18/20 = 9/10

1. Perimeter of the semicircle + Diameter of the circle
   = {tex}\pi (\frac{2.8}{2}){/tex} cm + 2.8 cm
   = {tex}\frac{22}{7} \times{/tex} 1.4 cm + 2.8 cm
   = 4.4 cm + 2.8 cm
   = 7.2 cm
2. Perimeter of the 3 sides of rectangle + Perimeter of semicircle
   = 2.8 cm + 1.5 cm + 1.5 cm + {tex}\pi (\frac{2.8}{2}){/tex} cm 
   = 5.8 cm + {tex}\frac{22}{7} \times{/tex} 1.4 cm 
   = 5.8 cm + 4.4 cm
   = 10.2 cm
3. Perimeter of semicircle + Perimeter of 2 sides of the triangle
   = 2 cm + 2 cm + {tex}\pi (\frac{2.8}{2}){/tex} cm 
   = 4 cm + {tex}\frac{22}{7} \times{/tex} 1.4 cm
   = 4 cm + 4.4 cm
   = 8.4 cm

Therefore, the ant would have to take a longer round for food pieces (b).

Mensuration is the branch of mathematics which deals with the study of Geometric shapes, their area, volume and related parameters.

Some important mensuration formulas are:

1. Area of rectangle (A) = length(l) × Breath(b)

2. Perimeter of a rectangle (P) = 2 × (Length(l) + Breath(b))

3. Area of a square (A) = Length (l) × Length (l)

4. Perimeter of a square (P) = 4 × Length (l)

5. Area of a parallelogram(A) = Length(l) × Height(h)

6. Perimeter of a parallelogram (P) = 2 × (length(l) + Breadth(b))

7. Area of a triangle (A) = (Base(b) × Height(b)) / 2

And for a triangle with sides measuring “a” , “b” and “c” , Perimeter = a+b+c

and s = semi perimeter = perimeter / 2 = (a+b+c)/2

And also: Area of triangle = 

This formulas is also knows as “Heron’s formula”.

8. Area of triangle(A) = 

Where A, B and C are the vertex and angle A , B , C are respective angles of triangles and  a , b , c are the respective opposite sides of the angles as shown in figure below:

area of triangle - mensuration

9. Area of isosceles triangle = 

Where a = length of two equal side , b= length of base of isosceles triangle.

10. Area of trapezium (A) = 

Where “a” and “b” are the length of parallel sides and “h” is the perpendicular distance between “a” and “b” .

11. Perimeter of a trapezium (P) = sum of all sides

12. Area of rhombus (A) =  Product of diagonals / 2

13. Perimeter of a rhombus (P) = 4 × l

where l = length of a side

14. Area of quadrilateral (A) = 1/2 × Diagonal × (Sum of offsets)

15.  Area of a Kite (A) = 1/2 × product of it’s diagonals

16. Perimeter of a Kite (A) = 2 × Sum on non-adjacent sides

17.  Area of a Circle (A) = 

Where r = radius of the circle and d = diameter of the circle.

18. Circumference of a Circle = 

r= radius of circle

d= diameter of circle

19. Total surface area of cuboid = 

where l= length , b=breadth , h=height

20. Total surface area of cuboid = 

where l= length

21. length of diagonal of cuboid = 

22. length of diagonal of cube = 

23. Volume of cuboid = l × b × h

24. Volume of cube = l × l × l

25. Area of base of a cone = 

26.  Curved surface area of a cone = C = 

Where r = radius of base , l = slanting height of cone

27. Total surface area of a cone = 

28. Volume of right circular cone = 

Where r = radius of base of cone , h= height of the cone (perpendicular to base)

29. Surface area of triangular prism = (P × height) + (2 × area of triangle)

Where p = perimeter of base

30. Surface area of polygonal prism = (Perimeter of base × height ) + (Area of polygonal base × 2)

31. Lateral surface area of prism = Perimeter of base × height

32. Volume of  Triangular prism = Area of the triangular base × height

33. Curved surface area of  a cylinder = 

Where r = radius of base, h = height of cylinder

34. Total surface area of a cylinder = 

35. Volume of a cylinder = 

36. Surface area of sphere = 

where r= radius of sphere, d= diameter of sphere

37. Volume of a sphere = 

38. Volume of hollow cylinder = 

where , R = radius of cylinder , r= radius of hollow , h = height of cylinder

39. Right Square Pyramid:

If a = length of base , b= length of equal side  ; of the isosceles triangle forming the slanting face , as shown in figure:

net diagram of right square pyramid

39.a Surface area of a right square pyramid = 

39.b Volume of a right square pyramid = 

40. Square Pyramid:

40.a. Johnson Pyramid:

Volume = 
Total Surface Area: 

40.b. Normal Square pyramid:

If a = length of square base and h = height of the pyramid then:
Volume = 
Total Surface Area = 

41. Area of a regular hexagon = 

42. area of equilateral triangle = 

43. Curved surface area of a Frustums = 

44. Total surface area of a Frustums = 

45. Curved surface area of a Hemisphere = 

46. Total surface area of a Hemisphere = 

47. Volume of a Hemisphere =  

48. Area of sector of a circle = 

where  = measure of angle of the sector , r= radius of the sector

{tex}\begin{array}{l}x^2+5x+6\\=x^2+3x+2x+6\\=x(x+3)+2(x+3)\\=(x+3)(x+2)\end{array}{/tex}

Suppose the actual speed of the train be x km/hr and the actual time taken be y hours. Then,
Distance covered =(xy) km ...........(i) [{tex}\therefore{/tex} Distance = Speed {tex}\times{/tex}Time]
If the speed is increased by {tex}6km/hr{/tex}, then time of journey is reduced by 4 hours i.e., when speed is {tex}(x + 6)km/hr{/tex}, time of journey is {tex}(y-4)hours.{/tex}
{tex}\therefore{/tex} Distance covered = {tex}(x + 6) (y - 4){/tex}
{tex}\Rightarrow{/tex}{tex}xy = (x + 6) (y - 4){/tex} [Using (i)]
{tex}\Rightarrow{/tex}{tex}-4x + 6y -24=0{/tex}
{tex}\Rightarrow{/tex}{tex}-2x + 3y -12=0{/tex} ..........(ii)
when the speed is reduced by 6 km/hr, then the time of journey is increased by 6 hours i.e.,
when speed is {tex}(x-6)km/hr{/tex}, time of journey is {tex}(y - 6)hours.{/tex}
{tex}\therefore{/tex} Distance covered ={tex}(x -6) (y + 6){/tex}
{tex}\Rightarrow{/tex} {tex}xy = (x - 6) (y + 6){/tex} [Using (i)]
{tex}\Rightarrow{/tex} {tex}6x - 6y -36 =0{/tex}
{tex}\Rightarrow{/tex} {tex}x - y -6=0 {/tex}...........(iii)
Thus, we obtain the following system of equations:
{tex}-2x + 3y -12 =0{/tex}
{tex}x - y - 6=0{/tex}
By using cross-multiplication, we have
{tex}\frac { x } { 3 \times - 6 - ( - 1 ) \times - 12 } = \frac { - y } { - 2 \times - 6 - 1 \times - 12 } = \frac { 1 } { - 2 \times - 1 - 1 \times 3 }{/tex}
{tex}\Rightarrow \quad \frac { x } { - 30 } = \frac { - y } { 24 } = \frac { 1 } { - 1 }{/tex}
{tex}\Rightarrow{/tex} {tex}x =30\ and\ y=24.{/tex}
Putting the values of x and y in equation (i), we obtain
Distance= {tex}(30\times24)km =720km.{/tex}
Hence, the length of the journey is {tex}720 km{/tex}.

When a circle is evenly divided into four sections by two perpendicular lines,

each of the four areas is a quadrant

Average {tex} = \frac{{sum\;of\;all\;observations}}{{total\;number\;of\;observations}}{/tex}

25555 ×  99 = 2529945

divide 
p4+p3-p2+1 by p-1

refer to the attachment given below:

Explanation: