Exercise 9.1 class 10.

Exercise 9.1

1. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30 degree (see Fig. 9.11).

Fig. 9.11

Solution:

From question it is clear that

Height of the pole AB =?

AC = 20 m

<C = 30⁰

We know that

sinC = AB/AC

Therefore, 

AB = AC×sinC

Now putting the value of AC and <C

or, AB = 20×sin30⁰

or, AB = 20×1/2

So, AB = 10 m Ans.

2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30 degree with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Solution:

From question it is clear that

BC = 8 m

Height of the tree = AB = AD + DB?

AD = DC

<C = 30⁰

Consider in right triangle DBC.

cosC = BC/DC

or, DC = BC/cos30⁰

or, DC = BC×cos30⁰

or, DC = 8×√3/2

or, DC = 4√3 m.

Again, consider in right triangle DBC.

tanC = DB/BC

or, DB = BC×tanC

Now, putting the value of BC and C.

DB = 8×tan30

or, DB = 4/√3

Height of the tree = DC + DB

or, AB = 4√3 + 4/√3

or, AB = 16/√3 m Ans.

3. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Solution:

From question it is clear that

Step slide for elder children

AB = 3 m.

AC = ?

∠C = 60⁰

Consider in right triangle ∆ABC.

sin60⁰ = AB/AC

∴ AC = AB/sin60⁰

Now, putting the value of AB and sin60⁰

∵ AC = 3×2/√3

AC = 2√3 m.

Step slide for the children below the age of 5 yrs.

DE = 1.5 m.

EF = ?

∠F = 30⁰

Consider in right triangle ∆DEF.

sin30⁰ = DE/EF

∴ EF = DE/sin30⁰

Now, putting the value of DE and sin30⁰

EF = 1.5×2 = 3 m.

Required answers

2√3 m & 3 m Ans.

4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Solution:

From question it is clear that

Height of the tower AB = ?.

BC = 30 m.

∠C = 30⁰

Consider in right ΔABC.

tan30⁰ = AB/BC

∴ AB = BC×tan30⁰

Now, putting the value of BC & tan30⁰

∴ AB = 30×1/√3

AB = 10√3 m Ans.

5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Solution:

From question and figure it is clear that

AB = Height of kite from the ground = 60 m.

∠C = 60⁰

AC = Length of string = ?

Now, consider in the right triangle ABC.

sin60⁰ = AB/AC

or, AC = AB/sin60⁰

Now, putting the value of AC and sin60⁰.

or, AC = 60/√3/2

or, AC = 60×2/√3

or, AC = 40√3 m Ans.

6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Solution:

From question and figure it is clear that

AB = Height of the building = 30 m.

Height of the boy = DE = 1.5 m.

∠D = 30⁰

∠F = 60⁰

He walked from D to F = DF =  ?

Now, consider in right Δ.ACD

tan30⁰ = AC/CD

∴ CD = AC/tan30⁰

Now, putting the value of AC & tan30⁰.

AC = AB - BC = AB - DE = 30 - 1.5 = 28.5

∴ CD = 28.5×√3

Now, consider in right ΔACF.

tan60⁰ = AC/CF

∴ CF = AC/tan60⁰

Now, putting the value of AC & tan60⁰.

∴ CF = 28.5/√3

DF = CD - CF

28.5√3 - 28.5/√3

or, DF = 28.5(3 - 1)/√3

or, DF = 28.5×2/√3

or, DF = 57/√3

or, DF = 19×3/√3

∴ DF = 19√3 m Ans.

6. 1.5 m लंबा एक लड़का 30 m ऊँचे एक भवन से कुछ दूरी पर खड़ा है।  जब वह ऊँचे भवन की ओर जाता है तब उसकी आँख से भवन के शिखर का उन्नयन कोण 30° से 60° हो जाता है। बताइए कि वह भवन की ओर कितनी दूरी तक चलकर गया है।

हल:

