Tag: Karnataka state syllabus mathematics class VIII

# Mensuration Exercise 15.5 – Class 10 – Solutions

### Mensuration Exercise 15.5 – Questions

1. A petrol tank is in the shape of a cylinder with hemisphere of same radius attached to both ends. If the total length of the tank is 6m and the radius is 1m, what is the capacity of the tank in litres.
2. A rocket is in the shape of a cylinder with a cone attached to one end and a hemisphere attached to the other. All of them are of the same radius of 1.5m. The total length of the rocket is 7m and height of the cup is 2m. Find the volume of the rocket.
3. A cup is in the form of a hemisphere surmounted by a cylinder. The height of the cylindrical portion is 8cm and the total height of the cup is 11.5cm. Find the TSA of the cup.
4. A storage tank consists of a circular cylinder with a hemisphere adjoined on either ends. The external diameter of the cylinder is 1.4 m and length is 8m, find the cost of painting it on the outside at the rate of Rs. 10 per m2.
5. A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 16 cm and its height is 15 cm. Find the cost of painting the toy at 7 per Rs. 100 cm2.
6. A circus tent is cylindrical up-to a height of 3m and conical above it. If the diameter of the base is 105m and the slant height of the conical part is 53m, find the total cost of canvas used to make the tent if the cost of the canvas per m2 is Rs. 100.

## Mensuration Exercise 15.5 – Solutions

1. A petrol tank is in the shape of a cylinder with hemisphere of same radius attached to both ends. If the total length of the tank is 6m and the radius is 1m, what is the capacity of the tank in litres.

Solution:

r = 1m

total length of the tank = 6m

h = 6 – (1 + 1) = 6 – 2 = 4 m

Volume of the tank = volume of the hemisphere + volume of the cylinder + volume of the hemisphere

= 2/3πr3 + πr2h + 2/3πr3

= 2/3 x 22/7­ x 13 + 22/7 x 12 x 4 + 2/3 x 22/7 x 13

= 16.76 m

Capacity of the water tank in litres = 16.76 x 1000 = 16761.9 l

1. A rocket is in the shape of a cylinder with a cone attached to one end and a hemisphere attached to the other. All of them are of the same radius of 1.5m. The total length of the rocket is 7m and height of the cup is 2m. Find the volume of the rocket.

Solution:

r = 1.5 m

Height of the cylinder, h = 7 – (1.5 + 2) = 7 – 3.5 = 3.5 m

height of the cone = 2m

Volume of the rocket = Volume of the cone + volume of the cylinder + volume of hemisphere

= 1/3πr2h+ πr2h + 2/3πr3

= 1/3 x 22/7 x (1.5)2 x 2 + 22/7 x (1.5)2 x 3.5 + 2/3 x  22/7 x (1.5)3

= 36.53 m3

1. A cup is in the form of a hemisphere surmounted by a cylinder. The height of the cylindrical portion is 8cm and the total height of the cup is 11.5cm. Find the TSA of the cup.

Solution:

total height of the cup = 11.5 cm

Height of the cylindrical portion = 8 cm

Height of the hemisphere = 11.5 – 8 = 3.5 cm

r = 3.5 cm

TSA of the cup = TSA of cylinder + TSA of hemisphere

= 2πrh + 2πr2

= 2 x 22/7 x 3.5 x 8 + 2 x 22/7 x 3.52

= 253 cm2

1. A storage tank consists of a circular cylinder with a hemisphere adjoined on either ends. The external diameter of the cylinder is 1.4 m and length is 8m, find the cost of painting it on the outside at the rate of Rs. 10 per m2.

Solution:

External diameter = 1.4 m

Radius = d/2 = 1.4/2 = 0.7 m

length = 8m

length of the hemisphere = 8 – (0.7+0.7) = 6.6 m

Total surface area of the storage tank = TSA of the cylindrical part + 2x TSA of  hemispherical part

= 2πr(r +h) + 2 x 3 πr2

= 2x22/7x(0.7)(0.7+6.6)+2x3x22/7 x (0.7)2

= 41.36 m2

Cost of polishing at the rate of  Rs. 10 per m2 = 41.34 x 10 = Rs. 413.6

1. A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 16 cm and its height is 15 cm. Find the cost of painting the toy at 7 per Rs. 100 cm2.

Solution:

diameter of the base of the cone = 16 cm

radius of the base of the cone = 8 cm

height = 15 cm

l2 = r2 + h2

= 82 + 152

= 64 + 25

= 289

l = 17 cm

Surface area of the toy = CSA of hemisphere + CSA of cone

= 2πr2 + πrl

= 2 x 22/7 x 82 + 22/7 x 8 x 17

= 829.71 cm2

Cost of painting the toy at Rs. 7 per 100 cm2 = 829.71×7/100 = Rs. 58.08

1. A circus tent is cylindrical up-to a height of 3m and conical above it. If the diameter of the base is 105m and the slant height of the conical part is 53m, find the total cost of canvas used to make the tent if the cost of the canvas per m2 is Rs. 100.

Solution:

d = 105 m

r = d/2 = 105/2 = 52.5 m

h = 3m

l = 53 m

l2 = r2 + h2

532 = 52.52 + h2

532 – 52.52 = h2

52.75 = h2

h = 7.26 m

Total canvas used = CSA of cylindrical part + CSA of conical part

= 2πrh + πrl

= 2 x 22/7 x 52.5 x 3 + 22/7 x 52.5 x 53

= 9735 m2

The total cost of canvas used to make the tent if the cost of the canvas per m2 is Rs. 100 = 9735 x 100 = Rs. 97,35,000

# Ratio and Proportion – Full Chapter – Class IX

After studying the chapter Ratio and Proportion and their general form; to understand and differentiate between different types of proportion; to acquire skills of writing proportion; to solve problems on time and work involving proportions; to apply proportion in day to day life situations.

## 2.4.1 Introduction to Ratio and Proportion

In a ratio a : b, the first term a is called the antecedent and the second term b is called the consequent. Ratio is an abstract quantity and has no unit. Ratio tells how many times the first term is there in the second term.

Example 1: In the adjacent figure, find the ratio of the shortest side of the triangle to the longest side.

Solution:

We see that the shortest side is of length 5 cm and the longest side is of length 13 cm. Hence the ratio is 5:13

Example : Suppose the ratio of boys to girls in a school of 720 students is 7 : 5. How many more girls should be admitted to make the ratio 1 : 1?

Solution:

For every 7 boys there are 5 girls. Thus out of 12 students, 7 are boys and 5 are girls. Hence the number of boys is 7/12 x 720 = 420.

The number of girls is 720 – 420 = 300. Now we want the ratio of boys to girls to be 1:1. This means the number of boys and girls must be same. Since the deficiency of  girls is 420 – 300 = 120, the school must admit 120 girls to make the ratio 1:1.

Example 5: Consider the ratio 12:5 If this ratio has to be reduced by 20% which common number should be added to both the numerator and denominator?

Solution:

Consider 12/5. This has to be reduced by 20%. This means we have to consider 80% of this number. Thus, we must get,

12/5 x 80/100 = 48/25

We have to find a such that,

12+a/5+a = 48/25

Cross multiplying

25(12 + a) = 48(5 + a)

48a – 25a = (25 x 12) – (48 x 5) =

23a = 60

a = 60/23

If we add 60/23 to both terms of 12 : 5 we get a ratio which is 20% less than the original ratio.

### Ratio and Proportion – Exercise 2.4.1

1.Write each of these ratios in the simplest form.

(i) 2:6

(ii)24:4

(iii) 14:21

(iv) 20: 100

(v) 18:24

(vi) 22:77

Solution:

(i) 2:6 = 1:3 (dividing both by 2)

(ii) 24:4 = 6:1 (dividing both by 2)

(iii) 14:21 = 2:3 (dividing both by 7)

(iv) 20:100 = 1:5 (dividing both by 2)

(v) 18:24 = 3:4 (dividing both by 6)

(vi) 22:77 = 2:7 (dividing both by 11)

2. A shop-keeper mixes 600 ml of orange juice with 900 ml of apple juice to make a fruit drink. Write the ratio of orange juice to apple juice in the fruit drink in its simplest form.

Solution:

Ratio of volumes of

Orange juice and apple juice O:A

= 600:900

= 6:9

= 2:3

3. a builder mixes 10 shovels of cement with 25 shovels of sand. Write the ratio of cement to sand.

Solution:

Ratio of cement to sand = 10 shovels :25 shovels

4.In a school there are 850 pupils and 40 teachers. Write the ratio of teachers to pupils.

Solution:

Number of teachers : Number of pupils

= 40 : 850 = 4:85

5. On a map, a distance of 5cm represent an actual distance of 15km. Write the ratio of the scale of the map.

Solution:

Let x be the number to be added them

(49 + x) = (68 + x) = 3:4

4(49+X) = (68 + X)3

196+4X = 204 + 3X

4X – 3X = 204 – 196

x = 8

## 2.4.2 Proportion – Ratio and Proportion

### Ratio and Proportion – Exercise 2.4.2

1. In the adjacent figure, two triangles are similar find the length of the missing side

Solution:

Let the triangles be ABC and PQR

BC/QR = AC/PR

5/X  = 13/39

13X = 5 X 39

X = 5X39/13  = 5 X3 = 15

1. What number is to 12 is 5 is 30?

Solution

Let x be the number

x:12 :: 5 : 30

30x = 12×5

x = 12×5/30   = 2

1. Solve the following properties:

(i). x : 5 = 3 : 6

(ii) 4 : y = 16 : 20

(iii) 2 : 3 = y : 9

(iv) 13 : 2 = 6.5 : x

(v) 2 : π = x : 22/7

Solution:

(i). x : 5 = 3 : 6

6x = 5 x 3

6x = 15

x = 15/6

(ii) 4 : y = 16 : 20

4×20 = 16y

y = 4×20/16

y = 5

(iii) 2 : 3 = y : 9

2×9 = 3y

y = 2×9/3 = 2×3 = 6

(iv) 13 : 2 = 6.5 : x

13x = 2 x 6.5

13x = 13

x = 13/13 = 1

(v) 2 : π = x : 22/7

2x22/7 = πx

x = (2x22/7) /π =(2x22/7) /(22/7)

x = 2

1. find the mean proportion to :

(i) 8, 16

(ii) 0.3, 2.7

(ii)162/3 , 6

(iv)  1.25, 0.45

Solution:

