There are four (\(\surd \)) marks against guava.

The other pair of opposite sides is \(\bar {BC} \ and\ \bar{DA}\) 

\(2\frac { 1 }{ 21 } \)

(i) By paper folding
(ii) By using ruler and compasses.

Two letters having horizontal line of symmetry are E and K.

Given, figure is covered by 10 full squares.
∴ Area=10 \(\times\) 1squnit =10sq units

In this algebraic expression, 2 is divided from product of 10 and b.

\(\because\) Time taken by Rohit = 25 min and Time taken by Shivam = 1h = 60 min
\(The\ required ratio=\frac{Time\ taken\ by\ Rohit}{Time\ taken\ by\ Shivam}=\frac{25}{60}=\frac{5}{12}=5:12\)
Hence, the required ratio is 5 : 12.

concave quadrilateral

To get the integer 5 more than -5, we start from -5 and move 5 steps to the right of -5 and reach at 0, as shown in the figure given below:

Hence, 5 more than -5 is O.

∵12904 ⟶ 13000[rounding off to nearest thousands]
2888 ⟶ 3000 [rounding off to nearest thousands]
∴ Estimated sum = 13000 + 3000 = 16000

We have, 2.25=2 ones + 2 tenths + 5 hundredths = 2 + 2 \(\times\) \(\frac { 1 }{ 10 } +\frac { 5 }{ 100 } =2+0.25\)
\(=2+\frac { 2 }{ 10 } +\frac { 5 }{ 100 } =2+\frac { 20+5 }{ 100 } =2+\frac { 25 }{ 100 } =2+\frac { 1 }{ 4 } =2\frac { 1 }{ 4 } =\frac { 9 }{ 4 } .\)

Here, we have 32
32 = 29 +3

390.

Given, length of a wooden board
\(=150\frac { 1 }{ 4 } \)cm \(=\frac { 601 }{ 4 } \) cm
Carpenter sawed off a piece of wooden board
\(=40\frac { 1 }{ 5 } \)cm \(=\frac { 201 }{ 5 } \)cm
\(\therefore\)  Length of the remaining piece \(=\frac { 601 }{ 4 } -\frac { 201 }{ 5 } \)
\(\because\) LCM of 4 and 5 = 20
\(\therefore \frac { 601 }{ 4 } \frac { 201 }{ 5 } =\frac { 601\times 5 }{ 4\times 5 } -\frac { 201\times 4 }{ 5\times 4 } \)
\(=\frac { 3005 }{ 20 } -\frac { 804 }{ 20 } \)
\(=\frac { 3005-804 }{ 20 } \)
\(=\frac { 2201 }{ 20 }\)cm
Hence the length of the remaining piece is\(\frac { 2201 }{ 20 } \)cm or \(110\frac { 1 }{ 20 } \)cm 

The given postal charges for three parcels are Rs20, Rs28, Rs 36.
The greatest denomination of stamps is the HCF of 20, 28 and 36.


HCF of 20 and 28 is 4.
Now, we.find the HCF of 4 and 36.
HCF of 4 and 36 is 4.
Hence, HCF of 20, 28 and 36 is 4.
\(\therefore\) Fatima would by stamp of denomination of Rs4.

Naresh walked in morning
= 2 km 35 m = 2 km + 35 m
\(=2\ km+35\times{1\over 1000}km\)        \(\begin{bmatrix} \because 1m={1\over 1000}km\end{bmatrix}\)
\(=2\ km+{35\over 1000}km\)
= (2 + 0.035) km = 2.035 km
Naresh walked in evening
=1 km 7 m = 1 km + 7 m
\(=1\ m+7\times{1\over1000}km\)   \(\begin{bmatrix} \because 1m={1\over 1000}km\end{bmatrix}\)
\(=1\ km+{7\over1000}km\)
= (1 + 0.007) km = 1.007 km
\(\therefore\)  Total distance
\(\quad 2.035\\+1.007\\\_\_\_\_\_\_\_\\\ \ \ 3.042\\\_\_\_\_\_\_\_\)
Hence, total distance walked by Naresh is 3.042 km.

The images of the point P, Q and R in the line n are P',Q' and R', respectively.

