NCERT Solutions – Construction

Q.1      Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
Sol.      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,A11A12andA12A13
4. Join A13B
5. From A5,drawA5P||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, ΔAPA5  ~ ΔABA3
⇒ APPB=AA5A5A13=58
⇒ AP : PB = 5 : 8

Q.2      Construct a triangle of sides 4 cm, 5 cm and 6cm and then a triangle similar to it whose sides are 23 of the corresponding sides of it.
Sol.          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 Δ ABC is the required triangle.

5. Below BC, make an acute angle CBX.
6. Along BX, mark off three points : B1,B2andB3 such that BB1=B1B2=B2B3.
7. Join B3C.
8. From B2,drawB2D||B3C, meeting BC at D.
9. From D, draw ED || AC, meeting BA at E. Then,
EBD is the required triangle whose sides are 23rd of the corresponding sides of Δ ABC.
               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 23 rd of the corresponding sides of Δ ABC.

Q.3      Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 75 of the corresponding sides of the first triangle.
Sol.         Steps of construction :

1. With the given data, construct Δ ABC in which BC = 7 cm , CA = 6 cm and AB = 5 cm
2. Below BC, make an acute ∠CBX.
3. Along BX, mark off seven points : B1,B2,B3,B4,B5,B6andB7 such that     BB1=B1B2=B2B3=B3B4=B4B5=B5B6=B6B7
4. Join B5C
5. From B7drawB7D||B5C, meeting BC produced at D.
6. From D, draw DE || CA, meeting BA produced at E.
Then EBD is the required triangle whose sides are 75 th of the corresponding sides of ΔABC.
               Justification :
Since DE || CA
Therefore,  ΔABC∼ΔEBDandEBAB=BDBC=DECA=75
              Hence, we get the new triangle similar to the given triangle whose sides are equal to 75 the of the corresponding sides of ΔABC.

Q.4      Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 112 times the corresponding sides of the isosceles triangle.
Sol.        Steps of Construction :
              1. Draw BC = 8 cm
2. Construct PQ the perpendicular bisector of line segment BC meeting BC at M.
3. Along MP cut off MA = 4 cm
4. Join BA and CA. Then ΔABC so obtained is the required ΔABC
5. Extend BC to D, such that BD = 12 cm.
6. Draw DE || CA, meeting BA produced at E.
Then ΔEBD is the required triangle.

               Justification :
Since DE || CA
Therefore, ΔABC  ~ ΔEBD
and EBAB=DECA=BDBC=128=32
Hence, we get the new triangle similar to the given triangle whose sides are 32,, i.e., 112 times of the corresponding sides of the isosceles ΔABC.

Q.5      Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC=600. Then construct a triangle whose sides are 34 of the corresponding sides of the triangle ABC.
Sol.       Steps of Construction :

1. With the given data, construct ΔABC in which BC = 6 cm, ∠ABC=600 and AB = 5 cm.
2. Below BC, make an acute ∠CBX.
3. Along BX, mark off 4 points : B1,B2,B3andB4 such that BB1=B1B2=B2B3=B3B4.
4. Join B4C.
5. From D, draw ED || AC, meeting BA at E.Then, EBD is the required triangle whose sides are 34 th of the corresponding sides of ΔABC.
                Justification :
Since, DE || CA Therefore  ΔABC ~ ΔEBD
and, EBAB=BDBC=DECA=34
Hence, we get the new triangle similar to the given triangle whose sides are equal to 34th of the corresponding sides of ΔABC.

