Question

Two circles touch each other externally.The sum of their areas is 74π cm² and the distance between their centres is 12 cm.Find the diameters of the circles.

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Answer 1

Answer:-

ᴛʜᴇ ᴅɪᴀᴍᴇᴛᴇʀ ᴏғ ᴛʜᴇ ᴄɪʀᴄʟᴇ ᴡɪʟʟ ʙᴇ 10 ᴄᴍ & 14 ᴄᴍ.

Solution :-

Let the radius of the circles be r1

and r2,

So, r +2 = 12 =

r2 = 12-r1

Sum of the areas of the circles =74π

=>πr1²+ πR1²= 74π

=>r12+r2²= 74

=> I2+ (12- r1)² = 74

=>r1²+144-24r1 + r1² = 74

=>2r1²-24r1 +70=0

=>r1²-12r1+35 = 0

=>(r1-7)(r1-5) = 0

=> (r1-7)(r1-5)=0

=> r1=7 or r1 =5

Ir r1= 7 cm, then r2=5 cm

If r1=5 cm, then r2 = 7 cm

So, the diameters of the circles will

10 cm and 14 cm.

Answer 2

Let's radius of circle is r1 and r2

r1 + r2 = 12

r1 = 12 - r2

sum of area of circles = 70 π

$\sf \pi {r_1}^{2} + \pi {r_2}^{2} = 70\pi \\ \\ \sf{r_1}^{2} + {r_2}^{2} = 70 \\ \\ \sf{(12 - r_2 )}^{2} + {r_2 }^{2} = 70 \\ \\ \sf144 + 24r_2 + {r_2 }^{2} + {r_2 }^{2} = 70 \\ \\ \sf2 {r_2}^{2} - 24r_2 + 70 \\ \\ \sf {r_2}^{2} - 12r_2 + 35 \\ \\\sf {r_2}^{2} - 7r_2 - 5r_2 + 35 \\ \\\sf r_2(r_2 - 7) - 5(r_2 - 7) \\ \\ \sf(r_2 - 7)(r_2 - 5) \\ \\ \large{\boxed{ \sf \green{r_2 = 7 }}}$

$\sf r_1 = 12 - r_2 \\ \\\sf r_1 = 12 - 7 \\ \\ \large\boxed{ \sf \green{ r_1 = 5}}$

Radius of circle are 7 cm and 5 cm