Equilateral Triangles - Intermediate Geometry

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Question

ΔABC is an equilateral triangle with side 17.

Find the area of ΔABC (to the nearest tenth).

Answer

Equilateral triangles have sides of equal length, with 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.

Now, the side of the original equilateral triangle (lets call it "a") is the hypotenuse of the 30-60-90 triangle. Because the 30-60-90 triange is a special triangle, we know that the sides are x, x $\sqrt{3}$ , and 2x, respectively.

Thus, a = 2x and x = a/2.
Height of the equilateral triangle = $\frac{a\sqrt{3}}{2}$

Given the height, we can now find the area of the triangle using the equation:
$Area = \frac{1}{2}bh=\frac{a^2\sqrt{3}}{4}$

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Question

ΔABC is an equilateral triangle with side 6.

Find the area of ΔABC (to the nearest tenth).

Answer

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.

Now, the side of the original equilateral triangle (lets call it "a") is the hypotenuse of the 30-60-90 triangle. Because the 30-60-90 triange is a special triangle, we know that the sides are x, x $\sqrt{3}$ , and 2x, respectively.

Thus, a = 2x and x = a/2.
Height of the equilateral triangle = $\frac{a\sqrt{3}}{2}$

Given the height, we can now find the area of the triangle using the equation:
$Area = \frac{1}{2}bh=\frac{a^2\sqrt{3}}{4}$

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Question

If the perimeter of an equilateral triangle is 54 inches, what is the area of the triangle in square inches?

Answer

The answer is $81\sqrt{3}$ .

To find the area you would first need to find what the length of each side is: 54 divided by 3 is 18 for each side.

Then you would need to draw in the altitude of the triangle in order to get its height. Drawing this altitude will create two 30-60-90 degree triangles as shown in the picture. The longer leg is $\sqrt{3}$ times the short leg. Thus the height is $9\sqrt{3}$ .

Next we plug in the base and the height into the formula to get

$\frac{1}{2}\cdot \left 18\right\cdot 9\sqrt{3}=81\sqrt{3}\ in^{2}$

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Question

What is the area of this triangle if ?

Answer

The formula for the area of an equilateral triangle with side length is

$Area=\frac{\sqrt{3}}{4}a^2$

So, since ,

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

$Area=\frac{\sqrt{3}}{4}*4=\sqrt{3}:cm^2$

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Question

If the sides of this triangle are doubled in length, what is the triangle's new area in terms of the original length of each of its sides, ?

Answer

The formula of the area of an equilateral triangle is $\frac{\sqrt{3}}{4}x^2$ if is a side.

Since the sides of our triangle have doubled, they have changed from to . We can substitute into the equation and solve for the triangle's new area in terms of :

$area=\frac{\sqrt{3}}{4}x^2=\frac{\sqrt{3}}{4}(2a)^2=\frac{\sqrt{3}}{4}4a^2=\sqrt{3}a^2$

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Question

Suppose we triple the sides of this equilateral triangle to . What is the area of the new triangle in terms of ?

Answer

The formula for the area of an equilateral triangle is $\frac{\sqrt{3}}{4}x^2$ if is the length of one of the triangle's sides.

In this problem, the length of one of the triangle's sides is being tripled, so we can substitute into the equation for and solve for the triangle's new area in terms of :

$Area=\frac{\sqrt{3}}{4}x^2=\frac{\sqrt{3}}{4}(3a)^2=\frac{\sqrt{3}}{4}9a^2=\frac{9\sqrt{3}}{4}a^2$

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Question

What is half the area of the above triangle if ?

Answer

The formula for the area of an equilateral triangle is $\frac{\sqrt{3}}{4}a^2$ . For this problem's triangle, , so we can substitute into the equation for and solve for the area of the triangle:

$Area=\frac{\sqrt{3}}{4}a^2=\frac{\sqrt{3}}{4}(4)^2=\frac{\sqrt{3}}{4}16=4\sqrt{3}:cm^2$

At this point, we need to divide by , since the problem asks for half of the triangle's area:

$\frac{4\sqrt{3}}{2}=2\sqrt{3}:cm^2$

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Question

If , what is the area of the triangle?

Answer

We know this triangle is equilateral since each of its sides has the same length, . The formula of the area of an equilateral triangle is $\frac{\sqrt{3}}{4}x^2$ if is the length of one of the triangle's sides.

Since our side length is , we can substitute that value into the equation for and solve for the area of the triangle:

$Area=\frac{\sqrt{3}}{4}x^2=\frac{\sqrt{3}}{4}(10)^2=\frac{\sqrt{3}}{4}100=25\sqrt{3}:cm^2$

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Question

If , what is the area of this equilateral triangle?

