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TG EAMCET 2 May 2025 Shift 2 Paper

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Question : 28 of 160

Marks: +1, -0

In a

\triangle ABC

r_1=4, r_2=8

r_3=24

a: b: c=

$4:7:9$
$2:3:5$
$1:2:6$
$6:2:1$

Solution:

We have,

r_1=4, r_2=8

and

r_3=24

\therefore \;4=\frac{\Delta}{s-a}, 8=\frac{\Delta}{s-b} \; \text{ and } \; 24=\frac{\Delta}{s-c}

\Rightarrow s-a=\frac{\Delta}{4} \dots (i)

s-b=\frac{\Delta}{8} \dots (ii)

\text{ And } \; s-c=\frac{\Delta}{24} \; \dots (iii)

On adding these three equations, we get

\;3s-(a+b+c)=\frac{\Delta}{4}+\frac{\Delta}{8}+\frac{\Delta}{24}

\Rightarrow 3s-2s=\frac{6\Delta+3\Delta+\Delta}{24}=\frac{10\Delta}{24} \Rightarrow s=\frac{5}{12}\Delta

From Eq. (i),

a=s-\frac{\Delta}{4}=\frac{5\Delta}{12}-\frac{\Delta}{4}=\frac{2\Delta}{12}=\frac{\Delta}{6}

From Eq. (ii),

b=s-\frac{\Delta}{8}=\frac{5}{12}\Delta-\frac{\Delta}{8}=\frac{7}{24}\Delta

From Eq. (iii),

c=s-\frac{\Delta}{24}=\frac{5}{12}\Delta-\frac{\Delta}{24}=\frac{9}{24}\Delta

\therefore a:b:c=\frac{\Delta}{6}:\frac{7}{24}\Delta:\frac{9}{24}\Delta=4:7:9

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