We name the dotted line in the figure as the line of symmetry.

A triangular prism looks like the shape of a Kaleidoscope. It has triangles as its bases.

Faces: 5.
Edges: 9.
Corners: 6.

Yes! Other pairs of adjacent sides are:
\(\bar {BC},\bar{CD};\bar{CD},\bar{DA};\bar{DA},\bar{AB}\)

Multiples of 5 = 5, 10, 15, 20, 25, 30, ...
Multiples of 15 = 15, 30, ...
First two common multiples = 15, 30

We have,10 g:40 g \(=\frac{10}{40}=1:4\)
and 25 kg: 100 kg \(=\frac{25}{100}=1:4\)
So, they are in proportion.

Now, mark the points P, Q and R on it according to question.

The pictorial representation of the numerical data using pictures or symbols is called the pictograph of the data.

Shurti is at the step x - 5.

2 cm

Here, 2 and 9 both are on the right of zero, bur 9 is farther from zero in comparison of 2. So, number 9 is right of number 2.

We know that, 1 kg = 1000 g = (1000 X 1000) mg [∵ 1 g = 1000 mg]
= 1000000 mg
∴ 1000000 milligrams make one kilogram

We have, \(\frac { 2 }{ 5 } \)
To make this fraction in tenths, we have to multiply by 2 in its numerator and denominator both.
\(\therefore \quad \frac { 2 }{ 5 } =\frac { 2\times 2 }{ 5\times 2 } =\frac { 4 }{ 10 } \)
Here, it has 4 tenths.
i.e. \(\frac { 4 }{ 10 } =0.4\)

Here, the given figure is divided into 4 equal parts and 3 parts out of 4 parts are shaded.
∴ The required fraction = \(\frac { 3 }{ 4 } \)

To find 2 + 6
Let us start from 2. Since, we have to add 6 to this number, we make 6 jumps to the right of 2. Each jump being equal to 1 unit. After six jumps, we reach at 8.

\(\therefore\)  2 + 6 = 8.

Let the fourth number be x.
Case I. Taking the correct number 12

∴ LCM=3 \(\times\) 5 \(\times\) 7 \(\times\) x
Case II. Taking 21 instead of 12

∴ LCM = 3 \(\times\) 5 \(\times\) 7 \(\times\) x
According to the question,
LCM is the same in both the cases.
∴  Fourth number = 4 x 7 = 28.

Here, 270.24 + 12.52+ 1.5 can be written as

270.24 + 12.52 + 1.50 = 284.26

We have,  \(4\frac{2}{3} \) and \(5\frac{3}{7}\)
Now, \(4\frac{2}{3}=\frac{4\times3+2}{3}=\frac{14}{3}\)
and \(5\frac{3}{7}=\frac{5\times7+3}{3}=\frac{38}{7}\)
Now, let us compare \(\frac{14}{3} and \frac{38}{7}\) by
cross - multiplication.
we have,

14 x 7 = 98  and 38 x 3 = 114
∵ 114>98
∴ \(\frac{38}{7}>\frac{14}{3}\)
So, \(5\frac{3}{7}>4\frac{2}{3}\)

P and R

50

In the given figures, we have,
\(\angle AOB\) is a right angle.

3n

The perimeter of the given figure is 28 cm

Given, statements can be represented using integer +35

The joins of the vertices, which are not adjacent i.e. opposite to each other, are called diagonals of quadrilateral. A rough sketch of a quadrilateral PQRS with its diagonals and their names are given below:

Here, PR and QS are its diagonals. They intersect each other at point O. So, point O is in the interior of the quadrilateral PQRS.

We have, 297144
(i) Divisibility by 2
\(\because\)Unit digit of number = 4, so 297144 is divisible by2.
(ii) Divisibility by 3
Sum of digits of given number = 2 + 9 + 7 + 1+ 4 + 4 = 27
\(\because\) 27 is divisible by 3, so 297144 is also divisible by 3.
Now, we see that 297144 is divisible by 2 and 3 both.
Hence, it is divisible by 6.

