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These NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 11 Constructions Exercise 11.2

Question 1.
Construct a triangle ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm.
Solution:

Steps of construction:
Step-I: Draw the base BC = 7 cm and at point B make an angle say ∠XBC = 75°.
Step-II: Cut the line segment BD = AB + AC = 13 cm from die ray BX.
Step-III: Join DC and draw perpendicular bisector of DC. Which intersect DB at A.
Step-IV: Join AC.
Step-V: Triangle ABC is the required triangle.

Question 2.
Construct a triangle ABC in which BC = 8 cm, ∠B = 45° and AB – AC = 3.5 cm.
Solution:

Steps of Construction:
Step-I: Draw the base BC = 8 cm and make ∠XBC = 45°.
Step-II: Cut the Sine segment BD = AB – AC = 3.5 cm from ray BX.
Step-III: Join DC and draw the perpendicular bisector of DC.
Step-IV: Let the perpendicular bisector of DC intersect XB at A.
Step-V: Join AC.
Step-VI: Triangle ABC is the required triangle.

Question 3.
Construct a triangle PQR in which QR = 6 m, ∠Q = 60° and PR – PQ = 2 cm.
Solution:

Steps of Construction:
Step-I: Draw the base QR = 6 cm and at point Q make an angle ∠XQR = 60°.
Step-II: Cutline segment QM = PR – PQ = 2 cm from the line XQ extended on the opposite side of line segment QR.
Step-III: Join MC and draw the perpendicular bisector of MR which intersect XQ at P.
Step-IV: Join PR.
Step-V: Triangle PQR is the required triangle.

Question 4.
Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.
Solution:

Steps of Construction:
Step-I: Draw the line segment PQ = 11 cm. (= XY + YZ + ZX)
Step-II: At P construct an angle of 30° and at Q, an angle of 90°.
Step-III: Bisect these angles. Let the bisector of these angles intersect at a point X.
Step-IV: Draw perpendicular bisectors of XP and XQ which intersect PQ at Y and Z respectively.
Step-V: Join XY and XZ.
Step-VI: Triangle XYZ is the required triangle.

Question 5.
Construct a right triangle whose base is 12 cm and the sum of its hypotenuse and other side is 18 cm.
Solution:

Steps of Construction:
Step-I: Draw the base BC = 12 cm and at a point B makes an angle say ∠XBC = 90°.
Step-II: Cut the line segment BD = 18 cm (= AB + AC).
Step-III: Join DC and make a ∠DCY = ∠BDC.
Step-IV: CY intersect BX at A.
Step-V: Join AC.
Step-VI: Triangle ABC is the required triangle.

These NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 11 Constructions Exercise 11.1

Question 1.
Construct an angle of 90° at the initial point of a given ray and justify the construction.
Solution:
Steps of construction:
1. Draw a straight line AB.
2. Taking A as centre, draw an arc, which intersects AB at a point D.

3. Now, taking D as a centre, the same radius AD, intersect at E and F on the succession arc. (here AD = \(\widehat{\text { DE }}\) = \(\widehat{\text { EF }}\))
4. From point E take a radius (which should be greater than \(\frac {1}{2}\) \(\widehat{\text { EF }}\)) then with the same radius from F intersect the arc at point C.
5. Now, join C to A.
6. The required angle ∠ABC = 90°.

Question 2.
Construct an angle of 45° at the initial point of a given ray and justify the construction.
Solution:

Steps of Constructions:
1. Follow the instruction of the previous question upto ∠ABC = 90°.
2. Take an arc from the points H and G each which Intersect at I.
3. Here ∠FBC is half of ∠ABC, FB is the angle bisector.
4. So, the required ∠FBC = 45°

Question 3.
Construct the angles of the following measurements:
(i) 30°
(ii) 22½°
(iii) 15°
Solution:
(i) Steps of Construction:

1. Take a straight line AB.
2. Draw an arc, taking A as centre which intersects AB at C.
3. From C take another arc \(\widehat{\mathrm{CD}}\) such that AB = \(\widehat{\mathrm{CD}}\)
4. From C and D take similar arc which intersects at E.
5. The required ∠FAB = 30°.

