
01. Real Numbers
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Lecture1.1

Lecture1.2

Lecture1.3

Lecture1.4

Lecture1.5

Lecture1.6

Lecture1.7

Lecture1.8

Lecture1.9


02. Polynomials
11
Lecture2.1

Lecture2.2

Lecture2.3

Lecture2.4

Lecture2.5

Lecture2.6

Lecture2.7

Lecture2.8

Lecture2.9

Lecture2.10

Lecture2.11


03. Linear Equation
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Lecture3.1

Lecture3.2

Lecture3.3

Lecture3.4

Lecture3.5

Lecture3.6

Lecture3.7

Lecture3.8

Lecture3.9


04. Quadratic Equation
8
Lecture4.1

Lecture4.2

Lecture4.3

Lecture4.4

Lecture4.5

Lecture4.6

Lecture4.7

Lecture4.8


05. Arithmetic Progressions
11
Lecture5.1

Lecture5.2

Lecture5.3

Lecture5.4

Lecture5.5

Lecture5.6

Lecture5.7

Lecture5.8

Lecture5.9

Lecture5.10

Lecture5.11


06. Some Applications of Trigonometry
7
Lecture6.1

Lecture6.2

Lecture6.3

Lecture6.4

Lecture6.5

Lecture6.6

Lecture6.7


07. Coordinate Geometry
17
Lecture7.1

Lecture7.2

Lecture7.3

Lecture7.4

Lecture7.5

Lecture7.6

Lecture7.7

Lecture7.8

Lecture7.9

Lecture7.10

Lecture7.11

Lecture7.12

Lecture7.13

Lecture7.14

Lecture7.15

Lecture7.16

Lecture7.17


08. Triangles
15
Lecture8.1

Lecture8.2

Lecture8.3

Lecture8.4

Lecture8.5

Lecture8.6

Lecture8.7

Lecture8.8

Lecture8.9

Lecture8.10

Lecture8.11

Lecture8.12

Lecture8.13

Lecture8.14

Lecture8.15


09. Circles
8
Lecture9.1

Lecture9.2

Lecture9.3

Lecture9.4

Lecture9.5

Lecture9.6

Lecture9.7

Lecture9.8


10. Areas Related to Circles
10
Lecture10.1

Lecture10.2

Lecture10.3

Lecture10.4

Lecture10.5

Lecture10.6

Lecture10.7

Lecture10.8

Lecture10.9

Lecture10.10


11. Introduction to Trigonometry
7
Lecture11.1

Lecture11.2

Lecture11.3

Lecture11.4

Lecture11.5

Lecture11.6

Lecture11.7


12. Surface Areas and Volumes
9
Lecture12.1

Lecture12.2

Lecture12.3

Lecture12.4

Lecture12.5

Lecture12.6

Lecture12.7

Lecture12.8

Lecture12.9


13. Statistics
12
Lecture13.1

Lecture13.2

Lecture13.3

Lecture13.4

Lecture13.5

Lecture13.6

Lecture13.7

Lecture13.8

Lecture13.9

Lecture13.10

Lecture13.11

Lecture13.12


14. Probability
9
Lecture14.1

Lecture14.2

Lecture14.3

Lecture14.4

Lecture14.5

Lecture14.6

Lecture14.7

Lecture14.8

Lecture14.9


15. Construction
7
Lecture15.1

Lecture15.2

Lecture15.3

Lecture15.4

Lecture15.5

Lecture15.6

Lecture15.7

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,drawA5PA13B, 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,PA5BA13
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,drawB2DB3C, 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 B7drawB7DB5C, 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 B4DB3C 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 semicircle.
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 semicircle.
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 semicircle.
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.
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