प्रश्न एवं चित्र से स्पष्ट है कि

AB = भवन की ऊंचाई = 30 मीटर।

DE = लड़के की ऊंचाई = 1.5 मीटर।

∠D = 30⁰

∠F = 60⁰

अब समकोण ∆ ACD में विचार करने पर

tan30⁰ = AC/CD

∴ CD = AC/tan30⁰

अब AC तथा tan30⁰ का मान रखने पर

CD = AC = (AB - BC)/1/√3

or, CD = (30 - 1.5)√3

or, CD = 28.5√3

इसी प्रकार समकोण ∆ ACF में विचार करने पर।

tan60⁰ = AC/CF

CF एवं tan60⁰ का मान रखने पर।

CF = AC/tan60⁰

or, CF = 28.5/√3

DF = CD - CF

DF = 28.5√3 - 28.5/√3

or, DF = 28.5(3 - 1)/√3

or, DF = 28.5×2/√3

or, DF = 57/√3

or, DF = 19×3/√3

or, DF = 19√3 m Ans.

7. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Solution:

From question and figure it is clear that

AD = Height of the tower =?

DB = Height of the building = 20 m.

∠ACB = 60⁰

∠DCB = 45⁰

Consider in right ΔABC.

tan60⁰ = AB/BC

Now, putting the values of tan60⁰.

∴AB = √3BC

Now, consider in right ΔDBC.

tan45⁰ = DB/BC

Now, putting the value of DB and tan45⁰.

∴ 1 = 20/BC.

∵ BC = 20 m.

Now, putting the value of BC to get the value of AB.

∴ AB = BC√3

∵ AB = 20√3

Height of the tower  = AD = AB - DB

∴ AD = 20√3 - 20 = 20(√3 - 1) 

= 20(1.732 -1) = 20×.732 = 14.640 m Ans.

7. भूमि के एक बिंदु से एक 20 m ऊँचे भवन के शिखर पर लगी एक संचार मीनार के तल और शिखर के उन्नयन कोण क्रमशः 45° और 60° है। मीनार की ऊँचाई ज्ञात कीजिए।

हल:

प्रश्न एवं चित्र से सपष्ट है कि

AD = मीनार की ऊंचाई जिसे हमे ज्ञात करना है।

DB = भवन की ऊंचाई = 20 मीटर।

∠ACB = 60°

∠DCB = 45°

अब समकोण ∆ABC में विचार करने पर

AB = BC×tan60° ------ (i)

अब समकोण ∆DBC में विचार करने पर

BC = DB×cot45°

अब DB एवं cot45° का मान रखने पर

BC = 20×1

BC = 20 मीटर।

अब BC का मान (i) में रखने पर

AB = 20√3 मीटर।

चित्र से स्पष्ट है कि मीनार की ऊंचाई

AD = AB – DB

or, AD = 20√3 –20

or, AD = 20(√3 –1)

or, AD = 20(1.732 –1)

or, AD = 20×0.732

∴AD = 14.64 मीटर।

अतः संचार मीनार की ऊंचाई = 14.64 मीटर उत्तर।

8. A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Solution;

From question and figure it is clear that

AD = Height of the statue = 1.6 metre.

DB = Height of the pedestal = ?

<ACB = 60°

<DCB = 45°

In right ∆ABC

BC = AB×cot60°

In right ∆DBC

BC = DB×cot45°

∴ AB×cot60° = DB×cot45°

or, (AD +DB)/√3 = DB

or, √3DB - DB = √3AD

DB = √3×1.6/(√3 -1)

Now, rationalise, the denominator of RHS.

DB = 0.8(3 + √3)

or, DB = 0.8×4.732

3.7856 metre Answer.

8. एक पेडस्टल के शिखर पर एक 1.6 m ऊँची मूर्ति लगी है। भूमि के एक बिंदु से मूर्ति के शिखर का उन्नयन कोण 60° है और उसी बिंदु से पेडस्टल के शिखर का उन्नयन कोण 45°  है। पेडस्टल की ऊँचाई ज्ञात कीजिए।

हल:

प्रश्न एवं चित्र से स्पष्ट है कि

AD = मूर्ति कि ऊंचाई = 1.6 मीटर।

<ACB = 60°

<DCB = 45°

DB = पेडस्टल कि ऊंचाई = ?