(ii) Steps of construction:

1. Follow the instruction of question (2) i.e. ∠FAB = 45°.
2. Take the same arc from points C and K, intersect at G.
3. The required ∠HAB = 22½°

(iii) Step of construction:

1. Follow the instruction of 3:
(i) ∠EAB = 30°.
2. Take a small arc from the points G and C that intersects at F.
3. The required angle ∠FAB = 15°.

Question 4.
Construct the following angles and verify by measuring them by a protractor:
(i) 75°
(ii) 105°
(iii) 135°.
Solution:
(i) Steps of construction:
1. Follow as in question 1. i.e. ∠EAC = 90°.
2. Take the same arc from points D and E which intersect at G.
3. The required ∠KAB = 75°.

(ii) Steps of construction:
1. Follow the instruction as in the previous question upto ∠EAB = 90°.
2. Take the same arc from points D and E intersect at F.
3. Join F to A.
4. The required ∠GAB = 105°.

(iii) Steps of construction:
1. Follow the instruction as ∠EAB = 90°.
2. Take the same arc from the points E and F, Intersect at G.
3. Join G to A.
4. The required ∠HAB = 135°.

Question 5.
Construct an equilateral triangle, given its side, and justify the construction.
Solution:
Steps of construction:

1. Take a straight line XY.
2. By measuring 6 cm on the scale, cut one point C on XY line the reverse the same to get B point. Here, BC = 6 cm
3. From B and C points draw an arc of the same length intersect at A.
4. We find here AB = BC = CA = 6 cm. Thus ABC is an equilateral triangle.

These NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.6 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.6

Question 1.
Subtract:
(a) ₹ 18.25 from ₹ 20.75
(b) 202.54 m from 250
(c) ₹ 5.36 from ₹ 8.40
(d) 2.051 km from 5.206 km
(e) 0.314 kg from 2.107 kg
Answer:
(a) 20.75 – 18.25 = ₹ 2.50
(b) 250 – 202.54 = 47.46 m
(c) 8.40 – 5.36 = ₹ 3.04
(d) 5.206 – 2.051 = 3.155 km
(e) 2.107 – 0.314 = 1.793

Question 2.
Find the value of:
Answer:
(a) 9.756 – 6.28 = 3.476
(b) 21.05 – 15.27 = 5.78
(c) 18.5 – 6.79 = 11.71
(d) 11.6 – 9.847 = 1.753

Question 3.
Raju bought a book of ₹ 35.65. He gave ₹ 50 to the shopkeeper. How much money did he get back from the shopkeeper?
Answer:
Total amount given to shopkeeper = ₹ 50
Cost of book = ₹ 35.65
Amount left = ₹ 50.00 – ₹ 35.65
= Rs 14.35
Therefore, Raju got back ₹ 14.35 from the shopkeeper.

Question 4.
Rani had ₹ 18.50. She bought one ice¬cream for ₹ 11.75. How much money does she have now?
Answer:
Total money = ₹ 18.50
Cost of Ice-cream = ₹ 11.75
Amount left = ₹ 18.50 – ₹ 11.75 = ₹ 6.75
Therefore, Rani has ₹ 6.75 now.

Question 5.
Tina had 20 m 5 cm long cloth. She cuts 4 m 50 cm length of cloth from this for making a curtain. How much cloth is left with her?
Answer:
Total length of cloth = 20 m 5 cm = 20.05 m
Length of cloth used = 4 m 50 cm = 4.50 m
Remaining cloth = 20.05 m – 4.50 m = 15.55 m
Therefore, 15.55 m of cloth is left with Tina.

Question 6.
Namita travels 20 km 50 m every day. Out of this she travels 10 km 200 m by bus and the rest by auto. How much distance does she travel by auto?
Answer:
Total distance travel = 20 km 50 m = 20.050 km
Distance travelled by bus
= 10 km 200 m = 10.200 km
Distance travelled by auto
= 20.050 – 10.200 = 9.850 km
Therefore, Namita travels 9.850 km by auto.

