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Writing Patterns in Geometry

In a triangle...

Question

In a triangle $A B C,$ if the radii of ex-circles $r_{1}, r_{2}$ and $r_{3}$ are given by $r_{1} = 8, r_{2} = 12$ and $r_{3} = 24,$ then:

in radius

r = 8

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side

a = 12

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side

b = 16

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side

c = 20

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Solution

The correct options are
B side $a = 12$
C side $b = 16$
D side $c = 20$
$a = \frac{r_{1} (r_{2} + r_{3})}{\sqrt{r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1}}}$
substituting the values for $r_{1} = 8, r_{2} = 12$ and $r_{3} = 24$ we get
$a = \frac{8 \times 36}{24} = 12$
$b = \frac{r_{2} (r_{3} + r_{1})}{\sqrt{r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1}}} = \frac{12 \times 32}{24} = 16$
$c = \frac{r_{3} (r_{1} + r_{2})}{\sqrt{r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1}}} = \frac{24 \times 20}{24} = 20$
Also, $r_{1} r_{2} r_{3} = r s^{2}$
$\Rightarrow r = \frac{r_{1} r_{2} r_{3}}{s^{2}}$
$= \frac{8 \times 12 \times 24}{{(24)}^{2}} = 4$

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In a triangle ABC, if the radii of ex-circles r1,r2 and r3 are given by r1=8,r2=12 and r3=24, then:

The correct options are B side a=12 C side b=16 D side c=20a=r1(r2+r3)√r1r2+r2r3+r3r1substituting the values for r1=8,r2=12 and r3=24 we geta=8×3624=12b=r2(r3+r1)√r1r2+r2r3+r3r1=12×3224=16c=r3(r1+r2)√r1r2+r2r3+r3r1=24×2024=20Also, r1r2r3=rs2⇒r=r1r2r3s2 =8×12×24(24)2=4

In a triangle $A B C,$ if the radii of ex-circles $r_{1}, r_{2}$ and $r_{3}$ are given by $r_{1} = 8, r_{2} = 12$ and $r_{3} = 24,$ then: