MTH2: CYCLIC QUADRILATERALS

This unit is about Cyclic quadrilaterals where its opposite angles are supplementary, the unit dwell on Mensuration.

CyclicQuadrilateral

A cyclic quadrilateral is a quadrilateral for which a circle can be circumscribed so that it touches each polygon vertex. A quadrilateral that can be both inscribed and circumscribed on some pair of circles is known as a bicentric quadrilateral.

The area of a cyclic quadrilateral is the maximum possible for any quadrilateral with the given side lengths. The opposite angles of a cyclic quadrilateral sum to pi radians (Euclid, Book III, Proposition 22; Heath 1956; Dunham 1990, p. 121). There exists a closed billiards path inside a cyclic quadrilateral if its circumcenter lies inside the quadrilateral (Wells 1991, p. 11).

The area is then given by a special case of Bretschneider’s formula. Let the sides have lengths abc, and d, let s be the semiperimeter

 s=1/2(a+b+c+d),
(1)

and let R be the circumradius. Then

A=sqrt((s-a)(s-b)(s-c)(s-d))
(2)
=(sqrt((ac+bd)(ad+bc)(ab+cd)))/(4R),
(3)

the first of which is known as Brahmagupta’s formula. Solving for the circumradius in (2) and (3) gives

 R=1/4sqrt(((ac+bd)(ad+bc)(ab+cd))/((s-a)(s-b)(s-c)(s-d))).
(4)

The diagonals of a cyclic quadrilateral have lengths

p=sqrt(((ab+cd)(ac+bd))/(ad+bc))
(5)
q=sqrt(((ac+bd)(ad+bc))/(ab+cd)),
(6)

so that pq=ac+bd.

In general, there are three essentially distinct cyclic quadrilaterals (modulo rotation and reflection) whose edges are permutations of the lengths abc, and d. Of the six corresponding polygon diagonals lengths, three are distinct. In addition to p and q, there is therefore a “third” polygon diagonal which can be denoted r. It is given by the equation

 r=sqrt(((ad+bc)(ab+cd))/(ac+bd)).
(7)

This allows the area formula to be written in the particularly beautiful and simple form

 A=(pqr)/(4R).
(8)

The polygon diagonals are sometimes also denoted pq, and r.

CyclicQuadRectangleCyclicQuadGrid

The incenters of the four triangles composing the cyclic quadrilateral form a rectangle. Furthermore, the sides of the rectangle are parallel to the lines connecting the mid-arc points between each pair of vertices (left figure above; Fuhrmann 1890, p. 50; Johnson 1929, pp. 254-255; Wells 1991). If the excenters of the triangles constituting the quadrilateral are added to the incenters, a 4×4 rectangular grid is obtained (right figure; Johnson 1929, p. 255; Wells 1991).

CyclicQuadPoints

Consider again the four triangles contained in a cyclic quadrilateral. Amazingly, the triangle centroids M_inine-point centers N_i, and orthocenters H_i formed by these triangles are similar to the original quadrilateral. In fact, the triangle formed by the orthocenters is congruent to it (Wells 1991, p. 44).

A cyclic quadrilateral with rational sides abc, and dpolygon diagonals p and qcircumradius r, and area a is given by a=25b=33c=39d=65p=60q=52r=65/2, and a=1344.

Let AHBO be a quadrilateral such that the angles ∠HAB and ∠HOB are right angles, then AHBO is a cyclic quadrilateral (Dunham 1990). This is a corollary of the theorem that, in a right triangle, the midpoint of the hypotenuse is equidistant from the three vertices. Since M is the midpoint of both right triangles DeltaAHB and DeltaBOH, it is equidistant from all four vertices, so a circle centered at M may be drawn through them. This theorem is one of the building blocks of Heron’s derivation of Heron’s formula.

CyclicQuadCircumcenter

An application of Brahmagupta’s theorem gives the pretty result that, for a cyclic quadrilateral with perpendicular diagonals, the distance from the circumcenter O to a side is half the length of the opposite side, so in the above figure,

 OM_(AB)=1/2CD=CM_(CD)=DM_(CD),
(9)

and so on (Honsberger 1995, pp. 37-38).

CyclicQuadOrthocenter

Let M_(AC) and M_(BD) be the midpoints of the diagonals of a cyclic quadrilateral ABCD, and let P be the intersection of the diagonals. Then the orthocenter of triangle DeltaPM_(AC)M_(BD) is the anticenter T of ABCD (Honsberger 1995, p. 39).

CyclicQuadCircles

Place four equal circles so that they intersect in a point. The quadrilateral ABCD is then a cyclic quadrilateral (Honsberger 1991). For a convex cyclic quadrilateral Q, consider the set of convex cyclic quadrilaterals Q_∥ whose sides are parallel to Q. Then the Q_∥ of maximal area is the one whose polygon diagonals are perpendicular (Gürel 1996).

           A VIDEO ABOUT CYCLIC  QUADRILATERALS

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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) = \frac{1}{2} a \times b \times \angle C = \frac{1}{2} b \times c \times \angle A = \frac{1}{2} a \times c \times \angle B

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 =  \sqrt{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)

 

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

 

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

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ASSIGNMENT : CYCLIC QUADRILATERALS ASSIGNMENT MARKS : 30  DURATION : 2 hours

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