Question 7.
Aakash bought vegetables weighing 10 kg. Out of this 3 kg 500 g in onions, 2 kg 75 g is tomatoes and the rest is potatoes. What is the weight of the potatoes?
Answer:
Weight of onions = 3 kg 500 g = 3.500 kg
Weight of tomatoes = 2 kg 75 g = 2.075 kg
Total weight of onions and tomatoes = 3.500 + 2.075 = 5.575 kg
Therefore, weight of potatoes = 10.000 – 5.575 = 4.425 kg
Thus, the weight of potatoes is 4.425 kg.

These NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.5

Question 1.
Find the sum in each of the following:
(a) 0.007 + 8.5 + 30.08
(b) 15 + 0.632+ 13.8
(c) 27.076 + 0.55 + 0.004
(d) 25.65 + 9.005 + 3.7
(e) 0.75 + 10.425 + 2
(f) 280.69 + 25.2 + 38
Answer:

Question 2.
Rashid spent ₹ 35.75 for Maths book and ₹ 32.60 for Science book. Find the total amount spent by Rashid.
Answer:
Money spent for Maths book = ₹ 35.75 Money spent for Science book = ₹ 32.60
Total money spent = ₹ 35.75 + ₹ 32.60
.’. Total amount of money spent by Rashid is = ₹ 68.35
Therefore, total money spent by Rashid is ₹ 68.35.

Question 3.
Radhika’s mother have her ₹ 10.50 and her father gave her ₹ 15.80. Find the total amount given to Radhika by the parents.
Ans. Money given by mother = ₹ 10.50
Money given by father = ₹ 15.80
Total money received by Radhika = ₹ 10.50 + ₹ 15.80 = ₹ 26.30
Therefore, the total money received by Radhika is ₹ 26.30.

Question 4.
Nasreen bought 3 m 20 cm cloth for her shirt and 2 m 5 cm cloth for her trouser. Find the total length of cloth bought by her.
Answer:
Cloth bought for shirt = 3 m 20 cm = 3.20 m
Cloth bought for trouser = 2 m 5 cm = 2.05 m
Total length of cloth bought by Nasreen = 3.20 + 2.05 = 5.25 m
Therefore, the total length of cloth bought by Nasreen is 5.25 m.

Question 5.
Naresh walked 2 km 35 m in the morning and 1 km 7 m in the evening. How much distance did he walk in all?
Answer:
Distance travelled in morning = 2 km 35 m = 2.035 km
Distance travelled in evening = 1 km 7 m = 1.007 km
Total distance travelled = 2.035 + 1.007 = 3.042 km
Therefore, the total distance travelled by Naresh is 3.042 km.

Question 6.
Sunita travelled 15 km 268 m by bus, 7 km 7 m by car and 500 m by foot in order to reach her school. How far is her school from her residence?
Answer:
Distance travelled by bus =15 km 268 m = 15.268 km
Distance travelled by car = 7 km 7 m = 7.007 km
Distance travelled on foot = 500 m = 0.500 km
Total distance travelled = 15.268 + 7.007 + 0.500 = 22.775 km Therefore, total distance travelled by Sunita is 22.775 km.

Question 7.
Ravi purchases 5 kg 400g rice, 2 kg 20g sugar and 10 kg 850g flour. Find the total weight of his purchases.
Answer:
Weight of Rice = 5 kg 400 g = 5.400 kg
Weight of Sugar = 2 kg 20 g = 2.020 kg
Weight of Flour = 10 kg 850 g = 10.850 kg
Total weight = 5.400 + 2.020 + 10.850
= 18.270 kg
Therefore, the total weight of Ravi’s purchase = 18.270 kg

These NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 8 Decimals Ex 8.4

Question 1.
Express as rupees using decimals:
(a) 5 paise
(b) 75 paise
(c) 20 paise
(d) 50 rupees 90 paise
(e) 725 paise
Answer:
(a) ∵ 1 paisa = ₹ \(\frac { 1 }{ 100 }\)
∴ 5 paise = \(\frac { 1 }{ 100 }\) × 5 = ₹ 0.05

(b) ∵ 1 paisa = \(\frac { 1 }{ 100 }\)
∴ 75 paise = \(\frac { 1 }{ 100 }\) × 75 = ₹ 0.75

