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Monday 6 May 2013

Solution

Q.1 ABC is an Isoceles triangle with AB=AC. A circle through B touching AC at the middle point intersects AB at P. Then AP:AB is:

Solution: First Draw the figure

Now The Informations are AB=AC and D is the middle point of AC so AD=DC=AC/2=AB/2 (Since AB=AC)

Since AD is the tangent to the Circle so 

AD^2= AP*AB
(AC/2)^2 = AP*AB
(AB/2)^2 = AP*AB
AB^2 / 4 = AP*AB
By Dividing both sides by AB^2 we get
1/4 = AP/AB

So the answer is AP:AB= 1:4

Q2. A,B,C and D purchase a gift worth Rs 60. A pays 1/2 of what others are paying, B pays 1/3rd of what others are paying and C pays 1/4th of what others are paying. What is the amount paid by D?

Solution: A+B+C+D=60 -----------> (Equation 1)

Now According to Question A= 1/2 (B+C+D) ------> B+C+D = 2A 
Put the value of B+C+D in Equation 1
A+2A =60 ------> 3A=60-------> A=20

Now again according to Question 
B=1/3 (A+C+D)--------> A+C+D= 3B
By Putting the Value of A+C+D in Equation 1
we get    B+3B= 60------> 4B=60 ---------> B=15

Now again accoring to Question 
C= 1/4 (A+B+D) ---------> A+B+D = 4C
By Putting the Value of A+B+D in equation 1 
We get C+4C = 60------> 5C= 60 ------> C=12

Since A+B+C+D= 60
20+15+12+D=60------> D=60-47=13

Answer : D=13

Thursday 2 May 2013

Solution of (The lengths of three median. Of a triangle are 9cm 12 cm and 15cm.the area of triangle is)

The lengths of three median. Of a triangle are 9cm 12 cm and 15cm.the area of triangle is:


In this triangle medians AD =12, BE =9 and CF=15.
Because the median cut each other on centroid G in the ratio of 2:1 so the segments AG= 8, GD=4, BG=6, GE=3, CG=10 AND GF= 5

Now we take Triangle ABG, in this Triangle we are seeing that one side AG=8 and other side BG= 6 so the third side AB should be 10 and the triangle ABG is a right angled triangle making an angle 90 degree at the point G.

Now take a bigger triangle ABE. In this triangle we know that AG is perpendicular to BE and so the area of the Triangle ABE should be

1/2 AG*BE= 1/2 * 8*9 = 36

Now we know that any median of triangle bisects the triangle into two triangle of equal areas so the median BE bisects the Triangle ABC into two triangles, ABE and BED, of Equal area so 

area of ABE = area of BED = 36 so the area of triangle ABC= area of ABE+ area of BED = 36+36= 72


Monday 15 April 2013

Some Important Theorems of Circle

THEOREMS: 

1. The angle at the centre is twice the angle at the circumference.



2.The angle in a semi-cicle is 90°.


3. Angles in the same segment are equal.


4. Opposite angles in a cyclic quadrilateral add up to 180°.

5. The lengths of the two tangents from a point to a circle are equal.

6.The angle between a tangent and a radius in a circle is 90°.

7. Alternate segment theorem:

The angle (α) between the tangent and the chord at the point of         contact (D) is equal to the angle (β) in the alternate segment*.


MENSURATION FORMULAS


Mensuration Formulas


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)
 A = l \times b

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

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

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

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

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

7. Area of a triangle (A) = (Base(b) × Height(b)) / 2
 A = \frac{1}{2} \times b \times h
Triangle
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 =  A = \sqrt{s(s-a)(s-b)(s-c)}
This formulas is also knows as “Heron’s formula”.

8. Area of triangle(A) = 1/2 ab sinC = 1/2 ac sinB = 1/2 bc sinA
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
area of triangle - mensuration

9. Area of isosceles triangle = \frac{b}{4}\sqrt{4a^2 - b^2}
Where a = length of two equal side , b= length of base of isosceles triangle.

10. Area of trapezium (A) = \frac{1}{2} (a+b) \times h
Where “a” and “b” are the length of parallel sides and “h” is the perpendicular distance between “a” and “b” .
Trapezium

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)
quadrilateral

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) =  \pi r^2 = \frac{\pi d^2}{4}
Where r = radius of the circle and d = diameter of the circle.

18. Circumference of a Circle =  2 \pi r = \pi d
r= radius of circle
d= diameter of circle

19. Total surface area of cuboid =  2 (lb + bh + lh)
where l= length , b=breadth , h=height

20. Total surface area of cuboid =  6 l^2
where l= length

21. length of diagonal of cuboid =  \sqrt{l^2+b^2+h^2}

22. length of diagonal of cube =  √3 l

23. Volume of cuboid = l × b × h

24. Volume of cube = l × l × l

25. Area of base of a cone = \pi r^2

26.  Curved surface area of a cone = C = \pi \times r \times l
Where r = radius of base , l = slanting height of cone

27. Total surface area of a cone =  \pi r (r+l)

28. Volume of right circular cone =  \frac{1}{3} \pi r^2 h
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 =  2 \pi r h
Where r = radius of base, h = height of cylinder

34. Total surface area of a cylinder =  2 \pi r(r + h)

35. Volume of a cylinder =  \pi r^2 h

36. Surface area of sphere =  4 \pi r^2 = \pi d^2
where r= radius of sphere, d= diameter of sphere

37. Volume of a sphere =  \frac{4}{3} \pi r^3 = \frac{1}{6} \pi d^3

38. Volume of hollow cylinder = \pi r h(R^2-r^2)
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
net diagram of right square pyramid
39.a Surface area of a right square pyramid =  a \sqrt{4b^2 - a^2}
39.b Volume of a right square pyramid =  \frac{1}{2} \times base \, \, area \times height

40. Square Pyramid:
40.a. Johnson Pyramid:
net diagram of johnson pyramid
net diagram of johnson pyramid
Volume = (1+ \sqrt{3})\times a^2
Total Surface Area: \frac{\sqrt{2}}{6} \times a^3
40.b. Normal Square pyramid:
If a = length of square base and h = height of the pyramid then:
Volume = V=\frac{1}{3}a^2h
Total Surface Area = a^2+a\sqrt{a^2+(2h)^2}

41. Area of a regular hexagon =  \frac{3\sqrt{3}a^2}{2}

42. area of equilateral triangle =  \frac{\sqrt{3}}{4} a^2

43. Curved surface area of a Frustums = \pi h (r_1 + r_2)          (h = lateral height)

44. Total surface area of a Frustums = \pi (r_1^2 + h(r_1+r_2) + r_2^2)  (h= lateral height)

45. Curved surface area of a Hemisphere =  2 \pi r^2

46. Total surface area of a Hemisphere =  3 \pi r^2

47. Volume of a Hemisphere =   \frac{2}{3} \pi r^3 = \frac{1}{12} \pi d^3

48. Area of sector of a circle =  \frac{\theta r^2 \pi}{360}
where  \theta  = measure of angle of the sector , r= radius of the sector