Class 6 Maths RS Aggarwal Solutions

RS Aggarwal Class 6 Solutions Chapter 15 Polygons Ex 15

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 15 Polygons Ex 15

Question 1.
Solution:
(a), (b), (d) and (f) are simple closed figures.

Question 2.
Solution:
(a), (b) and (c) are polygons.

Question 3.
Solution:
(i) two
(ii) triangle
(iii) quadrilateral
(iv) 3, 3
(v) 4, 4
(vi) closed figure.

 

Hope given RS Aggarwal Solutions Class 6 Chapter 15 Polygons Ex 15 are helpful to complete your math homework.

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RS Aggarwal Class 6 Solutions Chapter 14 Constructions (Using Ruler and a Pairs of Compasses) Ex 14B

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 14 Constructions (Using Ruler and a Pairs of Compasses) Ex 14B.

Other Exercises

• RS Aggarwal Solutions Class 6 Chapter 14 Constructions Ex 14A
• RS Aggarwal Solutions Class 6 Chapter 14 Constructions Ex 14B

Question 1.
Solution:
(1) 60°
Steps of construction :
(i) Draw a ray OA.

(ii) With centre O and with a suitable radius drawn an arc meeting OA at E.
(iii) With centre E and with same radius, draw another arc cutting the first arc at F.
(iv) Join OF and produce it to B Then ∠AOB = 60°
(2) 120°
Steps of construction :
(i) Draw a ray OA
(ii) With centre O and with a suitable radius draw an arc meeting OA at E
(iii) With centre E and with the same radius cut off the first arc firstly at F and then at G i.e. EF = FG.
(iv) Join OG and produce it to B.
Then, ∠AOB = 120°

(3) 90°
Steps of construction :
(i) Draw a ray OA

(ii) With centre O and a suitable radius draw an arc meeting OA at E.
(iii) With centre E and A with same radius cut off the arc first at F and then from F with same radius cut off arc at G.
(iv) With centres F and G with a suitable radius, draw two arcs intersecting each other at H.
(v) Join OH and produce it to B.
Then, ∠AOB = 90°.

Question 2.
Solution:
Steps of Construction :
(i) Draw a ray OA.

(ii) With O as centre and any suitable radius draw an arc above OA, cutting it at a point B.
(iii) With B as centre and same radius as before draw another arc to cut the previous arc at C.
(iv) Join OC and produce it to D. Then ∠AOD = 60° is the required angle. To bisect the angle ∠AOD, with B as centre and radius more than half BC draw an arc. With C as centre and the same radius draw another are cutting the previous arc at E. Join OE and produce it. Then, OE is the required bisector of ∠AOD.

Question 3.
Solution:
Steps of constructions :
(i) Draw a ray OA.
(ii) With centre O and a suitable radius draw an arc meeting OA at E.
(iii) With centre E and with same radius, cut the first arc firstly at F and then from F with same radius cut act at G.
(iv) With centres F and G, with suitable radius, draw arcs intersecting each other at H.

(v) Join OH intersecting the first arc at L and produce it to C.
(vi) With centre E and L and with suitable radius draw arcs intersecting each other at M.
(vii) Join OM and produce it to B.
Then ∠AOB = 45°

Question 4.
Solution:
(i) Steps of Construction :
1. Draw a ray OA.
2. With O as centre and any suitable radius draw an arc cutting OA at G.
3. With G as centre and same radius cut the arc at B and then B as centre and same radius cut the arc at C. Again, with C as centre and same radius cut the arc at D.

4. With C as centre and radius more than half CD draw an arc.
5. With D as centre and same radius draw another arc to cut the previous arc at E.
6. Join OE and produce it to F.
Then ∠AOF = 150°
(ii) Steps of Construction :
1. Draw a ray OA.
2. With O as centre and any suitable radius draw an arc above OA, cutting it at B.

3. With B as centre and same radius as before draw another arc to cut the previous arc at C. Join OC and prouce it to D.
4. Draw the bisector OE of ∠AQD. Then ∠AOE = 30°.
5. Draw the bisector OF of ∠AOE. Then ∠AOF = 15° is the required angle.
(iii) Steps of Construction :
1. Draw a ray OA.

