pka
Elite Member
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- Jan 29, 2005
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Why are you using an integral? If your lecturer rejects this answer, show the person this webpage and ask for some clarification.
The height of the triangle [imath]\Delta ABC[/imath] is [imath]\dfrac{\sqrt{15}}{2}[/imath]
The base of the tiangle [imath]\Delta ABC[/imath] is [imath]3[/imath].
Thus area is [imath]\mathcal{A}(\Delta ABC)=\dfrac{3\sqrt{15}}{4}[/imath]
The height of the triangle [imath]\Delta ABC[/imath] is [imath]\dfrac{\sqrt{15}}{2}[/imath]
The base of the tiangle [imath]\Delta ABC[/imath] is [imath]3[/imath].
Thus area is [imath]\mathcal{A}(\Delta ABC)=\dfrac{3\sqrt{15}}{4}[/imath]