अब समकोण ∆ABC से

BC = AB×cot 60° …….. (i)

अब पुनः समकोण ∆DBC से

BC = DB×cot 45° ……… (ii)

अब (i) तथा (ii) से

AB×cot 60° = DB×cot45°

हम जानते है कि 

cot 45° = 1

cot 60° = 1/√3

AB = AD + DB = 1.6 + DB

(1.6 + DB)/√3 = DB

1.6 + DB = √3DB

DB(√3 - 1) = 1.6

DB = 1.6/(√3 - 1)

अब दाहिने तरफ के हर का परिमेयकरण करने पर

DB = 1.6(√3 + 1)/2

DB = 0.8×4.732 (√3 = 1.732)

DB = 3.7856 मीटर उत्तर।

9. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Solution:

From question and figure it is clear that

AB = height of the tower = 50 metre.

∠BCA = 60°

CD = Height of the building = ?

∠DAC = 30°

Consider in right ∆BAC.

AC = AB×cot60° ………. (i)

Consider in right ∆DCA.

AC = CD×cot30° ………. (ii)

Now, from (i) & (ii).

CD = AB×cot60°/cot30°

Now, putting the value.

CD = 50/√3×√3

CD = 50/3

CD = 16.67 metre Answer.

9. एक मीनार के पाद-बिंआकृति   से एक भवन के शिखर का उन्नयन कोण 30º है और भवन के पाद-बिंदु से मीनार के शिखर का उन्नयन कोण 60º है। यदि मीनार 50m ऊँची हो, तो भवन की ऊँचाई ज्ञात कीजिए।

हल:

प्रश्न एवं चित्र से स्पष्ट है कि

AB = मीनार की ऊंचाई = 50 मीटर।

CD = भवन की ऊंचाई = ?

∠BCA = 60°

∠DAC = 30°

अब समकोण ∆BAC से

AC = AB×cot 60° ……….. (i)

अब समकोण ∆DAC से

AC = CD×cot 30° ……….. (ii)

अब (i) तथा (ii) से

AB×cot 60° = CD×cot 30°

CD = AB×cot 60°/cot 30°

CD = 50/√3×√3

CD = 50/3

CD = 16.67 मीटर उत्तर।

10. Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

Solution:

From question and figure it is clear that

AB = CD = Height of the pole

BC = Width of the road = 80 metre.

BE + EC = 80 metre.

∠AEB = 30°

∠DEC = 60°

Consider in right ∆ABE.

BE = AB×cot30° ……. (i)

Now, consider in right ∆DCE.

EC = CD×cot60° ……. (ii)

Now from (i) & (ii).

AB×cot30° + CD×cot60° = 80

AB(√3 + 1/√3) = 80 {AB = CD}

4AB = 80√3

∴ AB = 20√√3 metre.

BE = AB×cot30°

BE = 20√3×√3 = 60 metre.

EC = 80 - BE = 80 - 60 = 20 mette.

Required Answer.

20√3 metre, 60 metre, 20 metre Answer.

10. एक 80 m चौड़ी सड़क के दोनों ओर आमने-सामने समान लंबाई वाले दो खंभे लगे हुए हैं। इन दोनों खंभों के बीच सड़क के एक बिंदु से खंभों के शिखर के उन्नयन कोण क्रमशः 60° और 30° है। खंभों की ऊँचाई और खंभों से बिंदु की दूरी ज्ञात कीजिए।

हल:

प्रश्न एवं चित्र से स्पष्ट है कि

AB = CD = खंभे की ऊंचाई।

BE + EC = BC = 80 सड़क की चौडाई।

∠AEB = 30°

∠DEC = 60°

समकोण ∆ABE में विचार करने पर।

BE = AB×cot 30° ……… (i)

अब समकोण ∆DEC में विचार करने पर।

EC = CD×cot 60° ……… (ii)

अब (i) तथा (ii) से।

AB×cot30° + CD×cot60° = 80

AB(cot30° + cot60°) = 80

AB(√3 + 1/√3) = 80

4AB = 80√3

AB = 20√3

BE = AB×cot30°

BE = 20√3×√3

BE = 60 मीटर।

EC = 80 - BE = 80 - 60 = 20 मीटर।

अतः अभीष्ट उत्तर = 20√3 मीटर, 60 मीटर एवं 20 मीटर।

11. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joing this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.

from question and figire it is clear that

AB = height of the tower = ?