2. With O as centre and any suitable radius draw an arc above OA, cutting it at B.
3. With B as centre and same radius as before draw another arc to cut the previous arc at C. With C as centre and same radius draw the arc to cut it at D. Again with D as centre and same radius cut the arc at E.
4. Join OD and produce it to G. Then ∠AOG = 120°.
5. With D as centre and radius more than half DE draw an arc.
6. With E as centre and same radius draw another arc to cut the previous arc at F. Join OF.
7. Draw the bisector OH of ∠GOF. Then ∠AOH = 135° is the required angle.
(iv) Steps of Construction :
1. Draw a ray OA.

2. With O as centre and any suitable radius draw an arc above OA, cutting it at B.
3. With B as centre and same radius cut the previous arc at C and then with C as centre and same radius cut the arc at D.
4. With C as centre and radius more than half CD draw an arc.
5. With D as centre and same radius draw another arc to cut the previous arc at E.
6. Join OE. Then ∠AOE = 90°.
7. Draw the bisector OF of ∠AOE.
8. Draw the bisector OG of ∠AOF.
Then ∠AOG = \(22 \frac { 1 }{ 2 } \) ° is the required angle.
(v) Steps of Construction :
1. Draw a ray OA.

2. With O as centre and any suitable radius draw an arc cutting OA at B.
3. With B as centre and same radius cut the previous arc at C and then with C as centre and same radius cut the arc at D.
4. With C as centre and radius more than half CD draw an arc
5. With D as centre and same radius draw another arc to cut the previous arc at E.
6. Join OE. Also join OD and produce it to F.
7. Draw the bisector OG of ∠EOF Thus, ∠AOG = 105° is the required angle.
(vi) Steps of Construction :
1. Draw a ray OA.

2. With O as centre and any suitable radius draw an arc cutting OA at B.
3. With B as centre and same radius* cut the previous arc at C and then with C as centre cut the arc at D.
4. With C as centre and radius more than half CD draw an arc.
5. With D as centre and same radius draw another arc to cut the previous arc at E.
6. Join OE. Also join OC and produce it to G.
7. Draw the bisector OF of ∠EOG. Then, ∠AOF = 75° is the required angle.
(vii) Steps of Construction :
1. Draw a ray OA.

2. With O as centre and any suitable radius draw an arc above OA to cut it B.
With B as centre and same radius cut the previous arc at C and then with C as centre and same radius cut the arc at D.
4. With C as centre and radius more than half CD draw an arc.
5. With D as centre and same radius draw another arc to cut the previous arc at E.
6. Join OE. Then ∠AOE = 90°.
7. Draw the bisector OF of ∠AOE.
8. Draw the bisector OG of ∠EOF.
Then ∠AOG = \(67 \frac { 1 }{ 2 } \) ° is the required angle.
(viii) Steps of Construction :
1. Draw a ray OA.

2. With O as centre and any su itable radius draw an arc above OA to cut it at B.
3. With B as centre and same radius cut the previous arc at C and then with C as centre and same radius cut the arc at D.
4. With C as centre and radius more than half CD, draw an arc.
5. With D as centre and same radius draw another arc to cut the previous arc at E.
6. Join OE. Then ∠AOE = 90°.
7. Draw the bisector OF of angle ∠AOE. Then, ∠AOF = 45° is the required angle.

Question 5.
Solution:
Steps of Construction :
1. Draw a line-segment AB = 5 cm with the help of a rular.

2. With Aas centre and suitable radius draw an arc cutting AB at C.
3. With C as centre and same radius cut the previous arc at D and then with D as centre and same radius cut the arc at E
4. With D as centre and radius more than half DE draw an arc.
5. With E as centre and same radius draw another arc to cut the previous arc at F.
6. Join AF and produce it to G such that AG = 3.5 cm. Then ∠BAG = 90°.
7. With G as centre and radius equal to AB draw an arc. With B as centre and radius equal to AG draw another arc to cut the previous arc at H.
8. Join GH and BH. Then, AB HG is the required rectangle.

Question 6.
Solution:
Steps of Construction :
1. With the help of a ruler draw a line segment AB = 5 cm.

2. With A as centre and any suitable radius draw an arc cutting AB at C.
3. With C as centre and same radius cut the previous arc at D and then with D as centre and same radius cut the arc at E.
4. With D as centre and radius more than half DE draw an arc.
5. With E as centre and same radius draw another arc to cut the previous arc at F.
6. Join AF and produce it to G such that AG = 5 cm.
7. With G as centre and radius equal to AB draw an arc. With B as centre and same radius draw another arc to cut the previous arc at H.
8. Join GH and BH. Then, AB HG is the required square.

Hope given RS Aggarwal Solutions Class 6 Chapter 14 Constructions Ex 14B are helpful to complete your math homework.

