When given the lengths of two sides and the measure of the angle included by the two sides, the area formula is:

$\displaystyle A = \frac {1}{2}ab\sin C \\$

Plugging in the given values we are able to calculate the area.

$\displaystyle A = \frac {1}{2}\cdot 40\cdot 20 \cdot \sin 30^\circ= 200$

To find the area, use the formula associated with side, angle, side triangles which states,

 $\displaystyle area = \frac{ 1}{2} a b \sin C$

where $\displaystyle a$ and $\displaystyle b$ are side lengths and $\displaystyle C$ is the included angle.

In our case,

$\displaystyle a=8, b=17, C=90^\circ$.

Plug the values into the area formula and solve.

$\displaystyle area = \frac{ 1}{2} (8 )( 17 ) \cdot \sin( 90 ^ o ) = 68$

Use the area formula to find area that is associated with the side angle side theorem for triangles.

 $\displaystyle area= \frac{ 1}{2 } a b \sin C$

where $\displaystyle a$ and $\displaystyle b$ are side lengths and $\displaystyle C$ is the included angle.

Plugging these values into the formula above, we arrive at our final answer.

$\displaystyle area = \frac{ 1}{2} (11) (23 ) \sin (41^o ) \approx 82.99$

To solve, use the formula for area that is associated with the side angle side theorem for triangles,

$\displaystyle area = \frac{ 1}{2} ab \sin C$

where $\displaystyle a$ and $\displaystyle b$ are side lengths and $\displaystyle C$ is the included angle.

Here we are using $\displaystyle 25^o$ and not $\displaystyle 90^o$ since that is the angle between $\displaystyle 8$ and $\displaystyle 7$.

Therefore,

$\displaystyle a=8,b=7,C=25^\circ$.

Plugging the above values into the area formula we arrive at our final answer.

$\displaystyle area = \frac{ 1}{2} (8)(7) \sin (25^o ) \approx 11.83$

Find the area using the formula associated the side angle side theorem of a triangle,

$\displaystyle area = \frac{ 1}{2} ab \sin C$

where $\displaystyle a$ and $\displaystyle b$ are side lengths and $\displaystyle C$ is the included angle.

In this particular case,

$\displaystyle a=15, b=19, C=20^\circ$

therefore the area is found to be,

$\displaystyle area = \frac {1}{2} (15)(19) \sin (20^o ) \approx 48.74$.

Use the Heron's Formula:

$\displaystyle s=\frac{a+b+c}{2}$

$\displaystyle A=\sqrt{s(s-a)(s-b)(s-c)}$

Solve for $\displaystyle s$.

$\displaystyle s=\frac{a+b+c}{2}=\frac{2+3+4}{2}=\frac{9}{2}$

Solve for the area.

$\displaystyle A=\sqrt{s(s-a)(s-b)(s-c)}$

$\displaystyle A=\sqrt{\frac{9}{2}\left(\frac{9}{2}-2\right)\left(\frac{9}{2}-3\right)\left(\frac{9}{2}-4\right)}$

$\displaystyle A=\sqrt{\frac{9}{2}\left(\frac{5}{2}\right)\left(\frac{3}{2}\right)\left(\frac{1}{2}\right)}=\sqrt{\frac{135}{16}}=\frac{\sqrt{135}}{4}$

We can solve this question using Heron's Formula. Heron's Formula states that:

The semiperimeter is

$\displaystyle s=\frac{1}{2}(a+b+c)$

where $\displaystyle a$, $\displaystyle b$, $\displaystyle c$ are the sides of a triangle.

Then the area is

$\displaystyle area=\sqrt{s(s-a)(s-b)(s-c)}$

So if we plug in

$\displaystyle s=\frac{1}{2}(4+5+7)$

$\displaystyle s=\frac{1}{2}(16)$

$\displaystyle s=8$

So the area is

$\displaystyle area=\sqrt{8(8-4)(8-5)(8-7)}$

$\displaystyle area=\sqrt{96}$

$\displaystyle area=4\sqrt{6}$

We can solve this question using Heron's Formula. Heron's Formula states that:

The semiperimeter is

$\displaystyle s=\frac{1}{2}(a+b+c)$

where $\displaystyle a$, $\displaystyle b$, $\displaystyle c$ are the sides of a triangle.

Then the area is

$\displaystyle area=\sqrt{s(s-a)(s-b)(s-c)}$

So if we plug in

$\displaystyle s=\frac{1}{2}(6+5+9)$

$\displaystyle s=\frac{1}{2}(20)$

$\displaystyle s=10$

So the area is

$\displaystyle area=\sqrt{10(10-6)(10-5)(10-9)}$

$\displaystyle area=\sqrt{200}$

$\displaystyle area=10\sqrt{2}$

We can solve this question using Heron's Formula. Heron's Formula states that:

The semiperimeter is

$\displaystyle s=\frac{1}{2}(a+b+c)$

where $\displaystyle a$, $\displaystyle b$, $\displaystyle c$ are the sides of a triangle.

Then the area is

$\displaystyle area=\sqrt{s(s-a)(s-b)(s-c)}$

So if we plug in

$\displaystyle s=\frac{1}{2}(3+4+5)$

$\displaystyle s=\frac{1}{2}(12)$

$\displaystyle s=6$

So the area is

$\displaystyle area=\sqrt{6(6-3)(6-4)(6-5)}$

$\displaystyle area=\sqrt{36}$

$\displaystyle area=6$

We can solve this question using Heron's Formula. Heron's Formula states that:

The semiperimeter is

$\displaystyle s=\frac{1}{2}(a+b+c)$

where $\displaystyle a$, $\displaystyle b$, $\displaystyle c$ are the sides of a triangle.

Then the area is

$\displaystyle area=\sqrt{s(s-a)(s-b)(s-c)}$

So if we plug in

$\displaystyle s=\frac{1}{2}(5+12+13)$

$\displaystyle s=\frac{1}{2}(30)$

$\displaystyle s=15$

So the area is

$\displaystyle area=\sqrt{15(15-5)(15-12)(15-13)}$

$\displaystyle area=\sqrt{900}$

$\displaystyle area=30$