∠ACB =60°

∠ADC = 30°

DC = 20 metre.

CB = ?

Cosider in right ∆ADC.

DB = AB×cot30°

DB = AB×√3. ……… (i)

Again from right ∆ACB.

CB = AB×cot60° ……. (ii)

CB = AB/√3

Now, from (i) and (ii)

DB - CB = 20

AB√3 - AB/√3 = 20

2AB = 20√3

AB =10√3 metre.

CB = AB/√3

CB = 10 metre Answer.

Required Answers

10√3, 10 metre Ans.

11. एक नहर के एक तट पर एक टीवी टॉवरऊर्ध्वाधरतः खड़ा है। टॉवर के ठीक सामने दूसरे तट के एक अन्य बिंदु से टॉवर के शिखर का उन्नयन कोण 60° है। इसी तट पर इस बिंदु से 20 m दूर और इस बिंदु को मीनार के पाद से मिलाने वाली रेखा पर स्थित एक अन्य बिंदुु से टॉवर के शिखर का उन्नयन कोण 30° है। (देखिए आकृति 9.12)। टॉवर की ऊँचाई और नहर की चौड़ाई ज्ञात कीजिए।

आकृति 9.12

प्रश्न और चित्र से स्पष्ट है कि

AB = मीनार की ऊंचाई = ?

∠ACB =60°

∠ADC = 30°

DC = 20 metre.

CB = ?

समकोण ∆ADC में.

DB = AB×cot30°

DB = AB×√3. ……… (i)

पुनः समकोण त्रिभज ∆ACB.

CB = AB×cot60° ……. (ii)

CB = AB/√3

अब  (i) तथा  (ii)

useDB - CB = 20

AB√3 - AB/√3 = 20

2AB = 20√3

AB =10√3 metre.

CB = AB/√3

CB = 10 metre Answer.

अभिष्ट उत्तर

10√3, 10 metre Ans.

12. 7 m ऊँचे भवन के शिखर से एक केबल टॉवर के शिखर का उन्नयन कोण 60° है और इसके पाद का अवनमन कोण 45º है। टॉवर की ऊँचाई ज्ञात कीजिए।

हल:

प्रश्न एवं चित्र से स्पष्ट है कि

CD = भवन की ऊंचाई = 7 मीटर।

∠ADE = 60°

∠BDE = 45°= ∠CBD

AB = टावर कि उंचाई = AE + BE = AE + CD = AE+7

DE = BC।

अब समकोण त्रिभुज DCB में विचार करने पर।

BC = CD×cot 45°

BC = 7×1 = 7 = DE।

अब समकोण त्रिभुज AED में विचार करने पर।

AE = DE×tan 60°

AE = BC×√3

AE = 7√3 मीटर।

टावर कि ऊंचाई = AB = AE + 7 = 7√3 + 7।

AB = 7(√3 +1)

AB = 7×4.732 = 33.124 मीटर।

12. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Solution:

From question and figure it is clear that

CD = Height of the building = 7 metre.

∠ADE = 60°

∠BDE = 45°= ∠CBD

AB = Height of the tower = AE + BE = AE + CD = AE+7

DE = BC.

Now, consider in right triangle DCB.

BC = CD×cot 45°

BC = 7×1 = 7 = DE।

Now, consider in right triangle AED.

AE = DE×tan 60°

AE = BC×√3

AE = 7√3 metre.

Height of tower = AB = AE + 7 = 7√3 + 7.

AB = 7(√3 +1)

AB = 7×4.732 = 33.124 metre Answer.

13.समुद्र-तल से 75 m ऊँची लाइट हाउस के शिखर से देखने पर दो समुद्री जहाजों के अवनमन कोण 30° और 45° हैं। यदि लाइट हाउस के एक ही ओर एक जहाज दूसरे जहाज के ठीक पीछे हो तो दो जहाजों के बीच की दूरी ज्ञात कीजिए।

हल:

प्रश्न एवं चित्र से स्पष्ट है की

CD = लाइट हाउस की ऊंचाई = 75 मीटर।

∠EDA = ∠DAC = 30°

∠EDB = ∠DBC = 45°

AB = जहाजों के बीच की दूरी = ?