Answer

Given that the sides of our equilateral triangle are each long, we can just plug the value in to the formula for the area of an equilateral triangle and solve for the area of the triangle:

$Area=\frac{\sqrt{3}}{4}a^2$ if is a side of the triangle.

$Area=\frac{\sqrt{3}}{4}a^2=\frac{\sqrt{3}}{4}5^2=\frac{\sqrt{3}}{4}25=\frac{25\sqrt{3}}{4}:cm^2$

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Question

Find the area of an equilateral triangle with a perimeter of 24cm. Leave answer in simplest radical form.

Answer

To find the area of an equilateral triangle, one must find the base and the height.

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

All the sides of an equilateral triangle are congruent, so if the perimeter of the equilaterail triangle is 24, then each side must equal one third of that total which is 8cm.

This will produce a triangle that includes the following information below:

Dropping an altitude down the center of the equilateral triangle will result in two 30-60-90 triangles with a hypotenuse of 8.

In every 30-60-90 triangle the following formulas apply:

$longleg=shortleg\sqrt3$

When we plug in the given information on the triangle we get:

Dividing both sides by 2 gives the below result.

We can now plug this into the long leg formula to get the height of the triangle:

$longleg=4\sqrt3$

Now that we have all of the information needed to find the area we plug these values into the area formula.

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

$Area=\frac{1}{2}(8)(4\sqrt3)$

$Area=16\sqrt3$

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Question

An equilateral triangle has sides of length 6cm. If the height of the triangle is 4.5cm what is the area of the triangle?

Answer

To find the area of any triangle we can use the formula 1/2 (base x height) , that is the base times the height divided by two. It is important to remember any of the sides of an equaliateral triangle can be used as the base when the hieght is given. The area can be found by (6 x 4.5) divided by 2; which gives 13.5 square centimeters.

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Question

A circle with a radius of is inscribed in an equilateral triangle with side lengths of as shown in the figure below.

Find the area of the shaded region.

Answer

In order to find the area of the shaded region, we must first find the area of the circle and the area of the equilateral triangle.

Recall how to find the area of a circle:

$\text{Area of Circle}=\pi\times\text{radius}^2$

Plug in the given radius to find the area of the circle.

$\text{Area of Circle}=\pi\times 2^2=4\pi$

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

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(\text{side})^2$

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

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(6)^2=9\sqrt3$

In order to find the area of the shaded region, we will need to subtract the area of the circle from the area of the triangle.

$\text{Area of Shaded Region}=\text{Area of Triangle}-\text{Area of Circle}$

$\text{Area of Shaded Region}=9\sqrt3-4\pi=3.02$

Make sure to round to places after the decimal.

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Question

A circle with a radius of is inscribed in an equilateral triangle with side lengths of as shown in the figure below.

Find the area of the shaded region.

Answer

In order to find the area of the shaded region, we must first find the area of the circle and the area of the equilateral triangle.

Recall how to find the area of a circle:

$\text{Area of Circle}=\pi\times\text{radius}^2$

Plug in the given radius to find the area of the circle.

$\text{Area of Circle}=\pi\times 3^2=9\pi$

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

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(\text{side})^2$

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

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(9)^2=\frac{81\sqrt3}{4}$

In order to find the area of the shaded region, we will need to subtract the area of the circle from the area of the triangle.

$\text{Area of Shaded Region}=\text{Area of Triangle}-\text{Area of Circle}$

$\text{Area of Shaded Region}=\frac{81\sqrt3}{4}-9\pi=6.80$

Make sure to round to places after the decimal.

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Question

A circle with a radius of is inscribed in an equilateral triangle with side lengths of as shown in the figure below.

Find the area of the shaded region.

Answer

In order to find the area of the shaded region, we must first find the area of the circle and the area of the equilateral triangle.

Recall how to find the area of a circle:

$\text{Area of Circle}=\pi\times\text{radius}^2$

Plug in the given radius to find the area of the circle.

$\text{Area of Circle}=\pi\times 4^2=16\pi$

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

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(\text{side})^2$

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

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(12)^2=36\sqrt3$

In order to find the area of the shaded region, we will need to subtract the area of the circle from the area of the triangle.

$\text{Area of Shaded Region}=\text{Area of Triangle}-\text{Area of Circle}$

$\text{Area of Shaded Region}=36\sqrt3-16\pi=12.09$

Make sure to round to places after the decimal.

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Question

A circle with a radius of is inscribed in an equilateral triangle with side lengths of as shown in the figure below.

Find the area of the shaded region.

Answer

In order to find the area of the shaded region, we must first find the area of the circle and the area of the equilateral triangle.

Recall how to find the area of a circle:

$\text{Area of Circle}=\pi\times\text{radius}^2$

Plug in the given radius to find the area of the circle.

