To find the area of a triangle, we will use the following formula:

\(\displaystyle A = \frac{1}{2} \cdot b \cdot h\)

where b is the base and h is the height of the triangle.

Now, we know the base of the triangle is 9in. We know the height is two times the base. Therefore, the height is 18in. So, we can substitute. We get

\(\displaystyle A = \frac{1}{2} \cdot 9\text{in} \cdot 18\text{in}\)

\(\displaystyle A = \frac{1}{2} \cdot 18\text{in} \cdot 9\text{in}\)

\(\displaystyle A = 9\text{in} \cdot 9\text{in}\)

\(\displaystyle A = 81\text{in}^2\)

Write the formula for the area of a triangle.

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

Substitute the base and height into the equation.

\(\displaystyle A= \frac{20(40)}{2} = \frac{800}{2} = 400\)

The answer is:  \(\displaystyle 400\)

Write the formula for the area of a triangle.

\(\displaystyle A= \frac{bh}{2} = \frac{6\times 30}{2} = 3\times 30 = 90\)

The answer is:  \(\displaystyle 90\)

Write the formula for the area of a triangle.

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

Substitute the dimensions.

\(\displaystyle A =\frac{8(30)}{2} = 4(30) = 120\)

The answer is:  \(\displaystyle 120\)

Write the formula for the area of a triangle.

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

Substitute the base and the height into the equation.

\(\displaystyle A = \frac{(100)(300)}{2} = 100(150 ) = 15000\)

The answer is:  \(\displaystyle 15000\)

You want to build a garden in the shape of a right triangle. If the two arms will be 6ft and 8ft, how much area will the garden take up?

To find the area of a triangle, use the following formula.

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

Note that in a right triangle, our two arms correspond to our base and our height. Furthermore, it does not matter which is which, because when we multiply, order does not matter.

So, to find our area, simply plug in and simplify.

\(\displaystyle A=\frac{1}{2}(6ft)(8ft)=\frac{1}{2}(48ft^2)=24ft^2\)

So, our answer is 24 ft squared

Start by finding the length of a side of the triangle.

\(\displaystyle \text{side}=\frac{\text{Perimeter}}{3}\)

\(\displaystyle \text{side}=\frac{39}{3}=13\)

Next, recall how to find the area of an equilateral triangle.

\(\displaystyle \text{Area}=\frac{\sqrt3}{4}\text{side}^2\)

Plug in the length of the side to find the area.

\(\displaystyle \text{Area}=\frac{\sqrt3}{4}(13)^2=\frac{\sqrt3}{4}(169)=73.18\)

First we need to know that the formula for area of a triangle:

\(\displaystyle Area= \frac{1}{2}(Base\cdot Height)\)

We know that our base is 3.5 inches, and our height is twice that, which is 7 in.

Now we can plug in our base and height to the equation 

\(\displaystyle Area=\frac{1}{2}(3.5 in\cdot 7 in)\)

Multiply and solve

\(\displaystyle Area=\frac{1}{2}(24.5in^{2})\)

\(\displaystyle Area=12.25in^{2}\)

First we need to recall the formula for area of a triangle:

\(\displaystyle Area=\frac{1}{2}(base\cdot height)\)

We know that our base is 4cm, and our height is 3 times the length of the base, since 4x3=12 we know that our height is 12cm

Now we can plug in our numbers

\(\displaystyle Area=\frac{1}{2}(4cm\cdot 12cm)\)

First we multiply 4 and 12

\(\displaystyle Area=\frac{1}{2}(48cm^{2})\)

Next we distribute the fraction which is the same as dividing in half

\(\displaystyle Area=24cm^{2}\)

Notice our answer is in centimeters since we multiplied two terms measured in centimeters

For the triangle to be acute, all three angles must measure less than \(\displaystyle 90 ^{\circ }\). We can eliminate \(\displaystyle 32 ^{\circ }, 32 ^{\circ }, 116^{\circ }\) and \(\displaystyle 45 ^{\circ }, 45 ^{\circ }, 90^{\circ }\) for this reason. 

In an isosceles triangle, at least two angles are congruent, so we can eliminate \(\displaystyle 50 ^{\circ }, 60 ^{\circ }, 70 ^{\circ }\).

The degree measures of the three angles of a triangle must total 180, so, since \(\displaystyle 80 ^{\circ }+ 80 ^{\circ }+ 40 ^{\circ } = 200^{\circ }\), we can eliminate \(\displaystyle 80 ^{\circ }, 80 ^{\circ }, 40 ^{\circ }\).

\(\displaystyle 36^{\circ }, 72 ^{\circ }, 72 ^{\circ }\) is correct.