समकोण त्रिभुज DCA में

AC = DC×cot 30°

BC = DC×cot 45°

AB = AC - AB

AB = DC×cot 30° - DC×cot 45°

AB = 75√3 - 75

AB = 75(√3 -1)

AB = 75(1.732 -1)

AB = 75×.732

AB = 54.90 मीटर Answer.

13. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Solution:

From question and figure it is clear that

CD = Height of light house = 75 metre.

∠EDA = ∠DAC = 30°

∠EDB = ∠DBC = 45°

AB = Distance between the ship = ?

In right triangle DCA.

AC = DC×cot 30°

In right triangle DCB.

BC = DC×cot 45°

AB = AC - AB

AB = DC×cot 30° - DC×cot 45°

AB = 75√3 - 75

AB = 75(√3 -1)

AB = 75(1.732 -1)

AB = 75×.732

AB = 54.90 metre Answer.

14. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see Fig. 9.13). Find the distance travelled by the balloon during the interval.

Fig. 9.13

Solution:

From question and figure it is clear that

Distance travelled by the balloon during the interval = (88.2 -1.2)(cot30° - cot60°)

= 87×(√3 - 1/√3)

= 87(3 -1)/√3

= 87×2/√3

= 29×2×√3

= 54√3 metre Ans.

15. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Solution:

From question and figure it is clear that

Car moves from A to B in 6 seconds.

AC = CD×cot30°

BC = CD×cot60°

AB = AC - BC

Speed of the car = AB÷6 unit per second.

Speed of the car = CD(cot30° - cot60°)÷6

Time taken to move the car from A to C.

= AC÷Speed of the car.

= 6×(CD×cot30°)÷CD(cot30° - cot60°)

= 6×√3÷{√3 -(1÷√3)}

= 6×√3×√3÷(3 - 1)

= 18÷2 = 9 seconds.

15. एक सीधा राजमार्ग एक मीनार के पाद तक जाता है। मीनार के शिखर पर खड़ा एक आदमी एक कार को 30° के अवनमन कोण पर देखता है जो कि मीनार के पाद की ओर एक समान चाल से जाता है। छः सेकेंड बाद कार का अवनमन कोण 60° हो गया। इस बिंदु से मीनार के पाद तक पहुँचने में कार द्वारा लिया गया समय ज्ञात कीजिए।

हल:

प्रश्न एवं चित्र से स्पष्ट है कि

कार 6 सेकेंड में A से B तक जाती है।

AC = CD×cot30°

BC = CD×cot60°

AB = AC - BC

कार का चाल = AB÷6 ईकाई प्रति सेकेंड।

कार का चाल = CD(cot30° - cot60°)÷6

कार को A से C तक जाने में लगा समय.

= AC÷कार का चाल.

= 6×(CD×cot30°)÷CD(cot30° - cot60°)

= 6×√3÷{√3 -(1÷√3)}

= 6×√3×√3÷(3 - 1)

= 18÷2 = 9 सेकेंड.

Number Systems

1. Number Systems

Class : IX

Subject: Mathematics

NCERT Text book Solution.

EXERCISE 1.1

1. Is zero a rational number? Can you write it in the form p/q, where p and q are integers and q ≠ 0?

Solution:

Yes zero is a rational number.

It can be written in the form p/q which is 0/1.

2. Find six rational numbers between 3 and 4.

Solution:

Both 3 and 4 are multiplied and divided by (6 + 1 = 7)

3 = 3×7/7 = 21/7

4 = 4×7/7 = 28/7

Required rational numbers are

22/7, 23/7, 24/7, 25/7, 26/7 and 27/7 Ans.

3. Find five rational numbers between ⅗ and ⅘.

Solution:

Both ⅗ and ⅘ multiplied and divided by (5+1=6)

⅗ = 18/30

⅘ = 24/30

Required rational numbers are

19/30, 20/30, 21/30, 22/30 and 23/30 Ans.

4. State whether the following statements are true or false. Give reasons for your answers.

(i) Every natural number is a whole number.

(ii) Every integer is a whole number.