$\text{Area of Circle}=\pi\times 5^2=25\pi$

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

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(\text{side})^2$

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

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(20)^2=100\sqrt3$

In order to find the area of the shaded region, we will need to subtract the area of the circle from the area of the triangle.

$\text{Area of Shaded Region}=\text{Area of Triangle}-\text{Area of Circle}$

$\text{Area of Shaded Region}=100\sqrt3-25\pi=94.67$

Make sure to round to places after the decimal.

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Question

A circle with a radius of is inscribed in an equilateral triangle with side lengths of as shown in the figure below.

Find the area of the shaded region.

Answer

In order to find the area of the shaded region, we must first find the area of the circle and the area of the equilateral triangle.

Recall how to find the area of a circle:

$\text{Area of Circle}=\pi\times\text{radius}^2$

Plug in the given radius to find the area of the circle.

$\text{Area of Circle}=\pi\times 6^2=36\pi$

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

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(\text{side})^2$

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

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(24)^2=144\sqrt3$

In order to find the area of the shaded region, we will need to subtract the area of the circle from the area of the triangle.

$\text{Area of Shaded Region}=\text{Area of Triangle}-\text{Area of Circle}$

$\text{Area of Shaded Region}=144\sqrt3-36\pi=136.32$

Make sure to round to places after the decimal.

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Question

A circle with a radius of is inscribed in an equilateral triangle with side lengths of as shown by the figure below.

Find the area of the shaded region.

Answer

In order to find the area of the shaded region, we must first find the area of the circle and the area of the equilateral triangle.

Recall how to find the area of a circle:

$\text{Area of Circle}=\pi\times\text{radius}^2$

Plug in the given radius to find the area of the circle.

$\text{Area of Circle}=\pi\times 6^2=36\pi$

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

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(\text{side})^2$

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

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(20)^2=100\sqrt3$

In order to find the area of the shaded region, we will need to subtract the area of the circle from the area of the triangle.

$\text{Area of Shaded Region}=\text{Area of Triangle}-\text{Area of Circle}$

$\text{Area of Shaded Region}=100\sqrt3-36\pi=60.11$

Make sure to round to places after the decimal.

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Question

A circle with a radius of is inscribed in an equilateral triangle with side lengths of as shown by the figure below.

Find the area of the shaded region.

Answer

In order to find the area of the shaded region, we must first find the area of the circle and the area of the equilateral triangle.

Recall how to find the area of a circle:

$\text{Area of Circle}=\pi\times\text{radius}^2$

Plug in the given radius to find the area of the circle.

$\text{Area of Circle}=\pi\times 3^2=9\pi$

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

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(\text{side})^2$

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

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(14)^2=49\sqrt3$

In order to find the area of the shaded region, we will need to subtract the area of the circle from the area of the triangle.

$\text{Area of Shaded Region}=\text{Area of Triangle}-\text{Area of Circle}$

$\text{Area of Shaded Region}=49\sqrt3-9\pi=56.60$

Make sure to round to places after the decimal.

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Question

A circle with a radius of is inscribed in an equilateral triangle with side lengths of as shown by the figure below.

Find the area of the shaded region.

Answer

In order to find the area of the shaded region, we must first find the area of the circle and the area of the equilateral triangle.

Recall how to find the area of a circle:

$\text{Area of Circle}=\pi\times\text{radius}^2$

Plug in the given radius to find the area of the circle.

$\text{Area of Circle}=\pi\times 12^2=144\pi$

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

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(\text{side})^2$

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

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(40)^2=400\sqrt3$

In order to find the area of the shaded region, we will need to subtract the area of the circle from the area of the triangle.

$\text{Area of Shaded Region}=\text{Area of Triangle}-\text{Area of Circle}$

$\text{Area of Shaded Region}=400\sqrt3-144\pi=240.43$

Make sure to round to places after the decimal.

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Question

A circle with a radius of $\frac{1}{2}$ is inscribed in an equilateral triangle with side lengths of as shown by the figure below.

Find the area of the shaded region.

Answer

In order to find the area of the shaded region, we must first find the area of the circle and the area of the equilateral triangle.

Recall how to find the area of a circle:

$\text{Area of Circle}=\pi\times\text{radius}^2$

Plug in the given radius to find the area of the circle.

$\text{Area of Circle}=\pi\times( \frac{1}{2})^2=\frac{1}{4}\pi$

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

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(\text{side})^2$

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

$\text{Area of Equilateral Triangle}=\frac{\sqrt3}{4}(5)^2=\frac{25\sqrt3}{4}$

In order to find the area of the shaded region, we will need to subtract the area of the circle from the area of the triangle.

$\text{Area of Shaded Region}=\text{Area of Triangle}-\text{Area of Circle}$

$\text{Area of Shaded Region}=\frac{25\sqrt3}{4}-\frac{1}{4}\pi=10.04$

Make sure to round to places after the decimal.

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