(c) ∵ 1 paisa = X \(\frac { 1 }{ 100 }\)
∴ 20 paise = \(\frac { 1 }{ 100 }\) × 20 = ₹ 0.2

(d) ∵ 1 paisa = X \(\frac { 1 }{ 100 }\)
∴ ₹ 50 + 90 paise
= 50 + \(\frac { 1 }{ 100 }\) × 90 = ₹ 50.90

(e) ∵ 1 paisa = ₹ \(\frac { 1 }{ 100 }\)
∴ 725 paise = \(\frac { 1 }{ 100 }\) × 725
= ₹ 7.25

Question 2.
Express as metres using decimals:
(a) 15 cm
(b) 6 cm
(c) 2 m 45 cm
(d) 9 m 7 cm
(e) 419 cm
Answer:
(a) ∵ 1 cm = \(\frac { 1 }{ 100 }\) m
∴ 15 cm = \(\frac { 1 }{ 100 }\) × 15 = 0.15 m 100
(b) ∵ 1 cm = \(\frac { 1 }{ 100 }\) m
∴ 6 cm = \(\frac { 1 }{ 100 }\) × 6 = 0.06 m

(c) ∵ 1 cm = \(\frac { 1 }{ 100 }\) m
∴ 2 m 45 cm = 2 + \(\frac { 1 }{ 100 }\) × 45
= 2.45 m

(d) ∵ 1 cm = \(\frac { 1 }{ 100 }\) m
∴ 9 m 7 cm = 9 x \(\frac { 1 }{ 100 }\) × 9 = 9.07 m

(e) ∵ 1 cm = \(\frac { 1 }{ 100 }\) m
∴ 419 = \(\frac { 1 }{ 100 }\) × 419 = 4.19 m

Question 3.
Express as cm using decimals:
(a) 5 mm
(b) 60 mm
(c) 164 mm
(d) 9 cm 8 mm
(e) 93 mm
Answer:
(a) ∵ 1mm = \(\frac { 1 }{ 10 }\) cm
5mm = \(\frac { 1 }{ 10 }\) × 5 = 0.5 cm

(b) ∵ 1 mm = \(\frac { 1 }{ 10 }\) cm 10
∴ 60 mm = \(\frac { 1 }{ 10 }\) × 60 = 6 cm

(c) ∵ 1 mm = \(\frac { 1 }{ 10 }\) cm
∴ 164 mm = \(\frac { 1 }{ 10 }\) × 164 = 16.4 cm

(d) ∵ 1 mm = \(\frac { 1 }{ 10 }\) cm
∴ 9 cm 8 mm = 9 + \(\frac { 1 }{ 10 }\) × 8
= 9 + 0.8 cm = 9.8 cm

(e) ∵ 1 mm = \(\frac { 1 }{ 10 }\) cm
∴ 93 mm = \(\frac { 1 }{ 10 }\) × 93 cm

Question 4.
Express as km using decimals:
(a) 8m
(b) 88 m
(c) 8888 m
(d) 70 km 5 m
Answer:
(a) ∵ 1 m = \(\frac { 1 }{ 1000 }\) km
∴ 8m = \(\frac { 1 }{ 1000 }\) × 8 = 0.008 km

(b) ∵ 1 m = \(\frac { 1 }{ 1000 }\) km
∴ 88 m: 1 \(\frac { 1 }{ 1000 }\) × 88 = 0.088 km

(c) ∵ 1 m = \(\frac { 1 }{ 1000 }\) km
∴ 8888m = \(\frac { 1 }{ 1000 }\) × 88 = 8.888km

(d) ∵ 1m = \(\frac{1}{100}\)km
∴ 70km 5m = 70 + \(\frac{1}{100}\) × 5
= 70.005 km

Question 5.
Express as kg using decimals:
(a) 2g
(b) 100g
(c) 3750 g
(d) 5 kg 8 g
(e) 26kg 50g
Answer:
(a) 1 g = \(\frac{1}{1000}\) kg
∴ 2g = \(\frac{1}{1000}\) × 2 = 0.002kg

(b) ∵ 1g = \(\frac{1}{1000}\) kg
∴ 100g= \(\frac{1}{1000}\) × 100 = 0.1 kg

(c) ∵ 1g = \(\frac{1}{1000}\) kg
∴ 3750g = \(\frac{1}{1000}\) × 3750 = 3.750kg

(d) ∵ 1g = \(\frac{1}{1000}\) kg
∴ 5kg50g = 5 + \(\frac{1}{1000}\) × 8
= 5.008kg

(e) ∵ 1g = \(\frac{1}{1000}\) kg
∴ 26kg 50g = 26 + \(\frac{1}{1000}\) × 50
= 26.050kg

These NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Exercise 9.4

Question 1.
A survey of 120 school students was done to find which activity they prefer to do in their free time.

