Question

in given figure CE and DE are equal chords of a circle with centr o if angle AOB=90 ind ar(triangle CED)ar(triangle AOB)​

Answer 1

⋆ $\bold{\underline{\rm{\red{Given:-}}}}$

• CE AND DE are equal chords of a circle with center O.

• If angle AOB= 90 FIND ar[triangle CED] : ar[triangle AOB]

⋆ $\bold{\underline{\rm{\purple{Solution:-}}}}$

$\implies \; \sf{ \angle AOB}$ = 90°

★ CE and DE are equal chord.

We can take E = 90°

• Diameter (CD) = 2r

★ Let CE = DE = x

$\implies \sf{x^2 + x^2 = 4r^2} \\ \\ \\ \implies \sf{2x^2 + = 4r^2} \\ \\ \\ \implies \sf{x^2 = 2r^2} \\ \\ \\ \implies \sf{x = \sqrt{2}}$

$\rule{200}{2}$

$\implies \sf{ Area_{ \triangle CED } = \dfrac{1}{2} \times x \times x} \\ \\ \\ \implies \sf{ Area_{ \triangle CED } = \dfrac{1}{2} \times \sqrt{2} r \times \sqrt{2} r} \\ \\ \\ \implies \sf{ Area_{ \triangle CED } = r^2}$

$\implies \sf{ Area_{ \triangle AOB } = \dfrac{1}{2} \times r \times r} \\ \\ \\ \implies \sf{ Area_{ \triangle AOB } = \dfrac{r^2}{r}}$

$\rule{200}{2}$

Hence,

★The ratio of $\sf{ \triangle CED}$ and $\sf{ \triangle AOB}$

★ $\sf{r^2 : \dfrac{r^2}{r}}$

$\bold{\underline{\underline{\boxed{\sf{\red{\dag \; 2 : 1}}}}}}$

$\rule{200}{2}$

Answer 2

This is the ANSWER for the question given above.