How to find the area of an equilateral triangle - High School Math

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Question

A circle contains 6 copies of a triangle; each joined to the others at the center of the circle, as well as joined to another triangle on the circle’s circumference.

The circumference of the circle is $4\pi$

What is the area of one of the triangles?

Answer

The radius of the circle is 2, from the equation circumference $=2r\pi$ . Each triangle is the same, and is equilateral, with side length of 2. The area of a triangle $=\frac{1}{2}bh$

To find the height of this triangle, we must divide it down the centerline, which will make two identical 30-60-90 triangles, each with a base of 1 and a hypotenuse of 2. Since these triangles are both right traingles (they have a 90 degree angle in them), we can use the Pythagorean Theorem to solve their height, which will be identical to the height of the equilateral triangle.

$a^{2}+b^{2}=c^{2}$

We know that the hypotenuse is 2 so $2^{2}=4$ . That's our $c^{2}$ solution. We know that the base is 1, and if you square 1, you get 1.

Now our formula looks like this: $a^{2}+1=4$ , so we're getting close to finding .

Let's subtract 1 from each side of that equation, in order to make things a bit simpler: $a^{2}+0=3 \rightarrow a^{2}=3$

Now let's apply the square root to each side of the equation, in order to change $a^{2}$ into : $a=\sqrt{3}$

Therefore, the height of our equilateral triangle is $\sqrt{3}$

To find the area of our equilateral triangle, we simply have to multiply half the base by the height: $\frac{2}{2}\sqrt{3}=1*\sqrt{3}=\sqrt{3}$

The area of our triangle is $\sqrt{3}$

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Question

The area of square ABCD is 50% greater than the perimeter of the equilateral triangle EFG. If the area of square ABCD is equal to 45, then what is the area of EFG?

Answer

If the area of ABCD is equal to 45, then the perimeter of EFG is equal to x * 1.5 = 45. 45 / 1.5 = 30, so the perimeter of EFG is equal to 30. This means that each side is equal to 10.

The height of the equilateral triangle EFG creates two 30-60-90 triangles, each with a hypotenuse of 10 and a short side equal to 5. We know that the long side of 30-60-90 triangle (here the height of EFG) is equal to √3 times the short side, or 5√3.

We then apply the formula for the area of a triangle, which is 1/2 * b * h. We get 1/2 * 10 * 5√3 = 5 * 5√3 = 25√ 3.

In general, the height of an equilateral triangle is equal to √3 / 2 times a side of the equilateral triangle. The area of an equilateral triangle is equal to 1/2 * √3s/ 2 * s = √3s2/4.

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Question

What is the area of an equilateral triangle with a side length of 5?

Answer

Note that an equilateral triangle has equal sides and equal angles. The question gives us the length of the base, 5, but doesn't tell us the height.

If we split the triangle into two equal triangles, each has a base of 5/2 and a hypotenuse of 5.

Therefore we can use the Pythagorean Theorem to solve for the height:

$(\frac{5}{2})^2+b^2=5^2$

$\frac{25}{4}+b^2=25$

$b^2=\frac{75}{4}$

$b=\sqrt{\frac{3\times 25}{2\times 2}}=\frac{5\sqrt{3}}{2}$

Now we can find the area of the triangle:

$Area=\frac{1}{2}\times base\times height=\frac{1}{2}\times 5\times \frac{5\sqrt{3}}{2}=\frac{25\sqrt{3}}{4}$

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Question

What is the area of an equilateral triangle with sides 12 cm?

Answer

An equilateral triangle has three congruent sides and results in three congruent angles. This figure results in two special right triangles back to back: 30° – 60° – 90° giving sides of x - x √3 – 2x in general. The height of the triangle is the x √3 side. So Atriangle = 1/2 bh = 1/2 * 12 * 6√3 = 36√3 cm2.

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Question

An equilateral triangle has a perimeter of 18. What is its area?

Answer

Recall that an equilateral triangle also obeys the rules of isosceles triangles. That means that our triangle can be represented as having a height that bisects both the opposite side and the angle from which the height is "dropped." For our triangle, this can be represented as:

Now, although we do not yet know the height, we do know from our 30-60-90 regular triangle that the side opposite the 60° angle is √3 times the length of the side across from the 30° angle. Therefore, we know that the height is 3√3.

Now, the area of a triangle is (1/2)bh. If the height is 3√3 and the base is 6, then the area is (1/2) * 6 * 3√3 = 3 * 3√3 = 9√(3).

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Question

An equilateral triangle has a side length of $\sqrt{3}$ . What is the triangle's area?

Answer

The area of an equilateral triangle is found using the following formula.

$A=\frac{s^{2}\sqrt{3}}{4}$ where $s=\sqrt{3}$

$A=\frac{(\sqrt{3})^{2}*\sqrt{3}}{4}$

$A=\frac{3\sqrt{3}}{4}$

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Question

The length of one side of an equilateral triangle is ten. What is the area of the triangle?

Answer

To calculate the height, the length of a perpendicular bisector must be determined. If a perpendicular bisector is drawn in an equilateral triangle, the triangle is divided in half, and each half is a congruent 30-60-90 right triangle. This type of triangle follows the equation below.

$a^2+b^2=c^2\rightarrow (a)^2+(a\sqrt{3})^2=(2a)^2$

The length of the hypotenuse will be one side of the equilateral triangle.

.

The side of the equilateral triangle that represents the height of the triangle will have a length of $\small a\sqrt{3}$ because it will be opposite the 60o angle.