(i) 8, 16

Let x be the mean proportion to 8 and 16

Then 8/x = x/16

x2 = 8 x 16 = 128

x = √128 = √(64×2) = 8√2

(ii) 0.3, 2.7

Let x be the mean proportion to 0.3 and 2.7

Then 0.3/x = x/2.7

x2 = 0.3 x 2.7 = 0.81

x = √(0.81) = 0.9

(ii)162/3 , 6

Let x be the mean proportion to 162/3  and 6

Then (162/3) /x = x/6

x2 = 162/3  x 6 = 50/3 x 6 = 100

x = √100 = 10

(iv)  1.25, 0.45

Let x be the mean proportion to 1.25 and 0.45

Then 1.25/x = x/0.45

x2 = 1.25 x 0.45

x = √(1.25 x 0.45) = √(1.25 x 0.45)x√(100×100)/ √(100×100)

= √(125×45)/√(100×100) = √(25x5x5x9)/√(100×100)  = 5x5x3/10×10  = 3/4

1. Find the fourth proportion for the following:

(i) 2.8,  14, 3.5

(ii) 31/3, 12/3, 21/2

(iii)15/7, 23/4, 33/5

Solution:

(i) 2.8,  14, 3.5

Let x be the  fourth proportion

Then, 2.8 : 14  :: 3.5  : x

2.8x = 14×3.5

x  = 14×3.5/2.8 = 17.5

(ii) 31/3, 12/3, 21/2

Let x be the  fourth proportion

Then, 31/3: 12/3  :: 21/2: x

10/3 :5/3 : : 5/2 : x

10/3x = 5/3 x 5/2

10/3x = 25/6

x  = 25/6 x 3/10 = 75/60  = 15/12 = 5/4

(iii)15/7, 23/14, 33/5

Let x be the  fourth proportion

Then, 15/7: 23/14:: 33/5: x

12/7 :31/14 : : 18/5 : x

12/7 x = 31/14 x 18/5

12/7 x = 31×18/14×5

x  = 31×18/14×5 x 7/12 = 31×3/5×4 = 93/20 = 413/20

1. Find the third proportion to:

(i) 12, 16

(ii) 4.5, 6

(iii) 51/2 , 161/2

Solution:

(i) 12, 16

Let x be the third proportion

Then 16:12 :: x : 16

12x = 16 x16

x = 16×16/12 = 64/3  = 211/3

(ii) 4.5, 6

Let x be the third proportion

Then 6:4.5 :: x : 6

4.5x = 6 x6

x = 6×6/4.5 = 36/4.5  = 360/45  = 8

(iii) 51/2 , 161/2

Let x be the third proportion

Then 161/2: 51/2:: x : 161/2

11/2 x = 33/2 x 33/2

x = 33/2 x 33/2 x 2/11= 33×3/2  = 99/2  = 491/2

1. In a map 1/4 cm represents 25km, if two cities are 21/2c apart on the map, what is the actual distance between them?

Solution:

Let 21/2 cm represemts x km

1/4cm: 25km :: 21/2cm : x km

1/4 x x = 25 x 21/2

x/4 = 25 x5 /2

x = 25×5/2 x 4 = 25x5x2 = 250km

1. Suppose 30 out of 500 components for a computer were found defective. At this rate how many defective components would he found in 1600 components?

Solution:

Number of defective components in 500 components = 30

Let x be the number of defective components in 1600 components

then 30:500 :: x :1600

30×1600 = 500x

x = 30×1600/500 =96

## 2.4.3 Time and Work – Ratio and Proportion

### Ratio and Proportion – Exercise 2.4.3

1. Suppose A and B together can do a job in 12 days, while B alone can finish a job in 24 days. In how many days can A alone finish the work?

Solution:

Number of days in which A and B together can finish the work = 12 days

Number of days in which B alone can finish the work = 30

1/T = 1/m + 1/n

1/12 = 1/m + 1/30

1/m = 1/301/12 = 5-2/60 = 3/60 = 1/20

A can finish the work in 20 days.

Suppose A is twice as good a workman as B and together they can finish a job in 24 days. How many days A alone takes to finish the job?

Solution:

A is twice as good a workman as B

i.e if B can finish a work in t days A can finish it in 1/2 days

1/T = 1/m + 1/n

1/24 = 1/t/2 + 1/t = 2/t + 1/t = 3/t

1/24 = 3/t

t = 24 x 3 = 72

i.e, B takes 72days to finish the job

A takes 72/2 = 36 days to finish it

1. Suppose B is 60% more efficient them A. if A can finish a job in 15 days how many days B needs to finish the same job?

Solution:

A can finish a work in 15 days.

Work done A in 1 day = 1/15

B is 60% more efficient

Work done by B in 1 day

1/15 + 1/15 x 60/100

= 1/15 (1 + 60/100)

= 1/15 ( 8/5)

= 8/75

Number of days in which B alone can finish the work = 1/(8/75)  = 75/8 = 93/8 days

1. Suppose A can do a piece of work in 14 days while B can do it in 21 days. They begin together and worked at it for 6 days. Then A fell ill B had to complete the work alone. In how many days was the work completed?

Solution:

M = 14 days

N = 21 days

Part of work done in 6 days

= (1/14 + 1/21)6

= 6(3+2/42) = 5×6/42 = 5/7

Remaining part of the work = 1-5/7 = 2/7

Days taken by B to finish

2/7 part of the work = (2/7)/(1/21) = 2/7 x 21/7 = 6 days

Total number of days in which the work is completed = 6+6 = 12 days

1. Suppose A takes twice as much time as B and thrice as much time as C to complete a work. If all of them work together they can finish the work in 2 days. How much time B and C working together will take to finish it?

Solution:

If A alone takes to t1 days to do the work , B finishes it in t1/2  and C is t1/3 days

1/T = 1/t1 +1/t2 + 1/t3

= 1/t1 +1/(t1/2) + 1/(t1/3)

= 1/t1 +2/t1 + 3/t1

= 6/t1

1/T = 1/2

1/2 = 6/T1

i.e. t1 = 12 dyas

B takes 12/2 = 6days

C takes 12/3 = 4 days

Part of work done by B

In one day = 1/6

Part of work done by C in one day = 1/4

If B and C together takes t days to finish the work 1/T = 1/6 + 1/4 = 2+3/12 = 5/12

T = 12/5 = 2.4 days

# Cube Root – Ch.2 – cubes and cube roots – Class VIII

If N is number and n is another number such that N = n3, we say n is the cube root of N and write n =  cube root of N .

i.e.,

## Examples for Cube root

Example 12: Find the cube root of 216 by factorization.

Solution:

216 = 2 x (108) = 2 x 2 x 54 = 2 x 2 x 2 x 27 = 2 x 2 x 2 x 3 x 9 = 2 x 2 x 2 x 3 x 3 x 3

216 = (2 x 3) x (2 x 3) x (2 x 3)

216  = 6 x 6 x 6 = 63

Example 13: find the cube root of -17576 using factorization.

Solution:

-17576 = 2 x (-8788) = 2 x 2 x (-4394) = 2 x 2 x 2 x (-2197) = 2 x 2 x 2 x (-13) x (-13) x (-13)

= [2 x (-13)] x [2 x (-13)] x [2 x (-13)]

= (-26) x (-26) x (-26)

= (-26)3

Example 15: Find the cube root of 103823.

Solution:

Here unit in the digits place is 3. If n3 = 103823. Then, the unit in the digits place of n must be 7.

Let us split this as 103 and 823.

We observe that, 43 = 64 < 103 < 125 = 53.

Hence 403 = 64000 < 103823 < 125000 = 503. Hence n must lie between 40 and 50. Since the unit in the digits place is 7, therefore n must be 47.

473 = 103823.

## Cube root – Chapter 2 – Exercise 1.2.8

1. ### Find the cube root by prime factorization.

i) 10648

Solution:

10648 =  2 x 5324

= 2 x 2 x 2662

= 2 x 2 x 2 x 1331

= 2 x 2 x 2 x 11 x 121

= 2 x 2 x 2 x 11 x 11 x 11

= (2 x 11) x (2 x 11) x (2 x 11)

= 22 x 22 x 22

= 223.

ii) 46656

Solution:

46656 = 2 x (23328)

= 2 x 2x (11664)

= 2 x 2 x 2 x (5832)

= 2x2x2x2x(2916)

= 2x2x2x2x2x(1458)

= 2x2x2x2x2x2x(729)

= 2x2x2x2x2x2x9x(81) = 2x2x2x2x2x2x9x9x9 = (2x2x9)x(2x2x9)x(2x2x9) = 36 x 36 x36

= 363.

iii)15625

Solution:

15625 = 5 x (3125)

= 5 x 5 x (625)

= 5 x 5 x 5 x (125)

= 5 x 5 x 5 x 5 x (25)

= 5 x 5 x 5 x 5 x 5 x 5

= (5×5)x(5×5)x(5×5)

= (25) x (25) x (25)

= 253

1. ### Find the cube root of the following by looking at the last digit and using estimation.

i) 91125

Solution:

Here unit in the digits place is 5. If n3 = 91125. Then, the unit in the digits place of n must be 5.

Let us split this as 91 and 125.

We observe that, 43 = 64 < 91 < 125 = 53.

Hence 403 = 64000 < 91125 < 125000 = 503. Hence n must lie between 40 and 50. Since the unit in the digits place is 5, therefore n must be 45.

ii) 166375

Solution:

Here unit in the digits place is 5. If n3 = 166375. Then, the unit in the digits place of n must be 5.

Let us split this as 166 and 375.

We observe that, 53 = 125 < 166 < 216 = 63.

Hence 503 = 12500 < 166375 < 216000 = 603. Hence n must lie between 50 and 60. Since the unit in the digits place is 5, therefore n must be 55.

iii) 704969

Solution:

Here unit in the digits place is 9. If n3 = 704969. Then, the unit in the digits place of n must be 9.

Let us split this as 704 and 969.

We observe that, 83 = 512 < 704 < 729 = 93.

Hence 803 = 512000 < 704969 < 729000 = 903. Hence n must lie between 80 and 90. Since the unit in the digits place is 9, therefore n must be 89.

1. ### Find the nearest integer to the cube root of each of the following.

i) 331776

Solution:

For easy simplification let us split 331776 as 331 and 776,

63 = 216 < 331 < 343 = 73

603 = 216000 < 331776 < 343000 = 703.

Now it is closer to 703 than 603.

Let us go for more accurate, 693 = 328509 < 331776 < 34300 = 703.

Therefore , 331776 is closest to 693.

ii) 46656

Solution:

For easy simplification let us split 46656 as 46 and 656,

33 = 27 < 46 < 64 = 43

303 = 27000 < 46656 < 64000 = 403.

Now it is closer to 403 than 303.

The number closer to 40 than 30 are 36, 37, 38, 39. Let us go through one by one.

363 = 46656, satisfies the condition.

iii. 373248

Solution:

For easy simplification let us split 372248 as 373 and 248,

73 = 343 < 373 < 512 = 83

703 = 343000 < 373248 < 512000 = 803.

Now, it is closer to 703 than 803.

