Re: In triangle $A B C$, sides $A B$ and $A C$ are of lengths 10 and 12 un
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30 Aug 2025, 08:11
Given:
- Triangle $A B C$ with side lengths $A B=10$ units and $A C=12$ units.
- We want to find which of the given areas $(110,120,130)$ can be the area of triangle $A B C$.
The area of a triangle given two sides and the included angle $\theta$ is:
$$
\(\text { Area }=\frac{1}{2} \times A B \times A C \times \sin \theta=\frac{1}{2} \times 10 \times 12 \times \sin \theta=60 \sin \theta\)
$$
Since $\sin \theta$ ranges between 0 and 1 , the possible area values range between:
$$
\(0 \leq 60 \sin \theta \leq 60\)
$$
So the area cannot be more than 60 units.
Check the given options:
- 110 : too large (greater than 60 ), not possible.
- 120: too large, not possible.
- 130: too large, not possible.
Therefore, none of the given options can be the area of the triangle.
The correct answer is:
D) None of the options above.