$a=5\rightarrow a\sqrt{3}=5\sqrt{3}$

To calculate the area of the triangle, multiply the base (one side of the equilateral triangle) and the height (the perpendicular bisector) and divide by two.

$A=\frac{1}{2}bh=\frac{1}{2}(10)(5\sqrt3)=25\sqrt3$

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Question

What is the area of an equilateral triangle with side 11?

Answer

Since the area of a triangle is

$Area=\frac{1}{2}bh$

you need to find the height of the triangle first. Because of the 30-60-90 relationship, you can determine that the height is $5.5\sqrt{3}$ .

Then, multiply that by the base (11).

Finally, divide it by two to get 52.4.

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Question

What is the area of an equilateral triangle with a side length of $s=2\sqrt{3}$ ?

Answer

In order to find the area of the triangle, we must first calculate the height of its altitude. An altitude slices an equilateral triangle into two $30^{\circ}-60^{\circ}-90^{\circ}$ triangles. These triangles follow a side-length pattern. The smallest of the two legs equals and the hypotenuse equals . By way of the Pythagorean Theorem, the longest leg or $h=x\sqrt{3}$ .

Therefore, we can find the height of the altitude of this triangle by designating a value for . The hypotenuse of one of the $30^{\circ}-60^{\circ}-90^{\circ}$ is also the side of the original equilateral triangle. Therefore, one can say that $2x=s=2\sqrt{3}$ and $x=\sqrt{3}$ .

$h=x\sqrt{3}$

$h=\sqrt{3} * \sqrt{3}$

$\rightarrow3$

Now, we can calculate the area of the triangle via the formula $A=\frac{1}{2}bh$ .

$A= \frac{1}{2} * 2\sqrt{3} * 3$

$A=3\sqrt{3}$

$\rightarrow 5.20$

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Question

An equilateral triangle has a side length of $s=89\ cm$ find its area.

Answer

In order to find the area of the triangle, we must first calculate the height of its altitude. An altitude slices an equilateral triangle into two $30^{\circ}-60^{\circ}-90^{\circ}$ triangles. These triangles follow a side-length pattern. The smallest of the two legs equals and the hypotenuse equals . By way of the Pythagorean Theorem, the longest leg or $h=x\sqrt{3}$ .

Therefore, we can find the height of the altitude of this triangle by designating a value for . The hypotenuse of one of the $30^{\circ}-60^{\circ}-90^{\circ}$ is also the side of the original equilateral triangle. Therefore, one can say

that and .

$h=x\sqrt{3}$

$h=44.5cm\sqrt{3}$

Now, we can calculate the area of the triangle via the formula

$A=\frac{1}{2}bh$

$A= \frac{1}{2} 89cm * 44.5cm\sqrt{3}$

$A=3430 cm^{2}$

Now convert to meters.

$\rightarrow \frac{3430cm^{2}}{1}\frac{1m}{100cm}*\frac{1m}{100cm}= 0.343m^{2}$

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Question

Triangle A: A right triangle with sides length , , and .

Triangle B: An equilateral triangle with side lengths .

Which triangle has a greater area?

Answer

The formula for the area of a right triangle is $A = \frac{1}{2}bh$ , where is the length of the triangle's base and is its height. Since the longest side is the hypotenuse, use the two smaller numbers given as sides for the base and height in the equation to calculate the area of Triangle A:

$A = \frac{1}{2}(6)(8)$

The formula for the area of an equilateral triangle is $A = \frac{s^2\sqrt{3}}{4}$ , where is the length of each side. (Alternatively, you can divide the equilateral triangle into two right triangles and find the area of each). Triangle B's area is thus calculated as:

$A = \frac{(8)^2\sqrt{3}}{4}$

$A = \frac{64\sqrt{3}}{4}$

$A = 16\sqrt{3}$

To determine which of the two areas is greater without using a calculator, rewrite the areas of the two triangles with comparable factors. Triangle A's area can be expressed as $24= 8\sqrt{3}\sqrt{3}$ , and Triangle B's area can be expressed as $16\sqrt{3}= (8)(2)\sqrt{3}$ . Since is greater than $\sqrt{3}$ , the product of the factors of Triangle B's area will be greater than the product of the factors of Triangle A's, so Triangle B has the greater area.

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Question

Find the area of the following equilateral triangle:

Answer

The formula for the area of an equilateral triangle is:

$A=\frac{s^2\sqrt{3}}{4}$

Where is the length of the side

Plugging in our values, we get:

$A=\frac{s^2\sqrt{3}}{4}$

$A=\frac{(20m)^2\sqrt{3}}{4}=\frac{400m^2\sqrt{3}}{4}=100\sqrt{3}m^2$

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Question

Determine the area of the following equilateral triangle:

Answer

The formula for the area of an equilateral triangle is:

$A=\frac{s^2\sqrt{3}}{4}$ ,

where is the length of the sides.

Plugging in our value, we get:

$A=\frac{(8cm)^2\sqrt{3}}{4}=\frac{64cm^2\sqrt{3}}{4}=16cm^2\sqrt{3}$

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Question

Find the area of an equilateral triangle whose perimeter is

Answer

The formula for the perimeter of an equilateral triangle is:

Plugging in our values, we get

The formula for the area of an equilateral triangle is:

$A = \frac{s^2 \sqrt{3}}{4}$

Plugging in our values, we get

$A = \frac{(7cm)^2 \sqrt{3}}{4} = \frac{49cm^2 \sqrt{3}}{4}$

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