

therefore,Â

**2) To construct a triangle similar to a given triangle as per given scale factor which may be less than or may be greater than 1.**

*Construction:*

Draw a in which BC=6 cm, AB= 5Â cm, and AC= 4Â cm, Draw a triangle similar to with its sides equal to of the corresponding sides of .**Steps of Construction:
**

1:Â Draw a line segment BC = 6 cm

2: With B as centre and radius equal to 5 cm, draw an arc.

3: With C as centre and radius equal to 4 cm, draw an arc intersecting the previously drawn arc at A.

4: Join AB and AC, then is the required triangle.

5: Below BC, make an acute angle CBX.

6: Along BX, mark off three pointsÂ andÂ Â

7: JoinÂ

8: FromÂ , Draw , meeting at D.

9: From D, draw , meeting BA at E. Then,

EBD is the required triangle whose sides areÂ Â of the corresponding sides ofÂ

**Justification :
**SinceÂ

therefore,Â

AndÂ

Hence, we get the new triangle similar to the given triangle whose sides are equal toÂ Â of the

corresponding sides ofÂ

**3) To construct tangent at a point on a given circle.**

*Construction:*

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 length.**Steps of Construction:**

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 findÂ

**Justification :
**On joining OQ, We find thatÂ , asÂ is the angle in the semi-circle.

Therefore,Â .

Since OQ is the radius of the given circle, so PQ has to be a tangentÂ to the circle. Similarly, PR isÂ also a tangent to the circle.

**4) To construct the pair of tangents from an external point to a circle.**

*Construction:*

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.**Steps of Construction:**

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 beÂ andÂ .

4:Â WithÂ Â as centre andÂ as a radius, draw circle to intersect the circle atÂ andÂ .

5: joinÂ andÂ . ThenÂ andÂ are required tangents,

similarly, the tangentsÂ andÂ can be obtained.

**Justification :
**On joining Â we findÂ as an angle in semi circle.

Therefore,Â

since is a radius of the given circle, soÂ has to be a tangent to the circle.

similarly,Â ,Â andÂ are also tangent to the circle

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Nice

Nice

Nice for short revision of Construction.

pretty good notes........ worth studying for exams!!!!! ðŸ™‚

It's nice

good notes thanks