Q.6      Draw a triangle ABC with side BC = 7 cm, ∠B=45o,∠A=105o. Then construct a triangle whose sides are 43 times the corresponding sides of ΔABC.
Sol.        Steps of Construction :

1. With the given data, construct ΔABC in which BC = 7 cm ,
              ∠B=45o,∠C=180o−(∠A+∠B)
⇒ ∠C=180o−(105o+45o)
=180o−150o=30o
2. Below BC, make an acute ∠CBX
3. Along BX, mark off four points : B1,B2,B3andB4 such that  BB1=B1B2=B2B3=B3B4
4. Join B3C
5. From B4 draw B4D||B3C meeting BC produced at D.
6. From D, draw ED || AC, meeting BA produced at E.
Then EBD is the required triangle whose sides are 43 times of the corresponding sides of  ΔABC
             Justification :
Since, DE || CA Therefore, ΔABC ~ ΔEBD
and EBAB=BDBC=DECA=43
Hence, we get the new triangle similar to the given triangle whose sides are equal to 43 times of the corresponding sides of ΔABC

1. 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 lengths.
Sol.         Step 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 cm.
                 Justification :
On joining OQ, we find that ∠PQO=90o, 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.

Q.2      Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Sol.         Steps of Construction :-

1. Take a point O and draw two concentric circles of radii 4 cm and 6 cm respectively.
2. Mark a point P on the circumference of the bigger circle.
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 that PQ = PR = 4.8 cm (approx).
                 Justification : –
On joining OQ, we find that ∠PQO=90o, 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.

Q.3      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.
Sol.        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  M1andM2.
4. With M1 as centre and M1P as radius , draw a circle to intersect the circle at T1andT2
5. Join PT1andPT2. Then PT1andPT2 are the required tangents.
Similarly, the tangents QT3andQT4 can be obtained.
               Justification :
On joining OT1 we find ∠PT1O=90o, as an angle in the 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,QT3andQT4 are also tangents to the circle.

Q.4      Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60º
Sol.         Steps of Construction :
               1. With O as centre of radius = 5 cm, draw a circle.
2. Draw any diameter AOC.
3. Draw a radius OL such that ∠COL=60o (i.e., the given angle).

4. At L draw LM⊥OL
5. At A, draw AN⊥OA
6. These two perpendiculars intersect each other at P. Then PA and PL are the required tangents.
                 Justification :
Since, OA is the radius , so PA has to be a tangent to the circle.
Similarly, PL is the tangent to the circle.
∠APL=360o−∠OAP−∠OLP−∠AOL
= 360º – 90º – 90º – (180º – 60º)
= 360º – 360º + 60º = 60º
Thus tangents PA and PL are inclined to each other at an angle of 60º.

Q.5       Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Sol.         Steps of Construction :

1. Draw a line segment AB = 8 cm
2. Taking A as centre , draw a circle of radius 4 cm and taking B as centre, draw a circle of radius 3 cm.
3. Clearly, O is the centre of AB. With O as centre, draw a circle of radius OA or OB, intersecting circle with centre B at T1andT2, circle with centre A at T3andT4.
4. Join AT1,AT2,BT3andBT4. Then these are the required tangents.
                Justification :
On joining BT1, we find that, BT1A=90∘, as ∠BT1A is the angle in the semi- circle.
Therefore, AT1⊥BT1
Since, BT1 is the radius of the given circle, so AT1 has to be a tangent to the circle.
Similarly, AT2,BT3andBT4 are the tangents.

Q.6      Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B=90o. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Sol.        Steps of a Construction :

1. With the given data, draw a ΔABC, in which AB = 6 cm, BC = 8 cm and ∠B=90o.
2. Draw BD⊥AC.
3. Draw perpendicular bisectors of BC and BD. They meet at a point O.
4. Taking O as centre and OD as radius, draw a circle. This circle passes through B, C,D.
5. Join OA.
6. Draw perpendicular bisector of OA. This perpendicular bisector meets OA at M.
7. Taking M as centre and MA as radius, draw a circle which intersects the circle drawn in step 4 at points P and Q.
8. Join AP and AQ.
These are the required tangents from A.

Q.7      Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
Sol.        Steps of Construction :

1. Draw a circle with the help of a bangle.
2. Draw a secant ARS from an external point A. Produce RA to C such that AR = AC.
3. With CS as diameter, draw a semi-circle.
4. At the point A, draw AB⊥AS, cutting the semi- circle at B.
5. With A as centre and AB as radius, draw an arc to intersect the given circle, in T and T’. Join AT and AT’. Then, AT and AT’ are the required tangent lines.