The numbers which are closer to 70 than 80 are : 71, 72, 73, 74, 75

Let us go for more accurate, 713 = 357911 < 373248 < 373248 = 723.

Therefore, 373248 is closest to 723

Squares, Square roots, Cubes and Cube roots

# Perfect cubes – Chapter 2- Class VIII

Let,

1 = 1 x 1 x 1;

8 = 2 x 2 x 2;

27 = 3 x 3 x 3;

125 = 5 x 5 x 5;

We observe that, each number is written as a product of 3 equal integers.

We say that an integer N is a perfect cube if N can be written as a product of three equal integers. If N = m x m x m. we say N is the cube of m and write N = m3 (read as cube of m or simply m-cube)

Consider a few more examples:

(-4) x (-4) x (-4) = -64 = (-4)3

(-5) x (-5) x (-5) = -125 = (-5)3

We see that the negative numbers are also perfect cubes.

Example 7: Find the cube of 6.

Solution:

63 = 6 x 6 x 6 = 216

Example 8: What is the cube of 20?

Solution:

203 = 20 x 20 x 20 = 400 x 20 = 8000.

Example 9: If a cube has side length 10cm, what is its volume?

Solution:

Volume, V = 10 x 10 x 10 = 1000m3.

## Perfect cubes – Exercise 1.2.7

1. ### Looking at the pattern, fill in the gaps in the following:

 2 3 4 -5 —— 8 —— 23 = 8 33 = —- —- = 64 —- = —- 63= —– —- = —- — = -729

Solution:

 2 3 4 -5 6 8 -9 23 = 8 33 = 27 43 = 64 (-5)3 = -125 63= 216 83 = 512 (-9)3 = -729

1. ### Find the cubed of the first five odd natural numbers and the cubes of the first five even natural numbers. What can you say about parity of the odd cubes and even cubes?

Solution:

 Odd cubes Even cubes 13 = 1 23 = 8 33 = 27 43 = 64 53 = 125 63 = 216 73 = 343 83 = 512 93 = 729 103 = 1000

Parity:

Cubes of odd numbers is always odd.

Cubes of even numbers is always even.

1. ### How many perfect cubes you can find from 1 to 100 ? How many from – 100 to 100?

Solution:

We have 4 cubes from 1 to 100, those are,

13 = 1, 23 = 8, 33 = 27, 43 = 64. The next cube will be 53 = 125, which is larger than 100. so we have only 4 cubes from 1 to 100.

We have 8 cubes from -100 to 100. Those are, (-1)3 = 1, (-2)3 = 8, (-3)3 = 27, (-4)3 = 64 and 13 = 1, 23 = 8, 33 = 27, 43 = 64.

1. ### How many perfect cubes are there from 1 to 500? How many are perfect square among cubes?

Solution:

We have 7 cubes from 1 to 500, i.e., 13 = 1, 23 = 8, 33 = 27, 43 = 64, 53 = 125, 63 = 216, 73 = 343.

1. ### Find the cubes of 10, 30 , 100, 1000. What can you say about the zeros at the end?

Solution:

103 = 1000

303 = 27000

1003 = 1000000

10003 = 10000000000

The number of zero of a cube are 3 times, the no. of zero of numbers

1. ### What are the digits in the unit’s place of the cubes 1, 2, 3, 4, 5, 6, 7, 8 , 9, 10? Is it possible to say that a number is not a perfect cube by looking at the digit in unit’s place of the given number, just like you did for squares?

Solution:

 a a3 digit in the units place 1 1 1 2 8 8 3 27 7 4 64 4 5 125 5 6 216 6 7 343 3 8 512 2 9 729 9 10 1000 0

No, its not possible to tell a number is not a perfect cube by looking at the digit in unit’s place of the given number.

# Perfect squares near to a given number – chapter 2

### Examples to find perfect squares near to a given number

Example 5: If the area of a square is 90 cm2, what is its side-length rounded to the nearest integer?

Solution:

Since, area A = a2, we have a2 = 90.

But 81 < 90 < 100 and 81 is the nearer to 90 than 100. Hence, the nearest integer to √90 is √81 = 9.

Example 6: A square piece of land has area 112m2. What is the closest integer which approximates the perimeter of the land?

Solution:

If a is the side length of a square, its perimeter is 4a. We know that a2 = 112 m2.

Hence, (4a)2 = 16a2 = 16 x 112 = 1792

But 422 = 1764 < 1792 < 1849 = 432 .

!764 is nearer to 1792 than 1849. Therefore, the integer approximation for √1792 is 42. The approximate value of perimeter is 42cm.

## Perfect squares near to a given number – Exercise 1.2.6

1. Find the nearest integer to the square root of the following numbers:

i) 232

Solution:

We have, 152 = 225 and 162 = 256.

We know, 225 < 232 < 256

152 < 232 < 162

∴ Square root of 232 is nearest to 15

ii) 600

Solution:

We have, 242 = 576 and 252 = 625.

We know, 576 < 600 < 625

242 < 600 < 252

∴ Square root of 600 s nearest to 24.

iii) 728

Solution:

We have, 262 = 676 and 272 = 729

676 < 728 < 729

262 < 728 < 272

∴ Square root of 728 is nearest to 27.

iv) 824

Solution:

We have, 282 = 784 and 292 = 841

784 < 824 < 841

282 < 824 < 292

∴Square root of 824 is nearest to 29

v) 1729

Solution:

We have, 412 = 1681 and 422 = 1764

1681 < 1729 < 1764

412 < 1729 < 422

∴Square root of 1729 is nearest to 42

1. A piece of land is in the shape of a square and its area is 1000m2 . This has to be fenced using barbed wire. The barbed wire is available only in integral lengths. What is the minimum length of the barbed wire that has to be bought for this purpose.

Solution:

Area of the land = 1000m2

Area = a2 = 1000m2

Perimeter = 4a

Squaring , (Perimeter)2 = ( 4a)2 = 16a2 = 16 x 1000

(perimeter)2 = 16,000

We have, square of the perimeter i.e., 16000. now we have to find the square root of 16000. As 16000 does not have a perfect square root, let us find the square root nearest to square root of 16000.

We know, 1262  = 15876 and 1272 = 16129.

Then, 15876 < 16000 < 16129

1262 < 16000< 1272

∴ Nearest number is 126 But this is not enough to cover the hard.

∴Length of barbed wire required = 127 m.

1. A student was asked to find √961. He read it wrongly and found √691 to the nearest integer. How much small was his number forms the correct answer?

Solution:

We know, √ 961 = 31

It is given that student read it wrongly and found the result for √691. So, we have to find out the difference between  √961 and √691.

√676 < √691 < √729

262 < √691 < 272

26 is nearest number to √ 691

∴ difference = 31 – 26 = 5

# Square root of a perfect square by factorization – Chapter 2

## Examples for Square root of a perfect square by factorization

Example 1: Find the square root of 5929.

Solution:

Thus, 7 x 7 x 11 x 11 = 5929

Therefore, [group it]

(7 x 11) x (7 x 11) = 77 x 77 = 772 = 5929.

Example 2: Find the square root of 6724.

Solution:

Thus, 6724 = 2 x 2 x 41 x 41

6724 = (2 x 41) x (2 x 41)

= 82 x 82

Example 3: Find the smallest positive integer with which one has to divide 336 to get a perfect square.

Solution:

We observe that 336 = 2 x 2 x 2 x 2 x 3 x 7. Here both 3 and 7 occur only once. Hence we have to remove them to get a perfect square.

We divide 336 by 3 and 7

42.

The required least number is 21.

## Square root of a perfect square by factorization – Exercise 1.2.5

1. Find the square roots of the following numbers by factorization:

i) 196

Solution:

196 = 2 x 2 x 7 x 7

= (2 x 7) x (2 x 7)

= 14 x 14

196 = 142

ii) 256

Solution:

256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

= (2 x 2 x 2 x 2) x (2 x 2 x 2 x 2)

= 16 x 16

= 162

iii) 10404

Solution:

10404 = 2 x 2 x 3 x 3 x 17 x 17

= (2 x 3 x 17) x (2 x 3 x 17)

= (102) x (102)

= 1022

iv) 1156

Solution:

1156 = 2 x 2 x 17 x 17

= (2 x 17) x (2 x 17)

= 34 x 34

= 342

v) 13225

Solution:

13225 = 5 x 5 x 23 x 23

= (5 x 23) x (5 x 23)

= 115 x 115

= 1152

2. Simplify:

i) √100 + √36

Solution:

√100 + √36 = 10 + 6

= 16

ii) √(1360 + 9)

Solution:

√(1360 + 9) = √1369 = √(37 x 37) = 37

iii) √2704 + √144 + √289

Solution:

√2704 + √144 + √289 = √(52 x 52) + √(12 x 12) + √(17 x 17)

= 52 + 12 + 17

= 81

iv) √225 – √25

Solution:

√225 – √25 = √(15 x 15) – √(5 x 5)

= 15 – 5

= 10.

V) √1764 – √1444

Solution:

√1764 – √1444 = √(42 x 42) – √(38 x 38)

= 42 – 38

= 4

vi) √169 x √361

√169 x √361 = √(13 x 13) x √(19 x 19)

= 13 x 19

= 247

1. A square yard has area 1764m2. From a corner of this yard, other square part of area 784m2 is taken out for public utility. The remaining portion is divided in to 5 equal parts. What is the perimeter of each of these equal parts?

Solution:

Area of square = a2

Area of given square yard = 1764 m2

Area used for public utility = 784 m2

∴ Area of remaining portion = 1764 – 784 = 980 m2

If the area is divided into 5 parts

Then, the area of each Square = 980/5 = 196 m2

Length of each side = √ 196, a = 14m

Perimeter of square = 4a = 4 x 14 = 56m

4. Find the smallest positive integer with which one has to multiply each of the following numbers to get a perfect square:

i) 847

Solution:

847 = 11 x 11 x 7

Here 7 occur only once. Hence we have to multiply by 7 them to get a perfect square.

ii) 450

Solution:

450 = 5 x 5 x 3 x 3 x 2

Here 2 occur only once. So we have multiply by 2 to get a perfect square.

iii) 1445

Solution:

1445 = 17 x 17 x 5

Here 5 occur only once. So we have multiply by 5 to get a perfect square.

iv) 1352

Solution:

1352 = 2 x 2 x 13 x 13 x 2

Here 2 occur only three times. So we have multiply by 2 to get a perfect square.

5. Find the largest perfect square factor of each of the following numbers:

i) 48

Solution:

48 = 2 x 2 x 2 x 2 x 3 = (2 x 2) x (2 x 2) x 3

= 4 x 4 x 3 = 16 x 3 = 42 + 3

Therefore, 16 is the largest perfect square factor of 48.

ii) 11280

Solution:

11280 = 2 x 2 x 2 x 2 x 3 x 47

Therefore, 16 is the largest perfect square factor of 11280

iii) 729

Solution:

729 = 3 x 3 x 3 x 3 x 3 x 3

= (27) x (27)

= 272

Therefore, 729 is the largest perfect square factor of 729.

iv) 1352

Solution:

1352 = 2 x 2 x 2 x 13 x 13

= (2 x 13) x (2 x 13) x 2

= 26 x 26 x 2

Therefore, 676 is the largest perfect square of 1352.

1. Find a proper positive factor of 48 and a proper positive multiple of 48 which add up to a perfect square. Can you prove that there are infinitely many such pairs?

Solution:

The factor of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. Multiples of 48 are = 48, 96 , 144, 192 , 240, 280 ….

1) 48 + 1 = 49 = 72

2 ) 192 + 4 = 196 = 142

3)288 + 1 = 289 = 172

4 ) 96 + 4 = 100 = 102

5)96 + 48 = 144 = 122

6) 48 + 16 = 64 = 82

7)240 + 16 = 256 = 162

Yes, we can prove that there are infinitely many such fairs.

# 1.2.2 Perfect squares – Chapter 2 – Class VIII

Observe that, 1 = 1, 4 = 2 x 2, 9 = 3 x 3, 100 = 10 x 10. If a is an integer and b = a x a, we say b is a perfect square. Hence 1, 4, 9, 16, 25 are all perfect squares. Since 0 = 0 x 0, we see that 0 is a perfect square. If a is an integer, we denote a x a = a2. We read it as square of a or simply a square. Thus 36 = 62 and 81 = 92. Thus a perfect square is of the m2, where m is an integer.

For example, 4 = 2 x 2 and 4 = (-2) x (-2); in the second representation, we again have equal integers, but negative this time.

Thus a perfect square is either equal to 0 or must be a positive integer. It can be a negative integer.

Look at the following table

 a 1 2 3 8 -7 -12 20 -15 a2 1 4 9 64 49 144 400 225

We see that squares of 2, 8, -12, 20 are even numbers and the squares of 1, 3, -7, -15 are odd numbers.

### Statement 1: The Square of an even integer is even and the square of an odd integer is odd.

Consider first 10 perfect squares,

 12 22 32 42 52 62 72 82 92 102 1 4 9 16 25 36 49 64 81 100

If we observe that the units place in these squares are 1, 4, 9, 6, 5, 6, 9, 4, 1 and 0 in that order. Thus only the digits which can occupy digit’s place in perfect squares are 1, 4, 5, 6 and 9.

## Perfect squares – Exercise 1.2.2

1. Express the following statements mathematically:

i) square of 4 is 16

Solution:

42 = 16

ii) square of 8 is 64

Solution:

82 = 64

iii) square of 15 is 225

Solution:

152 = 225

1. Identify the perfect squares among the following numbers:

1, 2, 3, 8, 36, 49, 65, 67, 71, 81, 169, 625, 125, 900, 100, 1000, 100000

Solution:

Perfect squares among the above are:

1, 36, 49, 81, 169, 625, 125, 900, 100

1. Make a list of all perfect squares from 1 to 500

Solution:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484

1. Write the 3 digit numbers ending with 0, 1, 4, 5, 6 , 9 one for each digit but one of them is a perfect square.

Solution:

We can take 200, 201, 204, 205, 206 and 209. None of these is perfect square lies between 142 = 196 and 152 = 225.

We can also take 300, 301, 304, 305, 306 and 309. None of these is a perfect square lies between 172 = 289 and 182 = 324.

1. Find the numbers from 100 to 400 that end with 0, 1, 4, 5, 6 or 9, which are perfect squares.

Solution:

102 = 100, 112 = 121, 122 = 144, 132 = 169, 142 = 196, 152 = 225, 162 = 256, 172 = 289, 182 = 324, 192 = 361, 202 = 400

## 1.2.3 Some facts related to perfect squares – Perfect squares

There are some nice properties about perfect squares. Let us study them here:

1. Look at following table:
 a 4 10 20 25 100 300 1000 a2 16 100 400 625 10000 90000 1000000 The number of zeros at the end of a2 0 2 2 0 4 4 6

We observe that, the number of zeros at end of a square is always an even number.

STATEMENT 3: if a number has k zeros at the end, then its square ends in 2k zeros

Thus, if a number has odd number of zeros, it cannot be perfect square.

1. Look at the adjoining table:
 a a2 The remainder of a2when divided by 3 The remainder of a2when divided by 4 1 1 1 1 2 4 1 0 3 9 0 1 5 25 1 1 8 64 1 0 11 121 1 1 -6 36 0 0

We can see that, the remainder of a perfect square is either 1 or 0. we can also note that remainder of a perfect square when divided by 4 are either 0 or 1.

Since, when a number is divided by 3, the reminder used to be 0, 1 or 2. Similarly, when a number is divided by 4 the remainder used to be 0, 1, 2 or 3.

STATEMENT 4: The reminder of a perfect square when divided by 3 is either 0 or 1, but never be 2. The reminder of a perfect square when divided by 4 is either 0 or 1, but never be 2 and 3.

STATEMENT 5: When the product of 4 consecutive integers is added to 1, the resulting number is perfect square.

Ex:

(1 x 2 x 3 x 4) + 1 = 24 + 1 = 25 = 52

(5 x 6 x 7 x 8) + 1 = 1680 + 1 = 412

STATEMENT 6: The sum of the first n odd natural numbers is equal to n2, for every natural number n.

Ex:

1 = 1 = 1; 1 + 3 = 4 = 2; 1 + 3 + 5 = 9 = 3; 1 + 3 + 5 + 7 = 16 = 42  ; 1 + 3 + 5 + 7 + 9 = 25 = 52

STATEMENT 7: consider the number N = 1000….01, where zeros appear k times. (For example, for k = 6, you get N = 10000001; there are 6 zeros in the middle.) Then N2 = 1000…02000…01, where the number of zeros on both the sides of 2 is k.

Ex:

112 = 121

1012 = 10201

10012 = 1002001

100012 = 100020001 and so on.

STATEMENT 8: The sum of nth and (n+1)th triangular number is (n+1)2.

Ex:

The dots are now arranged in shapes. Now count the number of dots in each triangle. (Single dot is considered as the generate triangle.) They are 1, 3, 6, 10, 15, 21, 28, 36 and so on. These are called triangular numbers.

For nth triangular number, we form a triangle of dots with n- rows and each row contains as many points as index of that row.  If you want find the 8th triangular number, the number of points in the 8th triangle is

1 + 2 + 3 + 4+ 5 + 6 + 7 + 8 = 36

The first few triangular numbers are:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91.

Take any two consecutive triangular number and find their sum. For example : 10 + 15 = 25 = 52; 21 + 28 = 49 = 72. Which holds the statement 8.

## Perfect Squares – Exercise 1.2.3

1. Find the sum 1 + 3 + 5 + ……. +51 (the sum of all odd numbers from 1 to 51) without actually adding them.

Solution:

[We have 5 odd numbers from 1 to 10, therefore there are 5×5 = 25 odd numbers are there from 1 to 50, we know 51 is also an odd number…. So, there are 25+1 = 26 ]

There are 26 odd numbers are there from 1 to 51.

Thus, 1 + 3 + 5 + …….. + 51 = 262 = 676.

1. Express 144 as a sum of 12 odd numbers.

Solution:

We know that, the sum of the first n odd natural numbers is equal to n2, for every natural number n.

Therefore, the sum of first 12 odd natural numbers is equal to 122 = 144.

Thus,

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 22 = 122 = 144.

1. Find the 14th and 15th triangular numbers, and find their sum. Verify the statement 8 for this sum.

Solution:

We know, statement 8: The sum of nth and (n+1)th triangular number is (n+1)2.

Sum of 14th triangular number is,

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 = 105

Sum of 15th triangular number is,

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 = 120

Now, let us verify, statement 8: The sum of nth and (n+1)th triangular number is (n+1)2.

i.e., Sum of 14th triangular number(n) + sum of 15th triangular number(n+1) = 105 + 120

= 225 = 152 = (n+1)2, which holds the statement 8.

1. What are the remainders of a perfect square when divided by 5?

Solution:

 a a2 The remainders of a2 when divided by 5 1 1 1 2 4 4 5 25 0 6 36 1 -7 49 4 11 121 1

The remainders of perfect squares when divided by 5 are 0, 1 and 4.

### 1.2.4 Methods of squaring a number – Perfect Squares

Many times it is easy to find the square of a number without actually multiplying the number to itself.

Consider 42 = 40 + 2

Thus,

422 = (40+2)2

= 402 + 2(40)(2) + 22

Now it is easy to recognise 402 = 1600; 2x40x2 = 160; 22 = 4.

Therefore, 422 = 1600 + 160 + 4 = 1764.

Example: Find 892

Solution: 892 = (80 + 9)2

= 802 + 2 x 80 x 9 + 92

= 6400 + 1440 + 81

= 7921

### Perfect Squares –  Exercise 1.2.4

1. Find the squares of:

i) 31

Solution:

31 = 30 + 1

312 = (30 + 1)2

= 302 + 2 x 30 x 1 +1

= 900 + 60 + 1

= 961

ii) 72

Solution:

72 = 70 + 2

722 = (70 + 2)2

= 702 + 2 x 70 x 2 + 22

= 4900 + 280 + 4

= 5184

iii) 37

Solution:

37 = 30 + 7

372 = (30 + 7)2

= 302 + 2 x 30 x 7 + 72

= 900 + 420 + 49

= 1369

iii) 166

Solution:

166 = 160 + 6

1662 = (160 + 6)2

= 1602 + 2 x 160 x 6 + 62

= 25600 + 1920 + 36

= 27556

2. Find the squares of:

i) 85

Solution:

852 = (80 + 5)2

= 802 + 2 x 80 x 5 + 52

= 1600 + 800 + 25

=2425

ii) 115

Solution:

1152 = (100 + 15)2

= 1002 + 2 x 100 x 15 + 152

= 10000 + 3000 + 225

= 13225

iii) 165

1652 = (160 + 5)2

= 1602 + 2 x 160 x 5 + 52

= 25600 + 1600 + 25

= 27225

1. Find the squares of 1468 by writing this as 1465+3

14682 = (1465 + 3)2

= (1465)2 + 2 x 1465 x 3 + 32

= 2146225 + 8790 + 9

= 2155024

### 1.2.5 Square roots – Perfect Squares

Consider the following perfect squares:

1 = 12, 4 = 22, 9 = 32, 16 = 42, 49 = 72, 196 = 142.

In each case the number is obtained by the product of two equal numbers. Here we say 1 is the square root of 1; 2 is the square root of 4; Hence 3 is the square root of 9 and so on.

Suppose N is a natural number such that N= M2. The number M is called a square root of N.

We have seen earlier m2 = m x m = (-m) x (-m). Thus m2 has 2 roots m and –m. For example, 16 = 42 = (-4)2, thus both 4 and -4 are the roots of 16. Mathematically both 4 and -4 are accepted as the square root of 16. Therefore, whenever the word square root is used, it is always meant to be the positive square root. The square root on n is denoted by √N.

# Divisibility tests – Playing with Numbers

If a number ends with any of the digits 0, 2, 4, 6, or 8, you immediately say the number is divisible by 2. Why? We write any such number a as a = 10k + r, where r is the remainder when divided by 10. Hence r is one of the numbers 0, 2, 4, 6, 8. We know see that 10 is divisible by 2 and r is also divisible by 2. We conclude that 2 divides a.

## Divisibility tests – Playing with Numbers

1. ### Divisibility by 4 [Divisibility tests]

If a number is divisible by 4, it has to be divisible by 2. Hence the digit in the units place must be one of 0, 2, 4, 6, and 8. But look at the following numbers: 10, 22, 34, 46, and 58. We see that last digit in each of these numbers is as required, yet none of them is divisible by 4.  Thus, we can conclude that it is not possible to decide the divisibility on just reading the last digit. Perhaps, the last two digits may help.

If a number has two digits, we may decide the divisibility by actually dividing it by 4. All we need is to remember the multiplication table for 4. Suppose the given number is large, say it has more than 2 digits. Consider the numbers, for example, 112 and 122.  We see that 112 is divisible by 4. But 112 = 100 + 22; here 100 is divisible by 4 but 22 is not. Hence 122 is not divisible by 4.

We invoke the following fundamental principle on divisibility:

If a and b are integers which are divisible by an integer m ≠ 0, then m divides a+b, a-b and ab.

Now, let us see how does this help us to decide the divisibility of a large number by 4. Suppose we have number a with more than 2 digits. Divide this number by 100 to get a quotient q and remainder r; a = 100q + r, where 0 ≤ r < 100. Since 4 divides 100, you will immediately see that a is divisible by a 4 if and only if r is divisible by 4. But r is the number formed by the last two digits of a. Thus we may arrive at the following test:

STATEMENT 2: A number (having more than 2 digit ) is divisible by 4 if and only if the 2 digit number formed by the last two digits of a is divisible by 4.

Example 4: Check whether 12456 is divisible by 4.

Solution:

Here, the number formed by the last two digits is 56. This is divisible by 4 and hence so is 12456.

Example 5: Is the number 12345678 divisible by 4?

Solution:

The number formed by the last 2 digits is 78, which is not divisible by 4. Hence the given number is not divisible by 4.

1. ### Divisibility by 3 and 9 [Divisibility tests]

Consider the numbers 2, 23, 234, 2345, 23456, 234567. We observe that among these 6 numbers, only 234 and 234567 are divisible by 3. Here, we cannot think of the number formed by the last 2 digits or for those matter even three digits. Note that 3 divides 234, but it does not divide 34. Similarly, 3 divides 456 but it does not divide 23456.

STATEMENT 3: An integer a is divisible by 3 if and only if the sum of digits of a is divisible by 3. An integer b is divisible by 9 if and only if the sum of digits of b is divisible by 9.

Example 6: Check whether the number 12345321 is divisible by 3. Is it divisible by 9?

Solution:

The sum of digits is 1+2+3+4+5+3+2+1 = 21. Hence the number is divisible by 3, but not by 9. In fact 12345321 = (9 x 1371702) + 3.

Example 7: Is 444445 divisible by 3?

Solution:

The sum of digits is 25, which is not divisible by 3. Hence 444445 is not divisible by 3. Here the remainder is 1.

1. ### Divisibility by 5 and 10 [Divisibility tests]

Statement 4: An integer a is divisible by 5 if and only if it ends with 5 or 0. A number is divisible by 10 if and only if it ends with 0.

Example 8: How many numbers from 101 to 200 are divisible by 5?

Solution:

Write the numbers from 101 to 200 which end with 5 and 0. i.e.,

105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 155, 160, 165, 170, 175, 180, 185, 190, 195, 200. There are 20 such numbers.

Example 9: Is the number 12345 is divisible by 15?

Solution:

Note that 15 = 3 x 5. Here again the given number 12345 must be divisible by both 3 and 5.

The sum of the digits is 1+2+3+4+5 = 15, which is divisible by 3 and the given number 12345 ends with 5, so it’s also divisible by 5. Therefore, 12345 is divisible by 15.

Example 10: How many numbers from 201 to 250 are divisible by 5, but not by 3?

Solution:

The numbers which are divisible by 5 from 201 to 250 are 205, 210, 215, 220, 225, 230, 235, 240, 245, 250. Now let us compute the digital sum of these numbers:

205 ——-2+0+5 = 7, not divisible by 3

210 ——– 2+1 = 3, divisible by 3

215 ———2+1+5 = 8, not divisible by 3

220———–2+2+0 = 4, not divisible by 3

225———-2+2+5 = 9, divisible by 3

230———-2+3+0 = 5, not divisible by 3

235———-2+3+5 = 10, not divisible by 3

240———2+4+0 = 6, divisible by 3

245———-2+4+5 = 11, not divisible by 3

250———–2+5+0 = 7, divisible by 3

1. ### Divisible by 11 [Divisibility tests]

STATEMENT 5:

Given number n in decimal form, put alternatively – and + signs between the digits and compute them sum. The number is divisible by 11 if and only if this sum is divisible by 11. Thus a number is divisible by 11 and only if the difference between the sum of the digits in odd places and the sum of digits in even places is divisible by 11.

Example 11: Is the number 23456 divisible by 11?

Solution:

Observe that 2-3+4-5+6 = 4 and hence not divisible by 11. The test indicates that 23456 is not divisible by 11.

A palindrome is a number which leads the same from left to right or right to left. Thus a palindrome is a number n such that by reversing the digits of n, you get back n. For example, 232 is a 3 digit palindrome; 5445 is a 4 digit palindrome.

Example 12: Find all 3 digit palindromes which are divisible by 11.

Solution:

A 3 digit palindrome must be of the form aba, where a≠0 and b are digits. This divisible by 11 if and only if a-b+a = 2a-b is divisible by 11.

This is possible only if 2a-b = 0, 11 or -11. Since a≤1 and b≤9, we see that 2a-b ≥ 2(1)-(9) = 2-9 = -7 > -11. Hence, 2a-b = -11 is not possible.  Suppose 2a-b = 0. Then 2a = b. Thus, a = 1,b = 2 ; a = 2, b = 4; a=3, b = 6; and a = 4, b = 8 are possible.

We get the numbers, 121, 242, 363, 484.

For a = 6, b = 1, we see that 2a-b = 12-1 = 11 and hence divisible by for which 2a-b is divisible by 11. We get four more numbers 616, 737, 858 and 979.

Thus required numbers are 121, 242, 363, 484, 616, 737, 858 979.

Example 13: Prove that 12456 is divisible by 36 without actually dividing it.

Solution:

First notice that 36 = 4 x 9. So it is enough if we prove 12456 is divisible by 4 and 9 both. The last 2 didgits of the given number 12456 is 56 which is divisible by 4 and it is left to show that, the given number 12456 is divisible by 9. Let us find the sum of digits of the given number 12456, i.e., 1+2+4+5+6 = 18, which is divisible by 9. Thus, the number 12456 is divisible by 36.

## Divisibility tests – Playing with Numbers – Exercise 1.1.6

1. ### How many numbers from 1001 to 2000 are divisible by 4?

Solution:

We know that we have 25 numbers from 1001 to 1100. So we have 25X10 times =  250 numbers   from 1001 to 2000 are divisible  by 4.

1. ### Suppose a 3digit number abc is divisible by 3. prove that abc+bca+cab is divisible by 9.

Solution:

Let 3 digit number abc be 123, which is divisible by 3. Niw we have to prove, abc+bca+cab i.e., 123+231+312 is divisible by 9.

(1+2+3)+(2+3+1)+(3+1+2) = 6+6+6 = 18. Therefore abc+bca+cab is divisible by 9.

1. ### If 4a³b is divisible by 11, find all possible values of a+b.

Solution:

The number is divisible by 11 if and only if the sum is divisible by 11.

Therefore, the sum of 4-a+3-b must divisible by 11.

i.e., sum of 7-a-b = 7-(a+b) must be divisible by 11.

Thus, a+b = 18 and a+b = 7.

1. ### Prove that a 4-digit palindrome is always divisible by 11?

Solution:
Suppose X has the digits ‘abba’, then X can be expressed as:
X= 1000a + 100b + 10b + a

= 1001a + 110b

= 11(91a + 10b) where (91a+10b) is an integer.
Therefore, X is definitely divisible by 11.

Therefore, a 4 digit palindrome is always divisible by 11.

1. ### Use the digits 4,5,6,7,8 each once, construct a 5 digit number which is divisible by 132.

Solution:

We have 132 = 11 x 4 x 3. Now we have to construct a 5 digit number using 4,5,6,7,8 each once such that it must be divisible by 11, 4 and 3.

Step 1: The last 2 digits of a required 5 digit number must be 56, 76, 84 and 64. So as to get divided by 4.

Step 2: To get divided by 3, the sum of the required numbers must be divided by 3

4+7+8+5+6 = 30

Step 3: To get divided by 11 , the sum of the required numbers must be divisible by 11.

4-8+5-7+6 = 0, possible

5-8+4-7+6 = 0, possible

6-7+5-8+4= 0 possible

5-7+6-8+4 = 0 possible

# Mensuration – Class VIII

A solid occupies fixed amount of space. Solids occur in different shapes. Observe the following shapes.

These shapes (cuboid, cube, triangular prism, cylinder, cone etc) are known as three dimensional objects.

## Surface area of cuboid- Mensuration

Lateral Surface area of cuboid = 2h(l + b)

Total surface area of cuboid = 2(lb +bh + lh)

Here h – height ; l – length; b – breadth.

## Mensuration – Exercise 4.1.2

1. Find the total surface area of the cuboid with l = 4m, b = 3m and h = 1.5m

Solution:

Total surface area of cuboid = 2(lb +bh + lh)

We have, l = 4m; b = 3m; h = 1.5m

Therefore,

Total surface area of cuboid = 2[(4×3) +(3×1.5) + (4×1.5)]

= 2[ 12 + 4.5 + 6.0]

= 45 m²

2. Find the area of four walls of a room whose length 3.5m, breadth 2.5m and height 3m

Solution:

Area = 2h(l + b)

We have, l = 3.5m; b = 2.5m; h = 3m

Therefore,

Area = 2 x 3 [3.5 + 2.5]

= 6[ 6]

= 36 m²

3. The dimensions of a room are l = 8m, b = 5m, h = 4m. Find the cost of distempering its four walls at the rate of Rs. 40/m²

Solution:

We know that, l = 8m, b = 5m, h = 4m

Area = 2h(l + b)

We have, l = 8m; b = 5m; h = 4m

Therefore,

Area = 2 x 4 [ 8 + 5]

= 8[ 13]

= 104 m²

Thus, the cost of distempering its four walls at the rate of Rs. 40/m² = 104 x 40 = Rs. 4160.

4.A room is 4.8m long, 3.6m broad and 2m high. Find the cost of laying tiles on its floor and its four walls at the rate of Rs. 100/m²

Solution:

Area = 2h(l + b)

We have, l = 4.8m; b = 3.6m; h = 2m

Therefore,

Area = 2 x 2 [ 4.8 + 3.6]

= 4[ 8.4]

=  33.6 m²

Therefore, the area of four walls is 33.6 m²

Now, we have find the area of the floor = (l x b) = ( 4.8 x 3.6) = 17.28 m²

Thus, area of floor + area of 4 walls =  17.28m² + 33.6 m²  = 50.88m²

Thus, the cost of laying tiles on its floor and its four walls at the rate of Rs. 100/m²

= 50.88 x 100

= 50.88 Rs.

5. A closed box is 40cm long, 50cm wide and 60cm deep. Find the area of the foil needed for covering it.

Solution:

Total surface area of cuboid = 2(lb +bh + lh)

We have, l = 40 cm; b = 50 cm; h = 60 cm

Therefore,

Total surface area of cuboid = 2[(40 x 50) +(50 x 60) + (60 x 40)]

= 2[ 2000 + 3000 + 2400]

= 2[ 7400]

= 14800 cm²

6. The total surface area of a cube  is 384cm². Calculate the side of the cube.

Solution:

Total surface area of cube = 6(l²)

384 = 6(l²)

l² = 384/6 = 64 cm²

l = √64 = 8m

Thus, The side of the cube is 8m.

7. The L.S.A of a cube is 64m². Calculate the side of the cube.

Solution:

Lateral surface area of cube = 4(l²)

64 = 4(l²)

l² = 64/4 = 16

Therefore, l = 4m

Thus, the side of the cube is 4l.

8.Find the cost of whitewashing the four walls of a cuboidal room of side 4m at the rate of Rs. 20/m²

Solution:

Lateral surface area of the cuboidal room = 4(l²)

= 4 ( 4²)

= 4 ( 16)

= 64 m²

The cost of whitewashing the cuboidal room at the rate of Rs. 20/m² = 64 x 20 = Rs. 1280

9. A cubical box has edge 10 cm and another cuboidal box is 12.5cm long, 10cm wide and 8cm high.

(i) Which box has smaller total surface total surface area?

(ii) If each edge of the cube is doubled, how many times will its T.S.A increase?

Solution:

(i)

Cube:

Total surface area = 6l² = 6 x 10² = 6 x 100 = 600m²

Cuboid:

Total surface area = 2(lb + bh + lh)

= 2[(12.5 x 10) + (10 x 8) + (8 x 12.5)]

= 2 [ 125 + 80 + 100]

= 610 m²

Thus, Cube has a smaller total surface area.

(ii)

If each edge of the cube is doubled,

Then,

Cube:

Total surface area = 6l² = 6 x 20² = 6 x 400 = 2400m²

Cuboid:

Total surface area = 2(lb + bh + lh)

= 2[(25 x 20) + (20 x 16) + (16 x 25)]

= 2 [ 500 + 320 + 400]

= 2[1220]

= 2440 m²

Thus,

If each edge of the cube is doubled, then its T.S.A will increase 4 times.

## Volume of cubes and cuboids – Mensuration

Volume of cube = l³ , where l = length

Volume of cuboid = area of base x height;

where area of base = l x b, here l = length, b = breadth

## Mensuration – Exercise 4.1.3

1. Three metal cubes whose edges measure 3cm, 4cm and 5cm respectively are melted to form a single cube. Find (i) Side length (ii) total surface area of the new cube. What is the difference between the total surface area of the new cube and the sum of total surface areas of the original three cubes?

Solution:

Three metal cubes whose edges measure 3cm, 4cm and 5cm respectively are melted to form a single cube.

(i) Side length of the new cube = 3 + 4 + 5 = 12cm

(ii) total surface area of the new cube = 6 (l²) = 6(12²) = 6 x 144 = 864 cm²

Total surface area of the cube with edge 3cm = 6 (l²) = 6 x 9 = 54cm²

Total surface area of the cube with edge 4cm = 6 (l²) = 6 x 16 = 96cm²

Total surface area of the cube with edge 5cm = 6 (l²) = 6 x 25 = 150cm²

The sum of total surface area of the cubes = 54 + 96 + 150 = 300cm²

The difference between the total surface area of the new cube and the sum of total surface areas of the original three cubes = 864 cm² – 300 cm² = 564 cm²

2.Two cubes, each of volume 512cm³ are joined end to end. Find lateral and total surface areas of the resulting cuboid.

Solution:

since its a cube , Volume of cube = l³
512 = l³ ; l = 8
side is 8 cm
when you join two cubes , we will get a cuboid whose length is 16 cm , h= 8 , b=8
total s.area of cuboid = 2(lh+bh+lb)
= 2( 128+128+64)
= 640 cm sq
latral .s area = 2h(l+b)
= 2*8 (16+8)
= 384cm²

3. The length, breadth and height of a cuboid are in the ratio 6:5:3. If the total surface area is 504cm², find its dimension. Also find the volume of the cuboid.

Solution:

total surface area is 504cm²

The length, breadth and height of a cuboid are in the ratio 6:5:3

total surface area = 2(lb + bh + lh)

504 = 2(30x² + 15x² + 18x²)

504 = 2 x 63x²

504 = 126x²

x² = 504/126 = 4

x = 2cm

Therefore,

length= 6x = 12cm

height = 3x = 6cm

Volume of cuboid = area of the base x height = l x b x h = 12 x 10 x 6 = 720cm³

4. How many m³ of soil has to be excavated from a rectangular well 28m deep and whose base dimensions are 10m and 8m. Also find the cost of the plastering its vertical walls at the rate of Rs. 15/m²

Solution:

Volume of cuboid = area of the base x height = l x b x h = 28 x 10 x 8 = 2240cm³

Lateral surface area = 2h(l + b) = 2 x 28 (10 + 8) = 2 x 28 (18) = 1008

the cost of the plastering its vertical walls at the rate of Rs. 15/m² = 1008 x 15 = Rs. 15120

5. A solid cubical box of fine wood costs Rs. 256 at the rate Rs. 500/m³. Find its volume and length of each side.

Solution:

A solid cubical box of fine wood costs Rs. 256 at the rate Rs. 500/m³

Cost for one meter cube = Rs. 500

Original Cost = Rs. 256

Volume * cost for one meter cube = 256

volume = 256/500 = 0.512 m³

Volume of Cube = l³(l is the side)
l³=0.512m³

Now to convert it into cm³ we multiply 1000000 to it.
Thus, 0.512m³ = 0.512 x 1000000 = 512000cm³
So we have the volume as 512000cm³. Thus the length of the side of the cube = ∛512000
= 80cm

# Quadrilaterals – Full Chapter – Class VIII

A closed figure having four sides formed by joining four points, no three of which are collinear, in an order is called quadrilateral.

For example:

Quadrilateral is the union of four line segments that join four coplanar points; no three of which are collinear and each segment meet exactly two other lines, each at their end point.

A quadrilateral is convex if each of the internal angle is less than 180˚. Otherwise it is concave quadrilateral.

1. Then points A, B, C and D are called the vertices of the quadrilateral.
2. The segments AB, BC, CD and DA are the four sides of the quadrilateral.
3. The angles ∠DAB, ∠ABC, ∠BCD, and ∠CDA are the four angles of the quadrilateral.
4. The segments AC and BD are called as the diagonals of the quadrilateral.

### Adjecent sides and opposite sides:

1. The sides of a quadrilateral are said to be adjacent sides or consecutive sides. If they have a common end point. In the adjoining figure, AB and AD are adjacent or consecutive sides. Identify the other pair of adjacent sides.
2. Two sides of a quadrilateral are said to be opposite angles, if they do not contain a common side. Here ∠DAB and ∠BCD are opposite angles. Identify other pair of opposite angles.

1. The two angles of a quadrilateral are adjacent angles or consecutive angles. If they have a common side to them. Thus ∠DAB and ∠ABC are adjacent angles
2. Two angles of a quadrilateral are said to be opposite angles, if they do not contain a common side. Here ∠DAB and ∠BCD are opposite angles. Identify other pair of opposite angles.

Diagonal Property:

The diagonal AC divides the quadrilateral into two triangles, namely, triangles ABC and triangles ADC. Name the triangles formed when the diagonal BD is drawn.

## Angle Sum Property:

### Theorem 1: The sum of the angles of quadrilateral is 360˚

To prove: ∠A + B + C + D = 360˚

Construction: Draw the diagonal AC

Proof: In triangle ADC, 1 + 2 + 3 = 180˚(angle sum property)

In triangle ABC, 4 + 5 + 6 = 180˚

1 + 2 + 3 + 4 + 5 + 6 = 360˚

But , 1 + 4 = A and 3 + 6 = C. Therefore, A + D + B + C = 360˚

Thus, the sum of the angles of quadrilateral is 360˚

Example 1: The four angle of a quadrilateral are in the ratio 2:3:4:6. Find the measures of each angle.

Solution:

Given: The ratio of the angles is 2:3:4:6

To find: The measures of each angle.

Observe that 2 + 3 + 4 + 6 = 15. Thus 15 parts accounts for 360˚. Hence,

15 parts à 360˚

2 parts à (360˚/15) * 2 = 48˚

3 parts à (360˚/15) * 3 = 72˚

4 parts à (360˚/15) * 4 = 96˚

6 parts à (360˚/15) * 6 = 144˚

Thus the angles are 48˚, 72˚, 96˚, 144˚

Example 2: In a quadrilateral ABCD, ∠A and ∠C are of equal measure; ∠B is supplementary to ∠D. Find the measure of ∠A and ∠C.

Solution:

We are given B + D = 180˚. Using angle sum property of a quadrilateral, we get,

A + C = 360˚- 180˚ = 180˚

Since, A + C are of equal measure, we obtain A = C = 180˚/2 = 90˚

Example 3: Find all the angles in the given quadrilateral below:

Solution:

We know that, P + Q + R + S = 360˚. Hence,

x + 2x + 3 + x + 3x – 7 = 360˚

This gives, 7x – 4 = 360˚ or 7x = 364˚

Therefore, x = 364˚/7 = 52˚

We obtain P = 52˚; R = 52˚; Q = 2x + 3 = 107˚; S = 149˚. Check that, P + Q + R + S = 360˚

1. Two angles of a quadrilateral are 70° and 130° and the other two angles are equal. Find the measure of these two angles.

