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  • In triangle ABC ratio of sides a:b:c=3:4:5 and it has a incircle which has center at O then find ratio of area of <img alt="\Delta AOB:\Delta BOC:\Delta COA" src="https://entrancecorner.oncode

In triangle ABC ratio of sides a:b:c=3:4:5 and it has a incircle which has center at O then find ratio of area of \Delta AOB:\Delta BOC:\Delta COA 

Option: 1

5:4:3


Option: 2

25:16:9


Option: 3

3:4:5


Option: 4

9:16:25


Answers (1)

best_answer

 

 

Area of Triangle -

Area of Triangle

Recall that the area formula for a triangle is given as Area = ½ bh, where ‘b’ is base and ‘h’ is the height. For oblique triangles, we must find ‘h’ before we can use the area formula. 

 

   

\\\mathrm{Area=\frac{1}{2} \;base\times height}\\\\\mathrm{\quad\quad\;\; =\frac{1}{2}\mathit{\;b}\cdot{c\sin A}}

 

Area of triangle ABC is represented by Δ, Thus

\\\mathrm{Area\;of\;\Delta ABC= \Delta=\frac{1}{2}\mathit{\;b}\cdot{c\sin A}}\\\\\mathrm{\quad\quad\quad\quad\quad\quad\quad\quad\quad=\frac{1}{2}\mathit{\;a}\cdot{b\sin C}}\\\\\mathrm{\quad\quad\quad\quad\quad\quad\quad\quad\quad=\frac{1}{2}\mathit{\;c}\cdot{a\sin B}}

 

NOTE:

  1. Area of triangle in terms of sides (heron’s Formula)

\\\mathrm{\Delta=\frac{1}{2}\mathit{\;b}\cdot{c\sin A}=\frac{1}{2}\mathit{bc}\cdot2\sin \frac{A}{2}\cos\frac{A}{2}}\\\\\text{use half angle formula}\\\\\mathrm{\;\;\;=\frac{1}{2}\sqrt{\mathit{\frac{(s-b)(s-c)}{bc}}}\sqrt{\mathit{\frac{s(s-a)}{bc}}}}\\\\\mathrm{\;\;\;=\sqrt{\mathit{s(s-a)(s-b)(s-c)}}}

 

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Diagram-

By the above Diagram we can see 

\Delta AOB:\Delta BOC:\Delta COA=\frac{1}{2}\times3k \times r:\frac{1}{2}\times4k \times r:\frac{1}{2}\times5k \times r\\ =3:4:5

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jitender.kumar

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