The area of a triangle is given by the equation:

$\displaystyle A_{\Delta } = (1/2)*(base)*(height)$

Since the base leg of the given triangle is 4 cm, while the height is 3 cm, this gives:

$\displaystyle A_{\Delta }=(1/2)*3*4 = 6 cm^{2}$

The area of a triangle is given by the equation:

$\displaystyle A_{\Delta } = (1/2)*(base)*(height)$

Since the base leg of the given triangle is 4 cm, while the height is 3 cm, this gives:

$\displaystyle A_{\Delta }=(1/2)*4*6 = 12 cm^{2}$

The area of a triangle is given by the equation:

$\displaystyle A_{\Delta } = (1/2)*(base)*(height)$

Since the base leg of the given triangle is 4 cm, while the height is 3 cm, this gives:

$\displaystyle A_{\Delta }=(1/2)*3*7 = 10.5 cm^{2}$

The area of a triangle is given by the equation:

$\displaystyle A_{\Delta } = (1/2)*(base)*(height)$

Since the base leg of the given triangle is 4 cm, while the height is 3 cm, this gives:

$\displaystyle A_{\Delta }=(1/2)*10*6 = 12 cm^{2}$

The area of a triangle is given by the equation:

$\displaystyle A_{\Delta } = (1/2)*(base)*(height)$

Since the base leg of the given triangle is 4 cm, while the height is 3 cm, this gives:

$\displaystyle A_{\Delta }=(1/2)*4*3 = 6 cm^{2}$

The area of a triangle is given by the equation:

$\displaystyle A_{\Delta } = (1/2)*(base)*(height)$

Since the base leg of the given triangle is 4 cm, while the height is 3 cm, this gives:

$\displaystyle A_{\Delta }=(1/2)*20*10 = 100 cm^{2}$

The formula for the area of a triangle is

 $\displaystyle \frac{1}{2}bh$

where $\displaystyle b$ is the base of the triangle and $\displaystyle h$ is the height.

For the triangle shown, side $\displaystyle a$ is the base and side $\displaystyle b$ is the height.

Therefore, the area is equal to

 $\displaystyle \frac{1}{2}(7)(12)=42$

or, based on the units given, 42 square centimeters

An equilateral triangle has three congruent sides. The area of a triangle is given by $\displaystyle A = \frac{1}{2}bh$ where $\displaystyle b$ is the base and $\displaystyle h$ is the height.

The equilateral triangle can be broken into two $\displaystyle 30-60-90$ right triangles, where the legs are $\displaystyle 2$ and $\displaystyle x$ and the hypotenuses is $\displaystyle 4$.

Using the Pythagorean Theorem we get $\displaystyle 4^{2}=2^{2} + x^{2}$ or $\displaystyle x=2\sqrt{3}$ and the area is $\displaystyle A = \frac{1}{2}bh= \frac{1}{2}(4)(2\sqrt{3}) = 4\sqrt{3}$

In a $\displaystyle 30^{\circ }-60^{\circ }-90^{\circ }$, the shorter leg is half as long as the hypotenuse, and the longer leg is $\displaystyle \sqrt{3}$ times the length of the shorter. Since the hypotenuse is 8, the shorter leg is 4, and the longer leg is $\displaystyle 4\sqrt{3}$, making the area:

$\displaystyle A = \frac{1}{2} bh = \frac{1}{2} \cdot 4 \cdot 4\sqrt{3} = 8\sqrt{3}$

$\displaystyle Area \;\Delta= \frac{1}{2}(base)(height)$

$\displaystyle \Delta GHF=\frac{1}{2}(5)(12)=\frac{1}{2}(60)=30$