Solution:

Let A and B be x, C=70°

and D = 130°

A + B + C + D = 360° (Theorem 7)

x + x + 70° + 130° = 360°

2x = 360° − 200°

X = 160°/2 = 80°

Thus, ∠A = 80° and B = 80°

1. In the figure suppose P and Q are supplementary angles and R = 125°. Find the measures of S.

Solution:

P + Q = 180° (Supplementary angles)

P + Q + R + S = 360° (Theorem 7)

180° + 125° + S = 360°

S = 360°−(180° + 125°)

S = 360°−305°

S=55°

1. Three angles of a quadrilateral are in the ratio 2:3:5 and the fourth angle is

90°. Find the measures of the other three angles.

Solution:

Let the angles be 2x, 3x, 5x.

A + B + C + D = 360°

2x + 3x + 5x + 90° = 360°

10x = 360°−90°

x = 270°/10

x = 27°

∴ The angles are , A = 2x = 2×27 = 54°

B = 3x = 3×27 = 81

C = 5x = 5×27 = 135°

4.In the adjoining figure, ABCD is a quadrilateral such that D + C =100°. The bisectors of A and B meet at P. Determine APB.

Solution:

AP and BP are angular bisectors

D + C = 100°

To find: APB

Proof : A + B + C + D = 360° (Theorem 7)

A + B + 100° = 360°

A + B = 360° − 100°

Multipyling by 12 ,

12 × A + 12 × B = 12 × 260

a + b =130°

In Δ APB,

a + b + P = 180°

P = 180° − 130°

P = 50°

The set of fundamental have a pair of opposite sides which are parallel. Such a quadrilateral is known as trapezium.

For example:

Example 4: In the figure ABCD, suppose AB||CD; ∠A = 65° and ∠B = 75°. What is the measure of ∠C and ∠D?

Solution:

Observe that, ∠A + ∠D = 180° (a pair of adjacent angles of a trapezium is supplementary.)

Thus, 65° + ∠D = 180°

This gives, ∠D = 180° – 65° = 115°

Similarly, ∠B + ∠C = 180°, which gives 75° + ∠C = 180°. Hence,

∠C = 180° – 75° = 105°

Example 5: In an isosceles trapezium PQRS, ∠P and ∠S are in the ratio 1:2. Find the measure of all angles.

Solution:

In an isosceles trapezium base angles are equal; ∠P = ∠Q. Let ∠P = xº and ∠S =  2xº. Since, ∠P + ∠S = 180º (one pair of adjacent angles of a trapezium is supplementary). We get,

x + 2x = 180º

x = 180º/3 = 60º

Hence, ∠P = 60º and ∠S = 2 x 60º = 120º. Since, ∠P = ∠Q, we get, ∠Q = 60º. But, ∠Q + ∠R = 180º. Hence, we get,

∠R = 180º – 60º = 120º

1. In a trapezium PQRS, PQ || RS, and ∠P = 70° and ∠Q = 80° . Calculate the measure of ∠S and ∠R.

Solution:

∠P + ∠S = 180° (Supplementary angles)
∠S = 180° − 70° = 110°
∠Q + ∠R = 180° (Supplementary angles)
∠R = 180° −80° = 100°

2. In a trapezium ABCD with AB || CD. It is given that AD is not parallel to BC. Is ΔABC ≅ Δ ADC? Give reasons.

Solution:

In Δ ABC and Δ ADC,

(1) AC = AC (Common side)

(2) ∠BAC = ∠ACD (Alternate angles, AB || CD)
∴ Δ ABC ≅ Δ ADC It cannot be proved with the help of any beam postulated.

3. In the figure, PQRS is an isosceles trapezium; ∠SRP = 30°, and ∠PQS=40° Calculate the angles ∠RPQ and ∠RSQ.

Solution:

Data:PQRS is an isosceles trapezium,
PS = RQ and ∠P = ∠Q
∠SRP = 30° and ∠PQS = 40°
To find : ∠RPQ and ∠RSQ
Proof: ∠RPQ = ∠SRP = 30°
(Alternate angles, PQ || SR)
∠RSQ = ∠PRS = 40°
(Alternate angles, PQ || SR)

4. Proof that the base angles of an isosceles trapezium are equal.

Solution:

Data: ABCD is an isosceles trapezium.
To Prove: ∠A = ∠B
Construction: Join the diagonal BD and AC.
Proof : In Δ ACB and ΔBDA
(1) BC = AD (Isosceles trapezium)
(2)AC = BD (ISosceles trapezium)
(3) AB = AB (Common side)
∴ ACB ≅ Δ BDA (SSS postulate)
∴ ∠A = ∠B (Congruency property)

5. Suppose in a quadrilateral ABCD, AC= BD and AD = BC. Prove that ABCD is a trapezium.
Solution:

AC = BD and AD = BC
To prove: ABCD is trapezium
Proof: In Δ ADB and Δ BCA
(2) AC = BD (Data)

(3) AB = AB (Common side)
∴ Δ ADB ≅ Δ BCA (SSS postulate)
∴ ∠A = ∠B (Congruency property)
AC = BD (Data), AD = BC (Data)
∴ ABCD is an isosceles trapezium

Look at the following sets of quadrilaterals:

A quadrilateral in which both the pairs of opposite sides are parallel is called a parallelogram.

### Proposition 1: In parallelogram, opposite sides are equal and opposite angles are equal.

Proof:

Let ABCD is a parallelogram. Join BD. Then ∠1 = ∠2 and ∠3 = ∠4. In triangles ABD and CBD, we observe that,

∠1 = ∠2 ; ∠4 = ∠3 , BD(common)

Hence,  ΔABD ≅ Δ CDB (ASA postulate). It follows that,

AB = DC, AD = BC and ∠A = ∠C.

Similarly join AC, and we can prove that, ΔADC ≅ Δ CBA. Hence, ∠D =  ∠B

Example 6: The ratio of two sides of parallelogram is 3:\$ and its perimeter is 42cm. Find the measures of all sides of the parallelogram.

Solution:

Let the sides be 3x  and 4x. Then the perimeter of the parallelogram is 2(3x + 4x) = 2 x 7x = 4x.

The given data implies that, 42 = 14x, so, that, x = 42/14 = 3. Hence the sides of the parallelogram are 3 x 3 = 9cm and 3 x4 = 12cm

Example 7: In the adjoining figure, PQRS is a parallelogram. Find x and y in cm.

Solution:

In a parallelogram, we know, that the diagonals bisect each other. Therefore SO = OQ. This gives 16 = x + y. Similarly, PO = OR, so that 20 = y + 7. We obtain y = 20 – 7 = 13cm. Substituting the value of y in the first relation, we get 16 = x + 13. Hence x = 3cm

1.The adjacent angles of a parallelogram are in the ratio 2:1 Find the measures of all the angles.
Solution:

Let the angles be 2x and x
∠A + ∠B = 2x + x = 180° (Adjacent angles of parallelograms are supplementary)
2x + x = 180°
3x = 180°
X = 180°/3 = 60°
∴ ∠A = 2x = 60×2 = 120°
∴ ∠B = 60°
∴ ∠C = ∠A = 120°
(Opposite angles of parallelogram)
∠D = ∠B = 60°
(Opposite angles of parallelogram)

2. A field is in the form of a parallelogram, whose perimeter is 450 m and one of its sides is larger than the other by 75m. Find the lengths of all sides
Solution:

Perimeter = AD + DC + CB + BA
450 = x + x + 75 + x + x + 75
450 = 4x + 150
450 – 150 = 4x
300𝑥 = x
X = 75
∴ Side = 75m(x)
∴ Opposite side = x + 75 = 75 + 75 = 150m

3. In the figure, ABCD is a parallelogram. The diagonals AC and BD intersect at O; and ∠DAC = 40°, ∠CAB = 35°, and ∠DOC = 10°.

Solution:

Data: ABCD is a parallelogram AD || BC and DC || AB.
The diagonals AC and BD intersect at O
∠DAC = 40°, ∠CAB = 35° and ∠DOC = 110°
To find : (1) ∠ABO (2)∠ADC (3)∠ACB (4) ∠CBD
Proof: ∠DAC + ∠CAB = ∠A
40° + 35 = ∠A
∠A = 75°

∠C = ∠A = 75°
(Opposite angles of parallelogram are equal)
∠D + ∠A = 180° (Supplementary angles)
∠D = 180°−75° = 105°
∠B = ∠D = 105° (Opposite angles of parallelogram are equal)
∠DOC = ∠AOB = 110°
(Vertically opposite angles)
In Δ AOB, ∠A + ∠O + ∠B = 180°
(Sum of all angles of a Δ is 180°)
35° + 110° + ∠B = 180°
∠B = 180°-145°
(1) ∠ABO = 35°
(3)∠ACB = ∠CBD = 40°
∠CBD = 105°-35°
(4) ∠CBD = 70°

(1)∠ABO = 35°

(3)∠ACB = 40°

(4) ∠CBD = 70°

4. In a parallelogram ABCD, the side DC is produced to E and ∠BCE = 105°.
Calculate and ∠A, ∠B, ∠C and ∠D.

Solution:

∠BCD + ∠BCE = 180° (Linear pair)
∠BCD = 180° − 105° = 75°

In ABCD,
∠A = ∠C = 75° (Opposite angles of parallelogram)
∠ABC = ∠BCE = 105°
(Alternate angles, AB || DE)
∠D = ∠B = 105°
(Opposite angles of parallelogram)

5. Prove logically the diagonals of a parallelogram bisect each other. Show conversely that a quadrilateral in which diagonals bisect each other is a parallelogram.

Solution:

Data: ABCD is a parallelogram ( ║ln)
To Prove : AO = OC and DO = OB
Proof: In ΔABO and DOC
(1)AB = DC (Opposite sides of angles)
(2) ∠AOB = ∠DOC (V.O.A)
(3) ∠ABD = ∠BDC (Alternate angles AB ||CD)
∴ ΔABO ≅ ΔDOC (ASA portulate)
∴ AO = OC & DO = BO (Congruence properties)

6. In a parallelogram KLMN, ∟K = 60°. Find the measure of all the angles.
Solution:

∠M = ∠K = 180° (Supplementary angles)
∠N = 180°−60° = 120°
∠M = ∠K = 60°
(Opposite angles of parallelogram)
∠L = ∠N = 120°

7. Let ABCD be a quadrilateral in which ∠A = ∠C and ∠B = ∠D. Prove that ABCD is a parallelogram.
Solution:

Data: ∠A = ∠C and ∠D = ∠B
To Prove: ABCD is a parallelogram
Proof: ∠A + ∠B + ∠C + ∠D = 360°
x + y + x + y = 360°
2x + 2y = 360°
2(x + y) = 360°
x + y = 360°/2 = 180°
∴ ABCD is a parallelogram.

