Question

himmat hai to solve karke dikhao. ​

Answer 1

SOLUTION :-

=> Let, r1 and r2 be the the radius of two circles.

=> According to question ,

=> π(r1)²+ π(r2)² = 116π

=> (r1)² + (r2)² = 116 --------(1)

=> °.° r2-r1 = 6 -------(Given)

=> .°. r2 = 6 + r1

=> [ Put the value of r2 in equ. (1)]

=> (r1)² + (6+(r1)²) = 116

=> 2(r1)²+ 36 + (r1)² + 12r1 = 116

=> 2(r1)² + 12r1 -80 = 0

=> (r1)²+ 6r1 - 40 = 0

=> (r1)² + 10r1 - 4r1 - 40 = 0

=> r1(r1+10) -4(r1+10) = 0

=> (r1+10) ( r1-4)=0

=> r1 = -10(Neglect) or r1 = 4cm.

=> Where r1 = 4cm

=> r2 = 6 + r1

=> r2 = 6+4

=>.°. r2 = 10cm.

Hence, the radius of the circle is 4cm and 10cm.

Answer 2

$\huge\underline\mathbb {SOLUTION:-}$

Let, r and R be the radii of smaller and bigger circles respectively, then

• OO = R + r

$\implies$ R - r = 6 Cm ...... (i)

It is also given that:

Sum of the area of the two circles = 166 π Cm²

$\implies$πR² + πr² = 116 π

$\implies$π(R² + r²) = 116 π

$\implies$R² + r² = 116 ...... (ii)

We know that,

(R - r)² = R² + r² - 2Rr ..... (iii)

Putting the values of (i), (ii) and (iii),We get:

• (6)² = 116 - 2Rr

$\implies$ 36 = 116 - 2Rr

$\implies$ 2Rr = 80

$\implies$ Rr = 40

Now,

$\implies$(R + r)² = R² + r² + 2Rr

$\implies$116 + 2 × 40

[From (ii) R² + r² = 116]

$\implies$116 + 80

$\implies$196

$\implies$ R + r = 14 ..... (iv)

On solving (i) and (iv),We get:

• R = 10 Cm
• and r = 4 Cm