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RS Aggarwal Class 6 Solutions Chapter 11 Line Segment, Ray and Line Ex 11A

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 11 Line Segment, Ray and Line Ex 11A.

Other Exercises

• RS Aggarwal Solutions Class 6 Chapter 11 Line Segment, Ray and Line Ex 11A
• RS Aggarwal Solutions Class 6 Chapter 11 Line Segment, Ray and Line Ex 11B

Question 1.
Solution:
Line segment
(i) In figure (i) are \(\overline { XY } \) and \(\overline { YZ } \)
(ii) In figure (ii) \(\overline { AD } ,\overline { AB } ,\overline { AC } ,\overline { AE } ,\overline { BD } ,\overline { BC } ,\overline { CE } \)
(iii) In figure (iii) \(\overline { PQ } ,\overline { PR } ,\overline { PS } ,\overline { QR } ,\overline { QS } ,\overline { RS } \)

Question 2.
Solution:
(i) In fig. (i), line segments is \(\overline { AB } \), and
rays are \(\overrightarrow { AC }\) and \(\overrightarrow { BD }\).
(ii) In fig. (ii), line segments are
\(\overline { GE } ,\overline { GP } ,\overline { EP } \) and rays are
\(\overrightarrow { EF } ,\overrightarrow { GH } ,\overrightarrow { PQ }\)
(iii) In fig. (iii), line segments are \(\overline { OL } ,\overline { OP } \)
and rays are \(\overrightarrow { LM } ,\overrightarrow { PQ } \).

Question 3.
Solution:
(i) Four line segments are \(\overline { PR } ,\overline { QR } ,\overline { PQ } ,\overline { RS } \).
(ii) Four ray can be \(\overrightarrow { PA },\overrightarrow { QC }, \overrightarrow { RB } ,\overrightarrow { SD } \)
(iii) \(\overline { PR } ,\overline { QS } \) are two non-intersecting lines.

Question 4.
Solution:
(i) Three or more points in a plane are said to be collinear if they all lie on the same line.

(ii) In the figure given above, points A, B, C are collinear points.
We can draw exactly one line passing through three collinear points

Question 5.
Solution:
(i) Four pairs of intersecting lines are : (AB, PQ) ; (AB, RS) ; (CD, PQ) ; (CD, RS)
(ii) Four collinear points are : A, Q, S, B
(iii) Three non-collinear points are : A, Q, P
(iv) Three concurrent lines are : AB, PS and RS.
(v) Three lines whose point of intersection is P are : CD, PQ and PS.

Question 6.
Solution:
The lines drawn through given points A, B, C are as shown below. The names of these lines are AB, BC and AC.

Also it is clear that three different lines can be drawn.

Question 7.
Solution:
(i) In the the given figure, there are six line segments, namely AB, AC, AD, BD, BC, DC.

(ii) In the given figure, there are ten line segments, namely, AD, AB, AC, AO, OC, BC, BD, BO, OD, CD.

(iii) In the given figure, there are six line segments, namely AB, AF, BF, CD, DE, CE.

(iv) In the given figure, there are twelve line segments, namely, AB, BC, CD, AD, BF, CG, DH, AE, EF, FG, GH, EH.

Question 8.
Solution:
\(\overleftrightarrow { PQ } \) is a line

(i) False, as M does not lie on \(\overrightarrow { NQ } \)
(ii) True
(iii) True
(iv) True
(v) True

Question 9.
Solution:
(i) False
Point has no dimensions.
(ii) False
A line segment has a length.
(iii) False
A ray has no finite length.
(iv) False
If \(\overrightarrow { AB } \) and ray \(\overrightarrow { BA } \) have opposite directions.
(v) True
Length of \(\overline { AB } \) and \(\overline { BA } \) is same.
(vi) True
Line \(\overleftrightarrow { AB } \) and \(\overleftrightarrow { BA } \) are same.
(vii) True
Distance between A and B or B and A is same, so they determine a unique line segment.
(viii) True
Two lines intersect each other at one point.
(ix) False
Two intersecting planes intersect at one line not at one point.
(x) False
If A, B, C are collinear and points D, E are collinear then it is not necessary that there points A, B, C, D and E are collinear.
(xi) False
Infinite number of rays can be drawn with a given end point.
(xii) True
We can draw one and only one line passing through two given points.-
(xiii) True
We can draw infinite number of lines pass through a given point.