(iii) Every rational number is a whole number.

Solution:

(i)

True, Natural numbers are a part of whole numbers.

(ii)

False, Negative integers are not whole numbers.

(iii)

False, ½ is a rational number but not a whole number.

प्रश्नावली 1.1

1. क्या शून्य एक परिमेय संख्या है? क्या इसे आप p/q के रूप में लिख सकते हैं, जहाँ p और q पूर्णांक हैं और q ≠ 0 है?

हल:

हां, शून्य एक परिमेय संख्या है। इसे p/q के रूप में लिखा जा सकता है जो 0/1 है।

2. 3 और 4 के बीच में छः परिमेय संख्याएँ ज्ञात कीजिए।

हल:

3 और 4 दोनों के अंश एवं हर को 7 से गुना करने पर।

3 = 21/7

4 = 28/7

अतः 6 परिमेय संख्या है

22/7, 23/7, 24/7, 25/7, 26/7 एवं 27/7।

3. ⅗ और ⅘ के बीच पाँच परिमेय संख्याएँ ज्ञात कीजिये ।

हल:

⅗ एवं ⅘ दोनों के अंश एवं हर को 6 से गुना करने पर।

⅗ = 18/30

⅘ = 24/30

अतः 5 परिमेय संख्या है

19/30, 20/30, 21/30, 22/30 एवं 23/30

4. नीचे दिए गए कथन सत्य हैं या असत्य? कारण के साथ अपने उत्तर दीजिए।

(i) प्रत्येक प्राकृत संख्या एक पूर्ण संख्या होती है।

(ii) प्रत्येक पूर्णांक एक पूर्ण संख्या होती है।

(iii) प्रत्येक परिमेय संख्या एक पूर्ण संख्या होती है।

हल:

(i)

सत्य है, क्योंकि पूर्ण संख्या के समूह में प्राकृत संख्या शामिल है।

(ii)

असत्य है, क्योंकि ऋणात्मक पूर्णांक पूर्ण संख्या नही होता है।

(iii)

असत्य है, जैसे ½ परिमेय संख्या है लेकिन पूर्ण संख्या नही है।

EXERCISE 1.2

1. State whether the following statements are true or false. Justify your answers.

(i) Every irrational number is a real number.

Solution:

True, every irrational numbers are represented on a number line.

(ii) Every point on the number line is of the form, √m where m is a natural number.

Solution:

False, every point on the number is of the form of real number.

(iii) Every real number is an irrational number.

Solution:

False, every real number is either rational or irrational.

2. Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.

Solution:

The square root of all positive integers are not irrational.

Example

√4 is a rational number.

3. Show how √5 can be represented on the number line.

Solution:

Following steps are used to show √5 on number line.

Step 1.

Apply Pythagoras theorem to get base for unit perpendicular and for hypotenuse of √5.

b² = (√5)² - 1² = 4 = 2² 

Therefore, 

b = 2.

Step 1.

Draw perpendicular on 2 of unit that is P.

Step 2.

Join O to P.

Step 3.

Draw an arch of radius OP and centre O. The arch intersects the number line at Q. This represent √5 on number line.

Figure is given below.

Quadratic Equations

Let's Solve the Questions of

Class X.

4. Quadratic Equations.

Exercise 4.4 

Subject Mathematics 

Publication : NCERT.

1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

(i) 2x² – 3x + 5 = 0 

(ii) 3x² – 4√3x + 4 = 0

(iii) 2x² – 6x + 3 = 0

Solution:

(i)

a = 2, b = -3, c = 5

D = b² -4ac

or, D = -3² -4×2×5

or, D = 9 -40

∴ D = -31

D<0

Hence, roots are not real.

(ii)

a = 3, b =-4√3, c = 4

D = b² -4ac

D = -4√3² -4×3×4

D = 48 - 48

D = 0

D = 0

Hence, roots are real and equal.

x = (-b ±√D)÷2a

or, x = (-(-4√3) ±√0)÷2×3

or, x = 4√3 ÷ 6

x = 2√3/3 Ans.

