A geometry problem by Riel Diala

Let the radius of
Circle A be $a$ ,
Circle B be $b$ ,
and, that of Circle C be $c$ .
Now,
$a+b=24\rightarrow Eq.1$
$a+c=25\rightarrow Eq.2$
$b+c=19\rightarrow Eq. 3$
Adding $Eq.1$ and $Eq.2$ , we get,
$2a+b+c=24+25$
$\Rightarrow 2a+(b+c)=49$
$\Rightarrow 2a+19=49$ [From $Eq.3$ ]
$\Rightarrow 2a=30$
$\Rightarrow a=15$
So, Radius of Circle A is $\boxed {a=15}$

• $A_R + B_R = 24$
• $A_R + C_R = 25$
• $B_R + C_R = 19$

$\begin{aligned} (A_R + B_R) + (A_R + C_R) - (B_R + C_R) &= 24 + 25 - 19 \\ 2A_R &= 30 \\ A_R &= \boxed{15} \\ \end{aligned}$

Let the radius of circle A be a

Let the radius of circle B be b

Let the radius of circle C be c

eq1: a + b = 24

eq2: b + c = 19

eq3: a + c = 25

There are many ways in doing the "systems of equations" but here's one way:

From eq1, we find a formula for b in terms of a

b = 24 - a

Since we have an equation for b, we substitute it for b in eq2

(24-a) + c = 19

24 - a + c = 19

c - a = -5

and now we can find for c in terms of a

c = a - 5

Now we substitute the equation to c in eq3

a + (a - 5) = 25

a + a -5 = 25

2a - 5 = 25

2a = 30

a = 15

Therefore, the radius of circle A is 15 units.