Home
Class 11
MATHS
Internal bisectors of DeltaABC meet the ...

Internal bisectors of `DeltaABC` meet the circumcircle at points D, E and F then
area of `DeltaDEF` is

A

`2R^(2)cos^(2)((A)/(2))cos^(2)((B)/(2))cos^(2)((C)/(2))`

B

`2R^(2)sin((A)/(2))sin((B)/(2))sin((C)/(2))`

C

`2R^(2)sin^(2)((A)/(2))sin^(2)((B)/(2))sin^(2)((C)/(2))`

D

`2R^(2)cos((A)/(2))cos((B)/(2))cos((C)/(2))`

Text Solution

Verified by Experts

The correct Answer is:
D
Promotional Banner

Topper's Solved these Questions

  • PROPERTIES OF TRIANGLES

    AAKASH SERIES|Exercise ADDITIONAL PRACTICE EXERCISE (LEVEL II) (PRACTICE SHEET (ADVANCED) Straight Objective Type Questions)|10 Videos

  • PROPERTIES OF TRIANGLES

    AAKASH SERIES|Exercise ADDITIONAL PRACTICE EXERCISE (LEVEL II) (More than one correct answer Type Questions)|7 Videos

  • PROPERTIES OF TRIANGLES

    AAKASH SERIES|Exercise ADDITIONAL PRACTICE EXERCISE (LEVEL II) (LECTURE SHEET (ADVANCED)) (More than one correct answer Type Questions)|19 Videos

  • PLANES

    AAKASH SERIES|Exercise ADVANCED SUBJECTIVE TYPE QUESTIONS|30 Videos

  • PROPERTIES OF VECTORS

    AAKASH SERIES|Exercise PRACTICE EXERCISES|55 Videos

Similar Questions

Explore conceptually related problems

Internal bisectors of DeltaABC meet the circumcircle at points D, E and F then Ratio of area of triangle ABC and triangle DEF is

Internal bisectors of DeltaABC meet the circumcircle at points D, E and F then Length of side EF is

In DeltaABC with fixed length of BC , the internal bisector of angle C meets the side AB at D and the circumcircle at E . The maximum value of CD xx DE is

In DeltaABC the midpoints are D,E and F of the sides AB,BC and CA, then DeltaDEF:DeltaABC is

In a triangle ABC bisector of lfloor C meets the side AB at D and circum circle at E. The maximum value of CD×DE is

The internal angular bisectors of angles of a Delta^("le") ABC meets circum circle of Delta^("le") ABC at D,E,F respectively then ("Area of " Delta^("le") ABC)/("Area of " Delta^("le") DEF) =

In a DeltaDEF , A, B and C are mid points of EF,FD and DE respectively. IF the area of DeltaDEF is 14.4 cm^2 then find the area of DeltaABC .