Question 10.
Solution:
(i) definite
(ii) one
(iii) no
(iv) definite
(v) cannot. Ans.

 

Hope given RS Aggarwal Solutions Class 6 Chapter 11 Line Segment, Ray and Line Ex 11A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10D

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10D.

Other Exercises

• RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10A
• RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10B
• RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10C
• RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10D

OBJECTIVE QUESTIONS
Mark against the correct answer in each of the following :

Question 1
Solution:
(d) \(\\ \frac { 92 }{ 115 } \)

Question 2.
Solution:
(a) \(\\ \frac { 57 }{ x } \)
= \(\\ \frac { 51 }{ 85 } \)

Question 3.
Solution:
(a) \(\\ \frac { 25 }{ 35 } \)
= \(\\ \frac { 45 }{ x } \)

Question 4.
Solution:
(c) \(\\ \frac { 4 }{ 5 } \)
= \(\\ \frac { x }{ 35 } \)

Question 5.
Solution:
(b) \(\\ \frac { a }{ b } \)
= \(\\ \frac { c }{ d } \)
=>ad = bc

Question 6.
Solution:
(b) a : b :: b : c

Question 7.
Solution:
(b) \(\\ \frac { 5 }{ 8 } \) < \(\\ \frac { 3 }{ 4 } \) => 4 x 5 < 3 x 8 => 20 < 24

Question 8.
Solution:
Total amount = Rs 760
Ratio A : B = 8 : 11

Question 9.
Solution:
(d) largest = \(\\ \frac { 252\times 7 }{ 5+7 } \)
= \(\\ \frac { 252\times 7 }{ 12 } \)
= 21 x 7
= 147

Question 10.
Solution:
(b) largest = \(\\ \frac { 90\times 5 }{ 1+3+5 } \)
= \(\\ \frac { 90\times 5 }{ 9 } \)
= 50 cm

Question 11.
Solution:
(c) total strength of school
largest = \(\\ \frac { 840 }{ 5 } \) x (12 + 5)
= \(\\ \frac { 840\times 17 }{ 5 } \)
= 168 x 17
= 2856

Question 12.
Solution:
(b) Cost of 12 pens = Rs 138
Cost of 1 pen = Rs \(\\ \frac { 138\times 14 }{ 12 } \)
and cost of 14 pens = Rs 161

Question 13.
Solution:
(b) \(\\ \frac { 24\times 15 }{ 8 } \)
= 45 days

Question 14.
Solution:
(a) \(\\ \frac { 26\times 40 }{ 20 } \)
= 52 men

Question 15.
Solution:
(b) In 6 L of petrol, a car covers = 111 km
In 1 L, it will cover = \(\\ \frac { 111 }{ 6 } \) km
and in 10 L it will cover = \(\\ \frac { 111\times 10 }{ 6 } \) km
= 185 km

Question 16.
Solution:
(a) \(\\ \frac { 28\times 550 }{ 700 } \)
= 22 days

Question 17.
Solution:
Ratio in the angles of triangle
= 3 : 1 : 2

Question 18.
Solution:
(b) Ratio in length and breadth of a rectangle = 5 : 4
Width = 36 m
Length = \(\\ \frac { 5 }{ 4 } \) x 36 m
= 45m

Question 19.
Solution:
Bus covers in 3 hrs = 195 km
It will cover in 1 hr = \(\\ \frac { 195 }{ 3 } \) = 65 km

Question 20.
Solution:
1 dozen = 12 bars
Cost of 5 bars of soap = Rs.82.50

Question 21.
Solution:
Total pencils in 30 packets of 8 pencils
= 30 x 8= 240
and total pencils of 25 packets of 12
pencils = 25 x 12 = 300
Now cost of 240 pencils = Rs. 600
Then cost of 1 percent = Rs. \(\\ \frac { 600 }{ 24 } \)
and cost of 300 pencils = Rs. \(\\ \frac { 600\times 300 }{ 240 } \)
= Rs 750 (b)

Question 22.
Solution:
Journey of 75 km costs = Rs 215
Cost of 1 km = Rs \(\\ \frac { 215 }{ 75 } \)

Question 23.
Solution:
1st term = 12
2nd term = 21
fourth term = 14

Question 24.
Solution:
10 boys dig a patch in = 12 hrs
1 boy will dig it in = 12 x 10 hours

Hope given RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10D are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 6 Solutions Chapter 10 Ratio, Proportion and Unitary Method Ex 10C

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10C.

