Area of a triangle is:-  \(\displaystyle \frac{1}{2}* \textup{base} * \textup{vertical height}\)  

Area is in Square Units.

\(\displaystyle \frac{1}{2} * 18 *12 =\)

\(\displaystyle 9* 12 =\)

\(\displaystyle 108\ \textup{in}^2\)

Write the area for a triangle.

\(\displaystyle A=\frac{1}{2}\times B\times H\)

Substitute the base and height.

\(\displaystyle A=\frac{1}{2}\times 4\times (x+4) = 2(x+4) = 2x+8\)

Write the formula for the area of a triangle.

\(\displaystyle A=\frac{bh}{2}\)

Substitute the base and height.

\(\displaystyle A=\frac{5\times11}{2} = \frac{55}{2}\)

Write the area for a right triangle.

\(\displaystyle A=\frac{1}{2}bh\)

The height is unknown.  In order to solve for the height, use the Pythagorean Theorem to find the unknown length.

\(\displaystyle a^2+b^2=c^2\)

\(\displaystyle 4^2+b^2=5^2\)

\(\displaystyle b^2=5^2-4^2\)

\(\displaystyle b^2= 25-16\)

\(\displaystyle b^2 = 9\)

\(\displaystyle b=3\)

The height of the triangle is three.  Substitute both leg dimensions to find the area.

\(\displaystyle A=\frac{1}{2}(3\times 4) = 6\)

Write the formula to find the area of an equilateral triangle given the side.

\(\displaystyle A=\frac{\sqrt3}{4}s^2\)

Since all three sides are equal in an equilateral triangle, divide the perimeter by three to obtain the length of one side. The side length is one.

\(\displaystyle A=\frac{\sqrt3}{4}(1)^2 = \frac{\sqrt3}{4}\)

Write the formula to find the area of a triangle.

\(\displaystyle A=\frac{1}{2} bh\)

Substitute the base and height.  Simplify the expression.

\(\displaystyle A=\frac{1}{2} (x)(x^2+2) = \frac{1}{2} (x^3+2x) = \frac{x^3}{2}+x\)

Write the formula to find the area of a triangle.

\(\displaystyle A=\frac{1}{2}bh\)

Substitute the base and height.

From here multiply the monomial term, \(\displaystyle x^3\) to each term within the parentheses. Also, recall the rules of exponents. When like bases are multiplied together their exponents are added.

\(\displaystyle A=\frac{1}{2}(x^3)(x^3+1) = \frac{1}{2}(x^6+x^3) = \frac{x^6+x^3}{2}\)

First convert all the dimensions to yards.  There are 3 feet in 1 yard, and 12 inches in 1 foot.

\(\displaystyle 36\textup{ inches} \left(\frac{1\textup{ foot}}{12\textup{ inches}}\right)\left(\frac{1\textup{ yard}}{3\textup{ feet}}\right) = 1 \textup{ yard}\)

The side lengths of the triangle are 1 yard by 1 yard by 1 yard.  We can then see that this is an equilateral triangle.

Write the formula for the area of an equilateral triangle.

\(\displaystyle A=\frac{\sqrt3}{4}a^2\)

Substitute the side.

\(\displaystyle A=\frac{\sqrt3}{4}(1)^2 = \frac{\sqrt3}{4}\)

Write the formula for the area of a triangle.  The area is half the product of the base and the height.

\(\displaystyle A=\frac{bh}{2}\)

Substitute the base and height into the equation.

\(\displaystyle A=\frac{(6)(10)}{2} =\frac{60}{2} = 30\)

The area of the triangle is 30.

Write the area formula for triangles.

\(\displaystyle A=\frac{bh}{2}\)

The base is unknown.  To find this dimension, use the Pythagorean Theorem.

\(\displaystyle a^2+b^2 = c^2\)

Substitute the hypotenuse into \(\displaystyle c\) and the height into \(\displaystyle a\).  Solve for the base.

\(\displaystyle 6^2+b^2 = 10^2\)

Isolate the variable \(\displaystyle b^2\) by subtracting \(\displaystyle 6^2\) on both sides of the equation.

\(\displaystyle b^2 = 10^2-6^2\)

Simplify the squares on the right side of the equation.

\(\displaystyle b^2 = 100-36\)

Subtract the left side.

\(\displaystyle b^2 = 64\)

Square root both sides of the equation to eliminate the square root.

\(\displaystyle \sqrt{b^2} = \sqrt{64}\)

We will only consider the positive root because length cannot be a negative value.

\(\displaystyle b=8\)

The base is 8.  Substitute the base and height to find the area.

\(\displaystyle A=\frac{(8)(6)}{2} = \frac{48}{2} =24\)