s
| Company | Number of employee |
| Company A | 300 |
| Company B | 700 |
| Company C | 500 |
| Company D | 400 |
| Company E | 500 |
| Company F | 800 |
| Company G | 400 |
| Day | Number of Cycles |
| Monday | 20 |
| Tuesday | 25 |
| Wednesday | 15 |
| Thursday | 20 |
| Friday | 30 |
| Saturday | 25 |
Step 2: Now choose a convenient scale in order to easily preparing the pictograph. Like one
represents 5 cycles
Thus, the number of cycles sold by the shop on Monday
Dotted line in the above figures shows perimetre.
Perimetre of the triangle \[\text{ABC}=\text{AB}+\text{BC}+\text{CA}\]
Solution:
Perimetre of the triangle \[\text{PQR=PQ}+\text{QR}+\text{RP}\] \[=\text{6}.\text{5 cm}+\text{4 cm}+\text{7 cm}=\text{17}.\text{5 cm}\]
Perimetre of the triangle \[\text{ABC}=\text{3}\times \text{AB}\]
Solution:
Perimetre of an equilateral triangle \[=\text{3}\times \text{side}\]
In the triangle PQR Perimetre of the triangle \[~\text{PQR}=\text{3}\times \text{PQ}\] \[=\text{3}\times \text{3}.\text{5 cm}=\text{1}0.\text{5 cm}\]
Perimetre of the triangle \[\text{ABC}=\text{2}\times \text{AB}+\text{BC}\]
Solution:
Perimetre of an isosceles triangle\[=2\times \text{length}\] of one of the equal sides + length of unequal side In the triangle PQR Perimeter of the triangle \[\text{PQR}=\text{2}\times \text{PQ}+\text{QR}=\text{2}\times \text{5 cm}+\text{4 cm}=\]\[\text{14 cm}\]
LO, PQ and AD are heights of the triangles \[\Delta \text{LMN},\Delta \text{PQR}\] and\[\Delta \text{ABC}\]respectively.
BC, LN and PQ are the bases of the triangles \[\Delta \text{ABC},\text{ }\Delta \text{LMN},\]and\[\Delta \text{PQR}\]respectively.
Solution:
Area of the triangle\[\text{ABC}=\frac{1}{2}\times \text{Base}\times \text{Height}\] \[=\frac{1}{2}\times \text{BC}\times \text{AD}=\frac{1}{2}\times \text{7cm}\times \text{1}0\text{ cm}=\text{35 c}{{\text{m}}^{\text{2}}}\].
Area of the rectangle\[~\text{ABCD}=\text{AB}\times \text{BC}\]
AB or CD is the length of the rectangle ABCD
PR or QS is the breadth of the rectangle PQRS.
(c)
(d)
(e)
We recognize a number of objects by their shapes. Therefore, to know about the objects, being aware about their shape is very important. In this chapter we will study about the shapes of different geometrical figures.
In the above figure, point A shows the location of India, point B shows the location of Australia, point C shows the location of Russia, point D shows the location of China, point E shows the location of Iran, point F shows the location of Algeria.
Solution: Point (e) shows the location of Bangalore on the map.
Solution: \[\overline{\text{AB}},\text{ }\overline{\text{BC}},\text{ }\overline{\text{CD}},\text{ }\overline{\text{DA}},\text{ }\overline{\text{AC}},\text{ }\overline{\text{BD}},\text{ }\overline{\text{OA}},\text{ }\overline{\text{OB}},\text{ }\overline{\text{OC}},\text{ }\overline{\text{OD}}\]
In the given figure, 0 is the centre of the circle.
In the given figure, OA is the radius of the circle. Note: All the radius of a circle are equal in length.
In the figure, AB is the chord of the circle.
In the figure, AB is the diametre of the circle
Note: All the diametres of a circle are equal in length.
In the given circle, OM and ON are the two radius of the circle whereas MN is the diametre of the circle. We can see, MN = OM + ON ..........(i)
We know that all the radius of a circle is equal in length. Thus OM = ON ..........(ii)
So, we can write the equation (i) as followings: ' \[\text{MN}=\text{OM}+\text{OM or MN}=\text{ON}+\text{ON}\] Thus MN = 20M or MN = 2ON Thus, we see that the diametre of a circle is twice of the radius.
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