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Properties of Triangles

Section: Mathematics

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Question : 1 of 50

Marks: +1, -0

In a triangle ABC , if

A, B

and C are in arithmetic progression,

r_3=r_1 r_2

c=10

a^2+b^2+c^2=

[AP EAPCET 24 May 2025 S1]

$128$
$288$
$392$
$200$

Solution:

In a triangle

A, B

and

C

are in arithmetic progression.

So,

B=\frac{\pi}{3}

Now,

r r_3=r_1 r_2

\Rightarrow \frac{\Delta}{s} \cdot \frac{\Delta}{s-c}=\frac{\Delta}{s-a} \cdot \frac{\Delta}{s-b}

\Rightarrow \frac{1}{s(s-c)}=\frac{1}{(s-a)(s-b)}

\Rightarrow s^2-(a+b)s+ab=s^2-cs

\Rightarrow ab=(a+b-c)s

\Rightarrow ab=(a+b-c)\frac{a+b+c}{2}

\Rightarrow 2ab=(a+b)^2-c^2

\Rightarrow 2ab=a^2+b^2+2ab-c^2

\Rightarrow a^2+b^2=c^2

Also, using cosine rule,

b^2=a^2+c^2-2ac\cos B

\Rightarrow b^2=a^2+c^2-2ac\cos\left(\frac{\pi}{3}\right)

\Rightarrow b^2=a^2+c^2-ac

\Rightarrow c^2-a^2=a^2+c^2-ac

\Rightarrow 2a^2=ac\Rightarrow 2a=c

\Rightarrow a=\frac{c}{2}=\frac{10}{2}=5

So,

b^2=c^2-a^2=100-25=75

\Rightarrow b=5\sqrt{3}

Now,

a^2+b^2+c^2=25+75+100=200

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