There are many examples of such type. Five of them are as follows:

Digits in ascending order are 0,1,2,3,4,5,6,7,8,9 and digits in descending order are 9,8,7,6,5,4,3,2,1, 0. Also, we know that 0 can never be taken at the leftmost place.
Keeping the digit 4 at tens place, we can fill other places thousands, hundreds and ones place by remaining number.
Greatest number

[Keeping the greatest digits i.e. 9, 8 and 7 at thousands, hundreds and tens place]
Smallest number

[Keeping the smallest digits i.e. 9, 8 and 7 at thousands, hundreds and tens place].

(b) March (c) 1:2

Quantity of food eaten which turned into organism's own body = \(\frac{1}{10}\)of the total food Now, the quantity of food eaten which is not available for the next level =1-\(\frac{1}{10}\)
\(=\frac { 10-1 }{ 10 } =\frac { 9 }{ 10 } \)
Hence, the required fraction =\(\frac { 9 }{ 10 } \) 

(a) 18
(b) Rs 30000

Here, we will use the following steps of construction:
Step I Draw any angle XOY with vertex O.
                   
Step II Take a point A on OX and a point B on OY, such that OA = OB.
Step III Now, with O as centre, using compasses, draw an arc radius more than half of the length of \(\bar{OA}\).
Step IV With the same radius and with A as centre, draw another arcs using compasses, which cut the previous arc at C and D respectively.
Step V join \(\bar{CD}\) . It curs \(\bar{OA}\) at M therefore, CD is the perpendicular bisector of OA.
Step VI Now, with 0 as centre and radius equal to more than half of the length of OB, draw arcs.
Step VII With same radius and with B as centre, draw another arcswhich cut the previousarcsat E and F.
Step VIII join \(\bar{EF}\). It curs \(\bar{OB}\) at N. Therefore, EF is the perpendicular of \(\bar{OA}\).
Also, both perpendicular bisectors meet at A.
On measuring, we get PA = PB.

Reshma has cash in hand = Rs 5000
              Cost of one frock = Rs 1150.48
              Cost of one toy = Rs     540.52
              Cost of one bag = Rs   2160.70
                                            _____________
 \(\therefore\)   Total cost =                   Rs 3851.70
                                             _____________
Total cash in hand = Rs 5000.00
Total money spent = -Rs 3851.70
                                    ____________
\(\therefore\)    Balance          =     1148.30
                                    ____________
Hence, the money left with Reshma is Rs 1148.30.

4

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.

Let the age of Sunita's mother = 2x yr
Then, according to the question,
Sunita's age = \(\frac{2x}{2}\) = x [half of her mother's age
(i) After 4 yr, present age is added in given years.
Sunita's age = (x + 4) yr
∴ Her mother's age = (2x + 4) yr
(ii) Before 3 yr, given year is subtracted from present age.
∴  Sun ita's age = (x - 3) yr
and her mother's age = (2x - 3) yr

(a) Avneet lays 9 squares in the form of a square as shown in the figure, then side of the square
\(=\left( \frac { 1 }{ 2 } +\frac { 1 }{ 2 } +\frac { 1 }{ 2 } \right) m=\frac { 1+1+1 }{ 2 } =\frac { 3 }{ 2 } m\)
\(=\left( \frac { 1 }{ 2 } +1+1+\frac { 1 }{ 2 } +1+1+\frac { 1 }{ 2 } +1+1+\frac { 1 }{ 2 } +1+1 \right) m\)
∴ Perimeter of Avneet's arrangement
= 4 \(\times\) Length of a side = 4 \(\times\) \(\frac{3}{2}\) = 6 m
Hence, the perimeter of Avneet's arrangement is 6 m.
(b) Shari lays 9 squares in the form of a cross as shown in figure, then
Perimeter of Shari's arrangement = Sum of all sides

= AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL+ LA
\(=\frac { 1 }{ 2 } +\left( \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \right) +\left( \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \right) +\frac { 1 }{ 2 } +\left( \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \right) +\left( \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \right) +\frac { 1 }{ 2 } +\left( \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \right) +\left( \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \right) +\frac { 1 }{ 2 } +\left( \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \right) +\left( \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \right) \)
\(=\left( \frac { 1 }{ 2 } +1+1+\frac { 1 }{ 2 } +1+1+\frac { 1 }{ 2 } +1+1+\frac { 1 }{ 2 } +1+1 \right) m\)
\(=\left( \frac { 1 }{ 2 } +\frac { 1 }{ 2 } +\frac { 1 }{ 2 } +\frac { 1 }{ 2 } +8 \right) m=(1+1+8)m=10m\)
Hence, the perimeter of Shari's arrangement is 10m
(c) From above, it is clear that Shari's arrangement i.e, a cross has greater perimeter.
(d) Yes, there is a way shown alongside in which we get a greater perimeter.
Here, we have arrange the 9 squares in the form of a rectangle

Now, length of rectangle
\(=\left( \frac { 1 }{ 2 } +\frac { 1 }{ 2 } +\frac { 1 }{ 2 } +\frac { 1 }{ 2 } +\frac { 1 }{ 2 } +\frac { 1 }{ 2 } +\frac { 1 }{ 2 } +\frac { 1 }{ 2 } +\frac { 1 }{ 2 } \right) m=\frac { 9 }{ 2 } m\)
and breadth of rectangle = 1/ 2 m
∴ Perimeter of this rectangle = 2 \(\times\)(Length + Breadth
\(=2\times \left( \frac { 9 }{ 2 } +\frac { 1 }{ 2 } \right) =2\times \left( \frac { 9+1 }{ 2 } \right) \)
\(=2\times \frac { 10 }{ 2 } =10m\)
It is clear that, it has the perimeter equal to cross. So, we can say that cross has greater perimeter comparison to square.

According to the question,
Digit at ten's place = 4
Digit at unit's place = \(\frac{1}{4}\) of ten's place digit =\(\frac{1}{4}\times4=1\)
Digit at hundred's place = 0
Digit at thousand's place = 5 x Digit of unit's place
=5\(\times\)1=5
Digit at ten thousand's place = 2 x Digit of ten's place
=2\(\times\)4=8
∴ Required number = 85041

(a) Given, number of children, who got polio drops in March 2013 = 212583
Number of children, who got polio drops in next month = 216813
\(\therefore\) Difference of the number of children getting polio drops in two months
= 216813 - 212583 = 4230 children.
(b) Importance of pulse polio drops
The children are vaccinated for their protectior against deadly polio virus. All the babies below 5 yrs of age are given oral polio vaccine simultaneously, it helps to eradicate the virus.

100°

Draw a line and mark some points at equal distance on it as shown in the figure given below. Mark a point on it as zero. Points to the right of zero are positive integers and marked by +1, + 2, +3 erc., or simply 1, 2, 3 etc and points to the left of zero are negative integers and marked by -1, - 2, - 3 etc. Now, - 3 is a negative integer (since, - 3 has negative sign). So, move 3 points to the left of zero and represent it by point C. 7 is a positive integer (since + 7 has positive sign). So, move 7 points to the right of zero and represent it by point F. To mark - 4 on this line, move 4 points to the left of zero and represent it by point B. To mark - 8, on this line, move 8 points to the left of zero and represent it by point A. To mark - 1 on this line, move 1 point to the left of zero and represent it by point D. To mark 3 on this line, move 3 points to the right of zero and represent it by point E. Thus, we get the following representation of these integers on the number line:

(a) The following figure shows a closed curve, but not a polygon, because a polygon is made by line segments only.

(b) The following figure shows an open curve made up entirely of line segments.

(c) It is not possible, because a polygon of two sides cannot be drawn.

\(\frac { 6 }{ 2 } =\frac { 3 }{ 1 } \)=3:1

Perimeter = 4 x 2 = 8 m