In Delta ABC,AD and AE are bisectors of angle | vedclass

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(B) Given that $AE$ is the bisector of $\angle BAD$,therefore $\angle BAE = \angle EAD = 30^{\circ}$.Since $AD$ is the bisector of $\angle BAC$,$\angl

Vedclass · Accepted Answer

$54$

Vedclass · Answer

(B) Given that $AE$ is the bisector of $\angle BAD$,therefore $\angle BAE = \angle EAD = 30^{\circ}$.Since $AD$ is the bisector of $\angle BAC$,$\angle CAD = \angle BAD = \angle BAE + \angle EAD = 30^{\circ} + 30^{\circ} = 60^{\circ}$.In $\Delta AEC$,we have side $AE = 9 	ext{ cm}$ and $EC = 15 	ext{ cm}$.Assuming $\Delta AEC$ is a right-angled triangle at $A$ (based on the provided solution logic),we find the third side $AC$ using the Pythagorean theorem: $AC = \sqrt{EC^2 - AE^2} = \sqrt{15^2 - 9^2} = \sqrt{225 - 81} = \sqrt{144} = 12 	ext{ cm}$.The area of $\Delta AEC = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes AE 	imes AC = \frac{1}{2} 	imes 9 	imes 12 = 54 	ext{ cm}^2$.

In $\Delta ABC$,$AD$ and $AE$ are bisectors of $\angle BAC$ and $\angle BAD$ respectively. If $\angle BAE = 30^{\circ}$,$AE = 9 \text{ cm}$,and $EC = 15 \text{ cm}$,what is the area (in $\text{cm}^2$) of $\Delta AEC$?

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