How do you find an area from a equilateral triangle

2 answers:

5 0

$\sqrt{3}$ / 4 • s to the power of 2 The area of an equilateral triangle (all sides congruent) can be found using the formula

enot [183]3 years ago

4 0

<span>Equilateral triangles have sides of all equal length and angles of 60°. To find the area, we can first find the height. To find the height, we can draw an altitude to one of the sides in order to split the triangle into two equal 30-60-90 triangles.</span>

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In right ABC, AN is the altitude to the hypotenuse. FindBN, AN, and AC,if AB =2 5 in, and NC= 1 in.

Rama09 [41]

From the statement of the problem, we have:

• a right triangle △ABC,

• the altitude to the hypotenuse is denoted AN,

• AB = 2√5 in,

• NC = 1 in.

Using the data above, we draw the following diagram:

We must compute BN, AN and AC.

To solve this problem, we will use Pitagoras Theorem, which states that:

$h^2=a^2+b^2\text{.}$

Where h is the hypotenuse, a and b the sides of a right triangle.

(I) From the picture, we see that we have two sub right triangles:

1) △ANC with sides:

• h = AC,

• a = ,NC = 1,,

• b = NA.

2) △ANB with sides:

• h = ,AB = 2√5,,

• a = BN,

• b = NA,

Replacing the data of the triangles in Pitagoras, Theorem, we get the following equations:

$\begin{cases}AC^2=1^2+NA^2, \\ (2\sqrt[]{5})^2=BN^2+NA^2\text{.}\end{cases}\Rightarrow\begin{cases}NA^2=AC^2-1, \\ NA^2=20-BN^2\text{.}\end{cases}$

Equalling the last two equations, we have:

$\begin{gathered} AC^2-1=20-BN^2.^{} \\ AC^2=21-BN^2\text{.} \end{gathered}$

(II) To find the values of AC and BN we need another equation. We find that equation applying the Pigatoras Theorem to the sides of the bigger right triangle:

3) △ABC has sides:

• h = BC = ,BN + 1,,

• a = AC,

• b = ,AB = 2√5,,

Replacing these data in Pitagoras Theorem, we have: