Class 10 Maths Chapter 11 Constructions Notes

Constructions

Some Important Points

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: AA1,A1A2,A2A3,A3A4,A4A5,A5A6,A6A7,A7A8,A8A9,A9A10,A10A11,A11A12,A12A13
4: Join A13B
5: From A5, draw A5P∥A13B, 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 :
In ΔABA13, PA5∥BA13
therefore, ΔABA5∼ΔABA3
⇒APPB=AA5A5A13=58
⇒APPB=58

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 ΔABC in which BC=6 cm, AB= 5 cm, and AC= 4 cm, Draw a triangle similar to ΔABC with its sides equal to (2/3)th of the corresponding sides of ΔABC.Steps of Construction:

1: 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 ΔABC is the required triangle.
5: Below BC, make an acute angle CBX.
6: Along BX, mark off three points B1,B2 and B3 BB1=B1B2=B2B3
7: Join B3C
8: From B2, Draw B2D∥B3C, meeting BC at D.
9: From D, draw ED∥AC, meeting BA at E. Then,
EBD is the required triangle whose sides are (2/3)th of the corresponding sides of ΔABC

Justification :
Since DE∥CA
therefore, ΔABC∼ΔEBD
And EBAB=BDBC=DECA=23
Hence, we get the new triangle similar to the given triangle whose sides are equal to (2/3)th of the
corresponding sides of ΔABC

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 find PQ=PR=8

Justification :
On joining OQ, We find that ∠PQO=90∘, as ∠PQO is the angle in the semi-circle.
Therefore, PQ⊥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.

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 be M1 and M2.
4: With M1 as centre and M1P as a radius, draw circle to intersect the circle at T1 and T2.
5: join PT1 and PT2. Then PT1 and PT2 are required tangents,
similarly, the tangents QT3 and QT4 can be obtained.

Justification :
On joining  OT1 we find ∠PT1O=90∘ as an angle in semi circle.
Therefore, PT1⊥OT1
since OT1 is a radius of the given circle, so PT1 has to be a tangent to the circle.
similarly, PT2, QT3 and QT4 are also tangent to the circle