8. In a quadrilateral ABCD, suppose AB = CD and AD = BC. Prove that ABCD is a parallelogram.
Solution:

AB = DC, AD = BC
To Prove : ABCD is a parallelogram
Construction: Join DB

Proof: In Δ ABD and Δ DBC
(1) DB = DB (Common side)
(3) AB = DC (Data)
∴ ABD ≅ Δ DBC (SSS postulate)
∴ ∠DBA = ∠BDC (Congruency property)
But they are alternate angles
These we have AB ║ DC
But they are alternate angles
These we have DA ║ BC
∴ ABCD is a parallelogram.

## Special kinds of parallelograms -Quadrilaterals

### Rectangle:

A rectangle is a parallelogram whose all angles are right angles.

Diagonal properties of a rectangle:

(i) The diagonals of a rectangle are equal

(ii) The diagonals of a rectangle bisect each other.

Example 8: In a rectangle XYWZ, suppose O is the point of intersection of its diagonals. If ∠ZOW = 110º. Calculate the measure of ∠OYW.

Solution:

We know, ∠ZOW = 110º. Hence, ∠WOY = 180º – 110º = 70º (supplementary angles)

Now, OYW is an isosceles triangle, as OY = OW. Hence,

∠OYW = ∠OWY = (180º –  70º)/2  = 110º/2 = 55º

### Rhombus:

A rhombus is a parallelogram in which all the four sides are equal.

Example 10: The diagonals of a rhombus are 24cm and 10cm. Calculate the area of the rhombus.

Solution:

We are given that AC = 24cm; BD = 10cm. We know that the diagonals of a rhombus bisect each other at right angles. Let O be the point of intersection of these diagonals. Then we have AO = CO = 12cm and BO = DO = 5cm. We also know that AOD is right angled triangle. Hence the area of triangle AOD is

1/2 x OA x OD = 1/2 x 12 x 5 = 30 cm2

Since a rhombus has four congruent right triangles, is area is 4 x 30 = 120 cm2

### Square

A square is a parallelogram, in which

(i) in which all the sides are equal

(ii) each angle is right angle

(iii) diagonals are equal

(iv) diagonals bisect at right angles.

Example 12: A field is in the shape of a sqaure with side 20m. A pathway of 2m width is surrounding it. Find the outer perimeter of the pathway.

Solution:

Width of the pathway is 2m. Length of the side of the outer sqaure = (20 + 2 + 2) = 24m

Hence, perimeter 4 x 24 = 96m

### Kite:

Kite is a quadrilateral in which two of isosceles triangles are joined along the common base.

Example 14: In the figure PQRS is a kite; PQ = 3cm and QR = 6cm. Find the perimeter of PQRS.

Solution:

We have PQ = PS = 3cm, QR = SR = 6cm.

Hence, the perimeter = PQ + QR + RS + PS

3 + 6 + 6 + 3 = 18cm

1. The sides of the rectangle are in the ratio 2:1. The perimeter is 30 cm. Calculate the measure of all the sides.
Solution:

ABCD is a rectangle
Let AB:BC = 2:1
∴ AB = 2x and BC = x
Perimeter of the rectangle = 2(1+b)
2(1+b) = 30
2(2x+x) = 30
2×3x = 30
6x = 30
x = 306 = 5
2x = 2×x = 2×5 = 10
1x = 1×4 = 1×5 = 5
AB = 10m; BC = 5cm
CD = 10cm; DA = 5cm

2. In the adjacent rectangle ABCD, ∠OCD = 30°. Calculate ∠BOC. What type of triangle is BOC?

Solution:

ABCD is a rectangle

Diagonals AC and BD bisect each other at O and AC = BD.

AO = OC = BO = OD

OCD = ODC = 30°

On ΔCOD, ∠ODC + ∠OCD + ∠COD = 180˚

30° + 30° + ∠COD = 180°

60° + ∠COD = 180°

∴ ∠COD = 180˚ – 60˚ =120°

∠COD + ∠COB = 180˚

∠COB = 180˚ – 120˚

∠COB = 60°

∴ Δ BOC is an isosceles triangle

1. All rectangle are parallelograms, but all parallelograms are not rectangles. Justify the statements.

Solution:

All rectangles have all the properties of parallelograms but a parallelogram

May not have all the properties of a rectangle.

1. In a rectangle all the angles are right angle but in a parallelogram only opposite angles are equal.
2. In a rectangle the diagonals are equal but in a parallelogram diagonals are not equal.

1. Prove logically that the diagonals of a rectangle are equal.

Solution:

Data : ABCD is a rectangle.

AC and BD are the diagonals.

To Prove: AC = BD

Proof : In ΔABC and Δ ABD

AB = AB [Common side]

∴ ΔABC ≅ ΔABD [SAS]

∴ AC = BD [CPCT]

1. The sides of a rectangular park are in the ratio 4:3. If the area is 1728m2 , find the cost of fencing it at the rate of Rs.2.50/m.

Solution:

ABCD is a rectangular park let AB : BC = 4 : 3

∴ AB = 4x and BC = 3x

Area of rectangle = 1×b

4x × 3x = 1728

12x2 = 1728

X2 = 144

X = √144 = 12

Length = 4x = 4×12 = 48m

Breadth = 3x = 3×12 = 36m

Perimeter of a rectangle = 2(l + b)

2(48 + 36) = 2×84 = 168m

Cost of fencing = perimeter ×Rate

= 168 ×2.50

= Rs.420

1. A rectangular yard contains two flower beds in the shape of congruent isosceles right triangle. The remaining portion is a yard of trapezoidal.

Shape (see fig) whose parallel sides have lengths 15 m and 25 m. What fraction of the yard is occupied by the flower bed?

Solution:

ABCD is a rectangular yard

AEFB is a trapezium

AB ║ EF
AB = 25 m and EF = 15m

Area of rectangle = ½ × b = 25×5 = 125m²
Area of each flower bed = ½ ×b×h
= ½ ×5×5 = (25/2) m²
Area of both flower beds = 2× (25/2) = 25m²
Fraction of flower beds to yard = 25/125 = 1/5

7. In a rhombus ABCD ∠C = 70°. Find the other angle of the rhombus.

solution:

ABCD is a rhombus

∠C = 70°
But ∠A = ∠C = 70 ∴ ∠A = 70°

∠A + ∠B =180° (adjacent angles)
70 + ∠B = 180°
∠B = 180 − 70 = 110°
∠B = ∠D = 110°
∴ ∠A = 70°; ∠B = 110° ; ∠D = 110°

8. In a rhombus PQRS, ∠SQR = 40° and PQ = 3 cm. Find ∠SPQ, ∠QSR and the perimeter of the rhombus.

Solution:

PQRS is a rhombus
∠SQR = 40° and PQ = 3cm

∠SQR = 40° and PQ = 3cm
∠PQS = ∠SQR = 40°
∠PQS = 40° (diagonal bisector angles ) 3 cm
But ∠PQS = ∠QSP [PQ = PS]
∴ ∠QSP = 40°
In Δ PQS, ∠PQS + ∠QSP + ∠SPQ = 180°
40° + 40° + ∠SPQ = 180
∠SPQ = 180 −80
= 100°
∴ ∠SPQ = 100°
In a rhombus PQ = QR = RS = SP = 3cm
Perimeter of the Rhombus PQRS = 3×4 = 12cm

9. In a rhombus PQRS, if PQ = 3x -7 and QR = x + 3, find PS.

Solution:
In a rhombus all sides are equal
PQ = QR = RS = SP
∴ PQ = QR
3x -7 = x + 3
3x -x = 3 + 7
2x = 10
X = 5
PS = x+3 = 5+3 = 8 cm

10. Let ABCD be a rhombus and ∟ABC = 124°. Calculate ∟A, ∟D and ∟C.
In a rhombus, opposite angles are equal

Solution:
∠B = ∠D = 124°
∠A + ∠B =180°
Consecutive angles

∠A + 124° = 180°

∠A + 124° = 180°
∠A = 180-124=56°
∴ ∠A = 56° ; ∠B = 124° and ∠C = 56°

11. Rhombus is a parallelogram: justify.

Solution:
Rhombus has all the properties of parallelogram i.e
a) Opposite sides are equal and parallel.
b) Opposite angles are equal.
c) Diagonals bisect each other
∴ Rhombus is a parallelogram.

12. In a given square ABCD, if the area of triangle ABD is 36 cm2 , find
(i) the area of triangle BCD;
(ii) the area of the square ABCD.

Solution:

ABCD is a square
Bd is the diagonal
The diagonal divides the square into two congruent triangles
∴ Δ ABD ≅ Δ BCD D C
∴ Area of Δ ABD =area of Δ BCD
Area of Δ ABD= 36 cm² (given)
∴ Area of Δ BCD = 36cm²
Area of the square ABCD = Area of Δ ABD + Area of ΔBCD
=36cm² +36cm²
Area of the squre ABCD =72cm²

13. The side of a square ABCD is 5 cm and another square PQRS has perimeter equal to 40cm. Find the ratio of the area ABCD to the area of PQRS.

Solution:
Perimeter of the square ABCD = 4 ×side

= 4×5 = 20
Ratio of (perimeter of ABCD)/(Perimeter of PQRS) = 20/40 = 1/2 or 1:2
Area of ABCD = (side)² = (5)² = 25cm
Side of PQRS = perimeter/4 = 40/4 = 10cm
Area of PQRS = (side)² = (10)² = 100cm²
Ratio of (Area of ABCD)/(Area of PQRS) = 25/100 = 1/4 or 1:4

14. A square field has side 20m. Find the length of the wire required to fence it four times.

Solution:
Length of one side of the square = 20 Rs
Length of wire required to fence one round = 4×20
Length of wire required to fence four rounds = 80m
= 4×80m
= 320m

15. List out the differences between square and rhombus.

Solution:
Square
1. All the angles are equal
2. Diagonals are equal
3. Area = side × side=(s)²

### Rhombus

1. Opposite angles are equal

2. Diagonals are unequal

3. Area= 1/2 × Product of diagonals = 1/2 x d­1 d2

Example 16. Four congruent rectangles are placed as shown in the figure. Area of the outer square is 4 times that of the inner square. What is the ratio os length to breadth of the congruent rectangles?

Solution:

Let the length of rectangles be ‘a’ and breadth be ‘b’
Side of outer square –(a+b) units
Side of outer square (a-b) units
Area of outer square = 4 times area of inner square
Area of ABCD = 4(Area of PQRS)
(a + b)² = 4(a-b)

(a + b) = 2(a – b)

2a – 2b = (a + b)

2a – a = b + 2b

a = 3b

a/b = 1/3 or  1 : 3