# Chapter Notes: Constructions CBSE Class 10

## 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: $A{A_1},{A_1}{A_2},{A_2}{A_3},{A_3}{A_4},{A_4}{A_5},{A_5}{A_6},{A_6}{A_7},{A_7}{A_8},{A_8}{A_9},{A_9}{A_{10}},{A_{10}}{A_{11}},{A_{11}}{A_{12}},{A_{12}}{A_{13}}$
4: Join ${A_{13}}B$
5: From ${A_5}$, draw ${A_5}P\parallel {A_{13}}B$, 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 $\Delta AB{A_{13}}$$P{A_5}\parallel B{A_{13}}$

therefore, $\Delta AB{A_5} \sim \Delta AB{A_3}$
$\Rightarrow {{AP} \over {PB}} = {{A{A_5}} \over {{A_5}{A_{13}}}} = {5 \over 8}$
$\Rightarrow {{AP} \over {PB}} = {5 \over 8}$

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 $\Delta ABC$ in which BC=6 cm, AB= 5 cm, and AC= 4 cm, Draw a triangle similar to $\Delta ABC$ with its sides equal to ${(2/3)^{th}}$ of the corresponding sides of $\Delta ABC$. 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 $\Delta ABC$ is the required triangle.
5: Below BC, make an acute angle CBX.
6: Along BX, mark off three points ${B_1},{B_2}$ and ${B_3}$ $B{B_1} = {B_1}{B_2} = {B_2}{B_3}$
7: Join ${B_3}C$
8: From ${B_2}$, Draw ${B_2}D\parallel {B_3}C$, meeting $BC$ at D.
9: From D, draw $ED\parallel AC$, meeting BA at E. Then,
EBD is the required triangle whose sides are ${(2/3)^{th}}$ of the corresponding sides of $\Delta ABC$

Justification :
Since $DE\parallel CA$
therefore, $\Delta ABC \sim \Delta EBD$
And ${{EB} \over {AB}} = {{BD} \over {BC}} = {{DE} \over {CA}} = {2 \over 3}$
Hence, we get the new triangle similar to the given triangle whose sides are equal to ${(2/3)^{th}}$ of the
corresponding sides of $\Delta ABC$

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 $\angle PQO = {90^ \circ }$, as $\angle PQO$ is the angle in the semi-circle.
Therefore, $PQ \bot OQ$.
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 ${M_1}$ and ${M_2}$.
4: With ${M_1}$ as centre and ${M_1}P$ as a radius, draw circle to intersect the circle at ${T_1}$ and ${T_2}$.
5: join $P{T_1}$ and $P{T_2}$. Then $P{T_1}$ and $P{T_2}$ are required tangents,
similarly, the tangents $Q{T_3}$ and $Q{T_4}$ can be obtained.

Justification :
On joining  $O{T_1}$ we find $\angle P{T_1}O = {90^ \circ }$ as an angle in semi circle.
Therefore, $P{T_1} \bot O{T_1}$
since $O{T_1}$ is a radius of the given circle, so $P{T_1}$ has to be a tangent to the circle.
similarly, $P{T_2}$$Q{T_3}$ and $Q{T_4}$ are also tangent to the circle

• Anonymous

Nice

• Nice

• S.B. Verma

Nice for short revision of Construction.

• neha chunduri

pretty good notes........ worth studying for exams!!!!! 🙂

• Vaishnavi

It's nice

• shivang vivek

good notes thanks

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