Home
Class 12
MATHS
At the endpoint and midpoint of a circul...

At the endpoint and midpoint of a circular are AB, tangent lines are drawn, and the points, A and B are jointed with a chord. Prove that the ratio of the areas of the triangles thus formed tends to 4 as the arc AB decreases infinitely.

Text Solution

Verified by Experts

The correct Answer is:
4
Promotional Banner

Topper's Solved these Questions

  • LIMIT

    MOTION|Exercise EXERCISE-4|17 Videos

  • LIMIT

    MOTION|Exercise EXERCISE-2(LEVEL-2)|5 Videos

  • INVERSE TRIGONOMETRIC FUNCTIONS

    MOTION|Exercise Exercise -4 Level -II|7 Videos

  • MATRICES

    MOTION|Exercise Exercise - 4 (Level-II)|28 Videos

Similar Questions

Explore conceptually related problems

From a point P outside a circle with centre O , tangents PA and PB are drawn to the circle. Prove that OP is the right bisector of the line segment AB .

Through each vertex of a triangle, a line parallel to the opposite side is drawn. The ratio of the perimeter of the new triangle, thus formed, with that of the original triangle is

In a triangle ABC, a line is drawn from C which bisects AB at point D. Find the ratio of area of the triangles DBC and ABC.

ABCD is a trapezium with AB //DC. A line parallel to AC intersects AB at point M and BC at point N. Prove that: area of triangleADM=area of triangleACN.

In Delta ABC, D and E are the midpoint of AB and AC respectively. Find the ratio of the areas of Delta ADE and Delta ABC .

If A=(1,-3) and B=(4,3) are points on the curve y=x-(4)/(x) then the points on the curve at which the tangents are parallel to the chord AB are

From the point (4, 6) a pair of tangent lines are drawn to the parabola, y^(2) = 8x . The area of the triangle formed by these pair of tangent lines & the chord of contact of the point (4, 6) is

Points A and B are on a circle of radius 5 and AB=6. Point C is the midpoint of the minor arc AB length of the line segement AC, is:

Tangents are drawn from the point (-1,2) to the parabola y^(2)=4x The area of the triangle for tangents and their chord of contact is

MOTION-LIMIT-EXERCISE-3
  1. A circular are of radius 1 subtends an angle of x radians 0 < x < pi/2...

    Text Solution

    |

  2. A circular are of radius 1 subtends an angle of x radians 0 < x < pi/2...

    Text Solution

    |

  3. A circular are of radius 1 subtends an angle of x radians 0 < x < pi/2...

    Text Solution

    |

  4. At the end points A,B of the fixed segment of length L, lines are draw...

    Text Solution

    |

  5. At the endpoint and midpoint of a circular are AB, tangent lines are ...

    Text Solution

    |

  6. lim(x->0) [ ln(1+sin^2x) cot ln^2(1+x)]

    Text Solution

    |

  7. lim(x->oo)(a(2x^3-x^2)+b(x^3+5x^2-1)-c(3x^3+x^2))/(2(5x^4-x)-b x^4+c(4...

    Text Solution

    |

  8. find a&b if lim(x->oo) ((x^2+1)/(x+1)-ax-b)=0

    Text Solution

    |

  9. lim(x rarr oo) (sqrt(x^2-x+1)-a x-b)=0, then a+b=

    Text Solution

    |

  10. lim(x->0)[lim(x->oo)(e x p(xln(1+(a y)/x))-e x p(x In(1+(b y)/x)))/y]

    Text Solution

    |

  11. underset( x rarr 1) ( "Lim") ( tan "" ( pi x)/( 4))^(tan""( pix)/( 2))

    Text Solution

    |

  12. Let f(x)=(sin^-1(1-{x})*cos^-1(1-{x}))/(sqrt(2{x})(1-{x})), then find ...

    Text Solution

    |

  13. evaluate lim(x->a)1/((a^2-x^2)^2)((a^2+x^2)/(a x)-2sin((api)/2)sin((pi...

    Text Solution

    |

  14. Let x0=2(cos)pi/6and xn=sqrt(2+x(n-1)).n=1,2,3,....... find Lim(n->oo)...

    Text Solution

    |

  15. lim(x rarr 0)[ln(1+x)^(1+x)/(x^2)-1/x]

    Text Solution

    |

  16. f(x) is the function such that underset( x rarr 0 ) ("Lim") (f(x))/( x...

    Text Solution

    |

  17. The value of ("lim")(xvec1)(p/(1-x^p)-q/(1-x^q)),p ,q , in N , equal ...

    Text Solution

    |

  18. While finding the Sine of a certain angle x, an absent minded professo...

    Text Solution

    |

  19. While finding the Sine of a certain angle x, an absent minded professo...

    Text Solution

    |

  20. While finding the Sine of a certain angle x, an absent minded professo...

    Text Solution

    |