Solve fast: 21 Two circles touch internally The sum of their areas is 116 cm2 - Maths - Arithmetic Progressions

Question

Solve fast:

21 Two circles touch internally The sum of their areas is 116 π
cm2 and the distance between their centres is 6 cm Find
the radis of the circlm

22 If 1+4+7+10+ +x=287, find the value of x

Brijendra Pal · Accepted Answer

Hi, 22: 1 + 4 + 7 + 10 + + x = 287 Let there be n terms in the A P

21;-

Let the centre of the two circles be O1 and O2 and radii be r 1 = (O1A) and r 2 = (O2A) Both touch each other of A such that O1 O2 = 6 cm Now O1A – O2A = O1 O2 = 6 cm ⇒ r 1 – r 2 = 6 cm (1) $\Rightarrow (r_{1} - r_{2})^{2} = (6 cm)^{2} \Rightarrow r_{1}^{2} {+ r}_{2}^{2} - 2 r_{1} r_{2} = 36 \Rightarrow r_{1}^{2} {+ r}_{2}^{2} = 36 + 2 r_{1} r_{2}$ also sum of area of two circles = 116 π cm2 (2) $\Rightarrow π r_{1}^{2} + π r_{2}^{2} = 116 π \Rightarrow π (r_{1}^{2} + r_{2}^{2}) = 116 π \Rightarrow r_{1}^{2} + r_{2}^{2} = 116 (3)$ Equating (2) and (3) we get $36 + 2 r_{1} r_{2} = 116 \Rightarrow 18 + r_{1} r_{2} = 58 \Rightarrow r_{1} r_{2} = 58 - 18 = 40 \Rightarrow r_{2} = \frac{40}{r 1} (4)$ from (1) and (4) we get $r_{1} - \frac{40}{r_{1}} = 6 \Rightarrow r_{1}^{2} - 40 = 6 r_{1} \Rightarrow r_{1}^{2} - 6 r_{1} - 40 = 0 \Rightarrow r_{1}^{2} - 10 r_{1} + 4 r_{1} - 40 = 0 \Rightarrow r_{1} (r_{1} - 10) + 4 (r_{1} - 10) = 0 \Rightarrow (r_{1} + 4) (r_{1} - 10) = 0 \Rightarrow r_{1} = - 4 or r_{1} = 10$ Since radius cannot be negative Hence r 1 = 10 putting the value of r 1 in (4) we get $r_{2} = \frac{40}{10} = 4 cm$ Hence the radii of the circles are 4 cm and 10 cm

Solve fast: 21. Two circles touch internally. The sum of their areas is 116 π cm2 and the distance between their centres is 6 cm. Find the radis of the circlm. 22. If 1+4+7+10+.....+x=287, find the value of x.

Solve fast:

21. Two circles touch internally. The sum of their areas is 116 $π$ cm² and the distance between their centres is 6 cm. Find the radis of the circlm.

22. If 1+4+7+10+.....+x=287, find the value of x.