Question

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I want answer if both parts.... ​

Answer 1

QUESTION:-

SHOWN IN IMAGE.....

ANSWER:-

R= 25cm , r = 20.5cm

STEPS:-

1) Let the radii of the larger and smaller circles be R and r respectively. Then,

BD=9cm

⇒ 2R−2r=9

⇒ R−r=4.5

Join AE and DE. Let ∠CAE=θ Then, ∠AEC=90°−θ

Now, ∠AED=90 °

⇒∠AEC+∠DEC=90 °

⇒∠DEC=90 ° −(90 °−θ)=θ

Thus, in Δ ACE and DCE, we have

∠CAE=∠CED=θand∠ACE=∠ECD=90 °

So, by AA similarly criterion, we have

ΔACE ΔECD

⇒

$\frac{a \: c}{e \: c} = \frac{c \: e}{c \: d}$

⇒

$\frac{a \: c}{c \: f - e \: f} = \frac{c \: f - e \: f}{b \: c - b \: d}$

⇒

$\frac{r }{r - 5} = \frac{r - 5}{r - 9}$

( HERE I CAN 'T ABLE TO WRITE CAPITAL R SO here all r = R)

• ⇒R(R−9)=(R−5) ^2

• ⇒0=−R+25

⇒R=25cm

Substituting the value of R in (i), we get

• 25−r=4.5

⇒ r=20.5cm

Answer 2

QUESTION:-

SHOWN IN IMAGE.....

ANSWER:-

R= 25cm , r = 20.5cm

STEPS:-

1) Let the radii of the larger and smaller circles be R and r respectively. Then,

BD=9cm

⇒ 2R−2r=9

⇒ R−r=4.5

Join AE and DE. Let ∠CAE=θ Then, ∠AEC=90°−θ

Now, ∠AED=90 °

⇒∠AEC+∠DEC=90 °

⇒∠DEC=90 ° −(90 °−θ)=θ

Thus, in Δ ACE and DCE, we have

∠CAE=∠CED=θand∠ACE=∠ECD=90 °

So, by AA similarly criterion, we have

ΔACE ΔECD

⇒

\frac{a \: c}{e \: c} = \frac{c \: e}{c \: d}

ec

ac

=

cd

ce

⇒

\frac{a \: c}{c \: f - e \: f} = \frac{c \: f - e \: f}{b \: c - b \: d}

cf−ef

ac

=

bc−bd

cf−ef

⇒

\frac{r }{r - 5} = \frac{r - 5}{r - 9}

r−5

r

=

r−9

r−5

( HERE I CAN 'T ABLE TO WRITE CAPITAL R SO here all r = R)

⇒R(R−9)=(R−5) ^2

⇒0=−R+25

⇒R=25cm

Substituting the value of R in (i), we get

25−r=4.5

⇒ r=20.5cm

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