Maximizing Area

The figure shows this triangle with sides $10$ , $7$ and one unknown side.

We can see that the base of the triangle is $10$ . And the height is $7\sin x$ .

We know that the area of a triangle is $\frac { 1 }{ 2 } bh$ . On putting values of $b$ and $h$ , we get that the area of this triangle is $\frac { 1 }{ 2 } \left( 10 \right) \left( 7\sin { x } \right) =35\sin { x }$

To maximize this, we need to maximize $\sin x$ . We know that $\sin x$ lies from $-1$ to $1$ , so the maximum possible value is $1$ .

So the answer is $\left( 35 \right) \left( 1 \right) =\boxed { 35 }$

The area $A$ of a triangle with sides $a$ , $b$ and included angle $\theta$ is $A = \dfrac{1}{2}ab \sin \theta$ .

Then the area of the given triangle is $35\sin{\theta}$ .

We need to maximize this expression from $0<\theta<\pi$ , and we see that $\sin{\theta}$ is greatest when $\theta=\dfrac{\pi}{2}$ , so the maximum area is $35\sin{\dfrac{\pi}{2}}=\boxed{35}$

The area of the triangle will be 1/2(10)(7)sinC

The maximum value of sinC is 1; the range of values for sinC go from -1 to +1

The maximum value of the area of the the triangle is 1/2(10)(7)(1) = 35

If base is 10, the area will be maximum when side with length 7 is perpendicular fto the base and thus the area is 1/2 x7x10= 35

(Half of the base) multiplied by height. Either 5x7=35 or 10x3.5=35