Other Exercises

• RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10A
• RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10B
• RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10C
• RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10D

Question 1.
Solution:
Cost of 14 m of cloth = Rs. 1890
Cost of 1 m = Rs. \(\\ \frac { 1890 }{ 14 } \)
and cost of 6 m = Rs. \(\\ \frac { 1890\times 6 }{ 14 } \)
= Rs. 135 x 6
= Rs. 810

Question 2.
Solution:
Cost of 1 dozen or 12 soaps = Rs. 285.60
Cost of 1 soap = Rs. \(\\ \frac { 285.60 }{ 12 } \)
Cost of 15 soaps = Rs. \(\\ \frac { 285.60\times 15 }{ 12 } \)
= Rs. 357.00

Question 3.
Solution:
Cost of 9 kg of rice = Rs. 327.60
Cost of 1 kg = Rs. \(\\ \frac { 327.60 }{ 9 } \)
and cost of 50 kg = Rs. \(\\ \frac { 327.60\times 50 }{ 9 } \)
= Rs. 36.40 x 50
= Rs. 1820

Question 4.
Solution:
Weight of 22.5 metres of the iron rod: = 85.5 kg
Weight of 1 metre of the iron rod

Question 5.
Solution:
Quantity of oil in 15 tins = 234 kg
Quantity of oil in 1 tin = \(\\ \frac { 234 }{ 15 } \) kg
Quantity of oil in 10 tins

Question 6.
Solution:
Distance covered by the car in 12 litres of diesel = 222 kms
Distance covered by the car in 1 litre of diesel = \(\\ \frac { 222 }{ 12 } \) km

Question 7.
Solution:
Charges of 25 tonnes of weight = Rs. 540
charges of 1 ton = Rs.\(\\ \frac { 540 }{ 25 } \)

Question 8.
Solution:
Weight of copper in 4.5 g of alloy = 3.5g
Weight of copper in 1 g of alloy

Question 9.
Solution:
In Rs. 87.50, the inland letter are purchased = 35
In Re. 1, letters can be purchased
= \(\\ \frac { 35 }{ 87.50 } \)
and in Rs. 315, letters can be purchased

Question 10.
Solution:
4 dozen = 4 x 12 = 48 bananas
In Rs. 104, banana are purchased = 48

Question 11.
Solution:
In Rs. 22770, chairs are purchased =18
In Re. 1, chairs will be purchased
= \(\\ \frac { 18 }{ 22770 } \)
and in Rs. 10120, chairs will be
purchased = \(\\ \frac { 18\times 10120 }{ 22770 } \)
= 8

Question 12.
Solution:
(i) A car travels 195 km distance in = 3 hours
It will travel 1 km distance in = \(\\ \frac { 3 }{ 195 } \) hr.
and it will travel 520 km distance in
= \(\\ \frac { 3\times 520 }{ 195 } \)
= 8 hr
(ii) A car travels in 3hr = 195 km

Question 13.
Solution:
(i) A laborer earn in 12 days = Rs. 1980
He will earn in 1 day = Rs. \(\\ \frac { 1980 }{ 12 } \)
and he will earn in 7 days

Question 14.
Solution:
(i) Weight of 65 books = 13 kg
Then weight of 1 book = \(\\ \frac { 13 }{ 65 } \) kg

Question 15.
Solution:
Number of boxes needed for 6000 pens = 48
Number of boxes needed for 1 pen 48 = \(\\ \frac { 48 }{ 6000 } \)

Question 16.
Solution:
Clearly, less workers will build the wall in more days.
And, more workers will build the wall in less days.
24 workers can build the wall in 15 days
1 worker can build the wall in (15 x 24) days
(less worker, more days)
9 workers will build the wall in

Question 17.
Solution:
Men needed to finish a piece of work in 26 days = 40
Men needed to finish a piece of work in 1 day = 40 x 26 (less days, more men)

Question 18.
Solution:
Clearly, less men will take more days to consume the food.
And, more men will take less days to consume the food.
550 men have provisions for 28 days
1 men has provisions for (28 x 550) days [less men, more days]
700 men will have provisions for

Question 19.
Solution:
Clearly, less persons will consume the rice in more days.
And more persons will consume the rice in less days.
60 persons consume the bag of rice in 3 days.
1 person will consume the bag of rice in
(3 x 60) days (less persons, more days)
18 persons will consume the bag of rice

Hope given RS Aggarwal Solutions Class 6 Chapter 10 Ratio, Proportion and Unitary Method Ex 10C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.