Math, asked by frostshawn0, 9 months ago

In a right ∆ABC (angle B=90°), AB=5cm and BC=12cm. CD and AE are angle bisectors of angle C and A respectively and they intersect at I. Find thearea of ∆DIE.


amitnrw: Find inradius first

Answers

Answered by amitnrw
4

Given :  In a right ∆ABC (angle B=90°), AB=5cm and BC=12cm. CD and AE are angle bisectors of angle C and A respectively and they intersect at I.

To find : area of ∆DIE

Solution:

AB = 5 cm

BC = 12 cm

AC² = AB² + BC² = 5² + 12²  = 13²²

=> AC = 13

Area of ΔABC = (1/2) AB * BC =  (1/2) * 5 * 12 = 30 cm²

Area of ΔABC = (1/2)(AB + BC + AC) * r  =  (1/2)*(5 + 12 + 13)r  = 15r

( r = inradius)

15r = 30

=> r = 2

CD is angle bisector

=> AC/AD  = BC/BD

=> 13/(5 - BD) = 12/BD

=> 13BD = 60 - 12BD

=> BD = 12/5

AE is angle bisector

=> AB/BE  = AC/CE

=> 5/BE =  13/(12 - BE)

=> 60 - 5BE = 13BE

=> BE = 10/3

Area of ΔDIE =  Area of Δ DBI + Area of ΔEBI - Area of  Δ DBE

= (1/2) BD * r  + (1/2)BE * r  - (1/2)BD * BE

= (1/2) (12/5) * 2 + (1/2)(10/3)*2   -  (1/2)(12/5)(10/3)

= 12/5   + 10/3 - 4

= 86/15  - 4

= 26/15

area of ∆DIE = 26/15 cm²

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