It is clear that ㄥPQR = ㄥP'Q'R'

2160 sq m

18 kg

We know that,
Area of the rectangle = Length \(\times\) Breadth
Here, the area of the rectangle is 16 sq cm.
So, for getting the area 16 sq em, there are three cases.
Case I When length of the rectangle = 16 cm
and breadth of the rectangle = 1cm
∴ Area of the rectangle = Length \(\times\) Breadth
=16 cm \(\times\) 1 cm = 16sq cm
and perimeter of the rectangle
= 2 \(\times\) (Length + Breadth)
= 2 \(\times\)(16 cm + 1 cm) = ( 2 \(\times\) 17) cm= 34 cm
Case II When length of the rectangle = 4 cm
and breadth of the rectangle = 4 cm
∴ Area of the rectangle = Length\(\times\) Breadth
= 4 cm \(\times\) 4 cm 16 sq cm
and perimeter of the rectangle
= 2 \(\times\) (Length + Breadth)
= 2 \(\times\)(4 cm +4 cm) = (2 \(\times\) 8) cm = 16 cm
Case III When length of the rectangle = 8 cm
and breadth of the rectangle = 2 cm
∴ Area of the rectangle = Length\(\times\) Breadth
=8 cm \(\times\)2 cm =16sq cm
and perimeter of the rectangle
= 2 \(\times\) (Length + Breadth) = 2 \(\times\) (8 cm + 2 cm)
= (2\(\times\)10) cm = 20 cm
(a) The rectangle which made in Case I has the greatest perimeter.
(b) The rectangle which made in Case II has the least perimeter.
Now, given that the area of the rectangle is 24 sq cm.
Again, for getting the area 24 sq cm, there are three cases:
Case I When length of the rectangle = 24 cm
and breadth of the rectangle = 1 cm
∴ Area of the rectangle = Length \(\times\) Breadth
= 24 cm\(\times\) 1 cm = 24 sq cm
and perimeter of the rectangle
= 2 \(\times\) (Length + Breadth)
= 2 \(\times\) (24 cm + 1 cm) = (2 \(\times\) 25)cm = 50 cm
Case II When length of the rectangle = 12 cm
and breadth of the rectangle = 2 cm
∴Area of the rectangle = Length \(\times\)Breadth
= 12 cm \(\times\) 2 cm = 24 sqcm
and perimeter of the rectangle
= 2 \(\times\) (Length + Breadth)
= 2 \(\times\) (12 cm +2 cm)
= (2 \(\times\)14) cm = 28 cm
Case III When length of the rectangle = 6 cm
and breadth of the rectangle = 4 cm
∴ Area of the rectangle = Length \(\times\) Breadth
= 6 cm \(\times\) 4 cm = 24 sqcm
and perimeter of the rectangle
= 2 \(\times\) (Length + Breadth)
= 2 \(\times\) (6 cm +4 cm)
= 2 \(\times\) 10 cm = 20 cm
(a) The rectangle which made in Case I has the greatest perimeter.
(b) The rectangle which made in Case III has the least perimeter.
Yes, it is possible to predict the shape of the rectangle with greatest perimeter and with least perimeter.
e.g. A rectangle having area 16 sq ern and greatest perimeter 34 cm is of the shape 16 cm \(\times\) 1 cm
(i.e. length = 16 cm and breadth = 1 cm), also a rectangle
having area 24 sq cm and least perimeter 20 cm is of the shape 6 cm \(\times\) 4 cm (i.e. length = 6 cm and breadth = 4 cm).
The rectangle with the greatest length has the maximum perimeter and the rectangle with the smallest length has the least perimeter.

(a) Since, AC 丄 BD, it means \(\angle E = 90°\)
∴ \(\angle AEB, \angle BEC, \angle CED\) and \(\angle AED\) are right angles.
(b) Since, AE 丄 CE, it means \(\angle E = 90°\)
∴ \(\angle AEC\) is a right angle.
(c) Since, AC 丄 CD, it means \(\angle C= 90°\)
∴ \(\angle ACD\) is a right angle.
(d) Since, OP 丄 AB, it means \(\angle K = 90°\)
\(\therefore\ \ \ \ \ \angle AKO, \angle OKB, \angle BKP\) and \(\angle AKP\) are right angles.

(a) Non-parallel sides are AF and BE.
(b) Respect to teacher, happiness, beauty and knowledge.

Given,10 animals are represented by 1 symbol \(\bigotimes \).
\(\therefore\) 1animal is represented by 1/10symbol \(\bigotimes \)
For village A,
80 animals are represented by \(\frac{80\times 1}{10}\) = 8 symbols \(\bigotimes \)
For village B,
120 animals are represented by \(\frac{120\times 1}{10}\) = 12 symbols \(\bigotimes \)
For village C,
90 animals are represented by \(\frac{90\times 1}{10}\) = 9 symbols \(\bigotimes \)
For village D,
40 animals are represented by \(\frac{40\times 1}{10}\) = 4 symbols \(\bigotimes \)
For village E,
60 animals are represented by \(\frac{60\times 1}{10}\) = 6 symbols \(\bigotimes \)
Then we have the following pictograph.

