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3 years ago

What is the area of a triangle that has a base of 3 feet and a height of 6 feet?

Mathematics

1 answer:

Nitella [24]3 years ago

3 0

The formula of area of triangle is given by

$A= \frac{1}{2} b*h$

Where b is the base and h is the height .

In the given question, base ,b = 3 feet

Height, h= 6 feet

Substituting the values of b and h in the formula, we will get

$A = \frac{1}{2}*3*6 = 9 ft^2$

Correct option is C .

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Classify this triangle by its sides and angles.

FromTheMoon [43]

Answer:

choice 3) acute and equilateral

Step-by-step explanation:

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A segment with endpoints A(2,6) and C(5,9) is partitioned by a point B such that AB and BC form a 3:1 ratio. Find B

Aneli [31]

Answer:

The coordinate of point B is (4, 8.25)

Step-by-step explanation:

Here, we want to find the coordinates of point B

To do this, we are to use the section internal division formula as follows;

(x,y) = (nx1 + mx2)/(m + n) , (ny1 + my2)/(m + n)

In this case;

(x1,y1) = (2,6)

(x2,y2) = (5,9)

(m,n) = (3,1)

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(x,y) = (1(1) + 3(5))/(1 + 3) , (1(6) + 3(9))/3 + 1)

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3 years ago

The angle of elevation from me to the top of a hill is 51 degrees. The angle of elevation from me to the top of a tree is 57 deg

julia-pushkina [17]

Answer:

Approximately $101\; \rm ft$ (assuming that the height of the base of the hill is the same as that of the observer.)

Step-by-step explanation:

Refer to the diagram attached.

Let $\rm O$ denote the observer.
Let $\rm A$ denote the top of the tree.
Let $\rm R$ denote the base of the tree.
Let $\rm B$ denote the point where line $\rm AR$ (a vertical line) and the horizontal line going through $\rm O$ meets. $\angle \rm B\hat{A}R = 90^\circ$ .

Angles:

Angle of elevation of the base of the tree as it appears to the observer: $\angle \rm B\hat{O}R = 51^\circ$ .
Angle of elevation of the top of the tree as it appears to the observer: $\angle \rm B\hat{O}A = 57^\circ$ .

Let the length of segment $\rm BR$ (vertical distance between the base of the tree and the base of the hill) be $x\; \rm ft$ .

The question is asking for the length of segment $\rm AB$ . Notice that the length of this segment is $\mathrm{AB} = (x + 20)\; \rm ft$ .

The length of segment $\rm OB$ could be represented in two ways:

In right triangle $\rm \triangle OBR$ as the side adjacent to $\angle \rm B\hat{O}R = 51^\circ$ .
In right triangle $\rm \triangle OBA$ as the side adjacent to $\angle \rm B\hat{O}A = 57^\circ$ .

For example, in right triangle $\rm \triangle OBR$ , the length of the side opposite to $\angle \rm B\hat{O}R = 51^\circ$ is segment $\rm BR$ . The length of that segment is $x\; \rm ft$ .

$\begin{aligned}\tan{\left(\angle\mathrm{B\hat{O}R}\right)} = \frac{\,\rm {BR}\,}{\,\rm {OB}\,} \; \genfrac{}{}{0em}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}\end{aligned}$ .

Rearrange to find an expression for the length of $\rm OB$ (in $\rm ft$ ) in terms of $x$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{BR}}{\tan{\left(\angle\mathrm{B\hat{O}R}\right)}} \\ &= \frac{x}{\tan\left(51^\circ\right)}\approx 0.810\, x\end{aligned}$ .

Similarly, in right triangle $\rm \triangle OBA$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{AB}}{\tan{\left(\angle\mathrm{B\hat{O}A}\right)}} \\ &= \frac{x + 20}{\tan\left(57^\circ\right)}\approx 0.649\, (x + 20)\end{aligned}$ .