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.

The degree measures of a triangle total \(\displaystyle 180^{\circ }\), so

\(\displaystyle x + (x+24) + 2x = 180\)

\(\displaystyle 4x+24 = 180\)

\(\displaystyle 4x+24 -24 = 180 -24\)

\(\displaystyle 4x = 156\)

\(\displaystyle 4x \div 4 = 156 \div 4\)

\(\displaystyle x = 39\)

The degree measures of the interior angles of a triangle total \(\displaystyle 180 ^{\circ }\), so, if we let \(\displaystyle x^{\circ }\) be the measure of the unmarked angle, then

\(\displaystyle x + 41 + 72 = 180\)

\(\displaystyle x + 113 = 180\)

\(\displaystyle x = 67\)

Three angles with measures \(\displaystyle 67^{\circ}, y^{\circ}, y^{\circ}\) together form a straight angle, so

\(\displaystyle y+ y + 67= 180\)

\(\displaystyle 2y+ 67= 180\)

\(\displaystyle 2y+ 67- 67 = 180 - 67\)

\(\displaystyle 2y = 113\)

\(\displaystyle 2y \div 2 = 113 \div 2\)

\(\displaystyle y = 56 \frac{1}{2}\)

One of the angles of \(\displaystyle \Delta EFC\) - namely, \(\displaystyle \angle EFC\) - can be seen to be an obtuse angle, as it is wider than a right angle. This makes \(\displaystyle \Delta EFC\), by definition, an obtuse triangle.

One of the angles of \(\displaystyle \Delta CED\) - namely, \(\displaystyle \angle CDE\) - is marked as a right angle. This makes \(\displaystyle \Delta CED\), by definition, a right triangle.

A congruency statement about two triangles implies nothing about the relationship between two angles of one of the triangles, so \(\displaystyle \angle M \cong \angle N\) is not correct.

Also, letters in the same position between the two triangles refer to corresponding - and subsequently, congruent - angles. Therefore, \(\displaystyle \Delta MNO \cong \Delta PQR\) implies that:

\(\displaystyle \angle M \cong \angle P\)

\(\displaystyle \angle N \cong \angle Q\)

\(\displaystyle \angle O \cong \angle R\)

Of the given choices, only \(\displaystyle \angle M \cong \angle P\) is a consequence. It is the correct response.