(a) It is clear that from the pictograph, that 6 symbols represent animals of village E.
(b) It is clear that from the pictograph, that village B has the maximum number of animals i.e.120 animals.
(c) From the pictograph, number of animals in village A = 80
Number of animals in village C = 90
\(\therefore\)   90 > 80
\(\therefore\)   Village C has more animals

Given, initial temperature at 12:00 noon was +5°C. Since, the temperature increased by 3°C in first hour.
∴ Temperature at 1: 00 pm = 5°C + 3° C 8°C
Also, the temperature decreased by 1°C in the second hour.
∴ Temperature at 2: 00 pm = 8°C - 1°C= 7°C
Hence, the temperature at 2 : 00 pm is 7°C

0

After reading these number, we have
(i) 527864 Five lakhs twenty seven thousands eight hundreds sixty four.
(ii) 95432 Ninety five thousands four hundreds thirty two.
(iii) 18950049 One crore eighty nine lakhs fifty thousands forty nine.
(iv) 70002509 Seven crores two thousand five hundreds nine.
(a) By USIng placement boxes, gIven number can be written as
By using the commas, given numbers can be written as :

(b) Ascending order of these numbers are as follow:
95432 < 527864 < 18950049 < 70002509
Descending order of these numbers are as follow:
70002509> 18950049> 527864 > 95432.

The figure obtained is symmetric. It can be folded only in one way to produce two identical halves.

6cm²

\(4.50=4+\frac { 50 }{ 100 } =4+\frac { 1 }{ 2 } =4\frac { 1 }{ 2 } =\frac { 9 }{ 2 } \)

For Raju, cost of 10 pens = Rs.150
\(\therefore\) Cost of 1 pen \(=Rs.\frac{150}{10}=Rs.15\)
For Manish, cost of 7 pens = Rs.84
\(\therefore\) Cost of 1 pen \(=Rs.\frac{84}{7}=Rs.12\)
Here, Rs.12 < Rs.15
Hence, Manish got the pens cheaper.

On drawing, we get the following circle with O as centre and of any radius:
(i) Take a point A such that A lies on the circle.
(ii) Take a point B such that B lies in the interior of the circle.
(iii) Take a point C such that C lies in the exterior of the circle.
                     

5-x

Here, given production of dolls. Monday - 200, Tuesday - 500 Wednesday - 350, Thursday - 400
Let us assume that
100 may be represented by 
50 may be represented by 
Thus, the pictograph would be

The vertices of the figure are A. B, C,D, E and the line segments are \(\bar {AB},\bar {BC},\bar {CD},\bar {DE},\bar {EA},\bar {AC},\bar {AD}\).

We have, 30 and 415
30 = 1 x 30; 30 = 2 x 15; 30 = 3 x 10; 30 = 5 x 6
Factors of 30 are 1,2,3,5,6,10,15 and 30 and 415 = 1 x 415; 415 = 5 x 83
Factors of 415 are 1, 5, 83 and 415.
\(\because\)Common factors of 30 and 415 are 1 and 5.
So, 30 and 415 are not co-prime numbers.

To add -3 and -4. Mark -3 on the number line. Move 4 steps to left of -3, we reach at -7.

∴ -3+(-4)= -7

We have, \(\frac { 2 }{ 7 } ={{8}\over{\Box}}\)
∴ \(2\times \Box =7\times 8\)
So, \(2\times \Box =7\times 2\times 4=28\times 2\)   [∵ 8 = 2\(\times\)4]
On comparing we get \(\Box =28\)
Hence \(\frac { 2 }{ 7 } =\frac { 8 }{ \boxed { 28 } } \)

25000

The measure of angle (by using the protractor) and its classification is shown as below

Given, capacity of vessel = 4 L 500 mL
= 4 \(\times\)1000 + 500 = 4000 + 500 = 4500 mL. [1L= 1000mL]
and capacity of one glass = 25 mL
∴ Number of glasses of curd filled out of vessel
\(=\frac { Capacity\quad of\quad a\quad vessel }{ Capacity\quad of\quad a\quad glass } =\frac { 4500 }{ 25 } =180\)
Hence, curd can be filled in 180 glasses.

Length of the lace = 2(2 + 1) = 6 m