(iii)

a = 2, b = -6, c = 3

D = b² -4ac

or, D = -6² -4×2×3

or, D = 36 - 24

or, D = 12

D>0

Hence, roots are real and equal.

x = (-b ±√D)÷2a

x = (-(-6) ±√12)÷2×2

x = (6 ±2√3)÷4

x = (3 ±√3)÷2

∴ x = (3 +√3)/2 or (3 -√3)/2 Ans.

2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.

(i) 2x² + kx + 3 = 0 (ii) kx (x – 2) + 6 = 0

Solution:

2x² + kx + 3 = 0

a = 2, b = k , c = 3

For equal roots 

D = 0

b² - 4ac = 0

k² - 4.2.3 = 0

k² = 24

k = √24

k = ±2√6 Ans.

(ii)

kx(x - 2) + 6 = 0

kx² - 2kx + 6 = 0

a = k, b = -2k, c = 6

For equal roots

D = 0

b² - 4ac = 0

(-2k)² - 4.k.6 = 0

4k² - 24k = 0

4k(k - 6) = 0

Therefore,

k = 0 or 6 Ans.

3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.

Solution:

Let the breadth be x m.

Length = 2x

Area = Length ×breadth

800 = 2x × x

or, 800 = 2x²

or, 2x² -800 = 0

or, x² -400 = 0

a = 1, b = 0, c = -400

D = b² -4ac

or, D = 0² -4×1×-400

or, D = 1600

Here, D > 0

So, roots are real and unequal.

Hence design is possible.

Therefore, 

x = (-b ±√D)÷2a

or, x = (-0 ±√1600)÷2

or, x = ±40÷2

or, x = ±20

Breadth not be in negative.

Therefore,

x = 20 m.

Length = 2x = 2×20 = 40 m.

Required Answers

40 m & 20 m Ans.

3. क्या एक एेसी आम की बगिया बनाना संभव है जिसकी लंबाई, चौड़ाई से दुगुनी हो और उसका क्षेत्रफल 800 m²  हो? यदि है, तो उसकी लंबाई और चौड़ाई ज्ञात कीजिए।

हल:

मान लिया कि चौडाई = x मीटर।

∴ लंबाई = 2x मीटर।

हम जानते है कि

क्षेत्रफल = लंबाई×चौड़ाई

800 = 2x × x

or, 800 = 2x²

or, 2x² -800 = 0

or, x² -400 = 0

a = 1, b = 0, c = -400

D = b² -4ac

or, D = 0² -4×1×-400

or, D = 1600

D > 0

अतः बगिया बनना संभव है।

x = (-b ±√D)÷2a

or, x = (-0 ±√1600)÷2×1

or, x = ±40÷2

चौडाई ऋणात्मक नही हो सकता है।

अतः x = 20 मीटर।

लंबाई = 2x = 2×20 = 40 मीटर।

अभिष्ट उत्तर

40 मीटर एवं 20 मीटर Ans.

4. Is the following situation possible? If so, determine their present ages.

The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Solution:

Let the present age of one friend be x yrs.

Therefore,

Present age of second friend = 20 -x yrs.

4 yrs ago

(x -4)×(20 -x -4) = 48

or, (x -4)×(16 -x) = 48

or, 20x -x² -64 = 48

or, x² -20x + 112 = 0

a = 1, b = -20, c = 112

D = b² -4ac

or, D = (-20)² -4×1×112

or, D = 400 - 448

or, D = -48

Here D is less than 0

Hence, under this situation  age cannot be determined. 

5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m² ? If so, find its length and breadth.

Solution:

Let the length of the park is = x metre. 

Breadth = 80/2 -x 

or, Breadth = (40 -x)

Area = Length× breadth

400 = x(40 -x)

or, 400 = 40x - x²

ot, x² -40x + 400 = 0

a = 1, b = -40, c = 400

or, D = b² -4ac

or, D = -40² -4×1×400

or, D = 1600 -1600

or, D = 0

It is not possible to design rectangular park but possible to design square park.

x = -b ÷ 2a

or, x = -(-40) ÷ 2×1

or, x = 40 ÷ 2

or, x = 20 meter.

Breadth = 40 -x = 40 -20 = 20 meter

Hence, both Length and breadth are equal. 

So, design of square park is possible instead of rectangular park.