Draw a bar graph to illustrate the above data taking scale of 1 unit length = 5 students.
Which activity is preferred by most of the students other than playing?
Answer:
Steps :
(i) Draw two perpendicular lines OX and OY.
(ii) Draw vertical bars (rectangles) of same width keeping equal spacing between them.
(iii) ∵ 1 unit length = 5 students,
∴ Heights of bars will be
For playing = 45 ÷ 5 = 9 units
For reading story books = 30 ÷ 5 = 6 units
For watching TV = 20 ÷ 5 = 4 units
For listening to music = 10 ÷ 5 = 2 units
For painting = 15 ÷ 5 = 3 units
The required graph is as under :

Reading story books is preferred by most of the students other than playing.

Question 2.
The number of Mathematics books sold by a shopkeeper on six consecutive days is shown below:

Draw a bar graph to represent the above information choosing the scale of your choice.
Answer:
Steps:
(i) Draw two perpendicular lines OX and OY.
(ii) Draw bars (rectangles) of equal width on OX having same spacing between them.
(iii) Taking an appropriate scale (here 1 unit length = 10 books) fix the heights of various bars, such as:
For Sunday = 65 ÷ 10 = 6.5 units
For Monday = 40 ÷ 10 = 4 units
For Tuesday = 30 ÷ 10 = 3 units
For Wednesday = 50 ÷ 10 = 5 units
For Thursday = 20 ÷ 10 = 2 units
For Friday = 70 ÷ 10 = 7 units
∴ The required graph is as under :

Question 3.
Following table shows the number of bicycles manufactured in a factory during the years 1998 to 2002. Illustrate this data using a bar graph. Choose a scale of your choice.

(a) In which year was the maximum number of bicycles manufactured?
(b) In which year was the minimum number of bicycles manufactured?
Answer:
Steps :
(i) Draw two perpendicular lines OX and OX
(ii) Draw bars of equal width on OX having same spacing between them.
(iii) For fixing heights of various bars, choose an appropriate scale (here: 1 unit length = 200 bicycles) so that, we have:
∴ For 1998: 800 ÷ 200 = 4 units
For 1999: 600 ÷ 200 = 3 units
For 2000: 900 ÷ 200 = 4.5 units
For 2001: 1100 ÷ 200 = 5.5 units
For 2002: 1200 ÷ 200 = 6 units

(a) The maximum number of bicycles manufactures in 2002.
(b) The minimum number of bicycles manufactured in 1999.

Question 4.
Number of persons in various age groups in a town is given in the following table.

Draw a bar graph to represent the above information and answer the following
questions. (take 1 unit length = 20 thousands)
(a) Which two age groups have same population?
(b) All persons in the age group of 60 and above are called senior citizens. How many senior citizens are there in the town?
Answer:
Steps :
(i) Draw two perpendicular lines OX and OY.
(ii) Draw bars of equal width on OX having same spacing between them.
(iii) Since the scale is given:
1 unit length = 20 thousand person
∴ The heights of various bars are given below such as :
For 1 – 14 : 200000 ÷ 20000 = 10 units
For 15 – 29 : 160000 ÷ 20000 = 8 units
For 30 – 44 : 120000 ÷ 20000 = 6 units
For 45 – 59 : 120000 ÷ 20000 = 6 units
For 60 – 74 : 80000 ÷ 20000 = 4 units
For 75 and above
: 40000 ÷ 20000 = 2 units
Now, the required bar graph is as follows

(a) Group 30 – 44 and group 45 – 59 have same population.
(b) 80,000 + 40,000 = 1,20,000 senior citizens are there in the town.