Equilateral Triangles - PSAT Math

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.

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).

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.

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.

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).

There are 2 components to this problem. The first, and easier one, is recognizing how much of the perimeter the equilateral triangles take up—since there are 40 triangles in total, there must be 40 – 4 = 36 of these triangles. By observation, each contributes only 1 side to the overall perimeter, thus we can simply multiply 36(5) = 180" contribution.

The second component is the corner triangles—recognizing that the congruent sides are adjacent to the 5-inch equilateral triangles, and the congruent angles can be found by

180 = 30+2x → x = 75°

We can use ratios to find the unknown side:

75/5 = 30/y → 75y = 150 → y = 2''.

Since there are 4 corners to the square rug, 2(4) = 8'' contribution to the total perimeter. Adding the 2 components, we get 180+8 = 188 inch perimeter.

An altitude drawn in an equilateral triangle will form two 30-60-90 triangles. The height of equilateral triangle is the length of the longer leg of the 30-60-90 triangle. The length of the equilateral triangle's side is the length of the hypotenuse of the 30-60-90.

The ratio of the length of the hypotenuse to the length of the longer leg of a 30-60-90 triangle is 2:$\sqrt{3}$

The length of the longer leg of the 30-60-90 triangle in this problem is 2$\sqrt{3}$

Using this ratio, we find that the length of this triangle's hypotenuse is 4. Thus the perimeter of the equilateral triangle will be 4 multiplied by 3, which is 12.

There are 2 components to this problem. The first, and easier one, is recognizing how much of the perimeter the equilateral triangles take up—since there are 40 triangles in total, there must be 40 – 4 = 36 of these triangles. By observation, each contributes only 1 side to the overall perimeter, thus we can simply multiply 36(5) = 180" contribution.

The second component is the corner triangles—recognizing that the congruent sides are adjacent to the 5-inch equilateral triangles, and the congruent angles can be found by

180 = 30+2x → x = 75°

We can use ratios to find the unknown side:

75/5 = 30/y → 75y = 150 → y = 2''.

Since there are 4 corners to the square rug, 2(4) = 8'' contribution to the total perimeter. Adding the 2 components, we get 180+8 = 188 inch perimeter.

An altitude drawn in an equilateral triangle will form two 30-60-90 triangles. The height of equilateral triangle is the length of the longer leg of the 30-60-90 triangle. The length of the equilateral triangle's side is the length of the hypotenuse of the 30-60-90.

The ratio of the length of the hypotenuse to the length of the longer leg of a 30-60-90 triangle is 2:$\sqrt{3}$

The length of the longer leg of the 30-60-90 triangle in this problem is 2$\sqrt{3}$

Using this ratio, we find that the length of this triangle's hypotenuse is 4. Thus the perimeter of the equilateral triangle will be 4 multiplied by 3, which is 12.

One meter, 100 centimeters, and 10 decimeters are all equal to the same quantity. This makes the triangle equilateral and, subsequently, acute.

A right triangle has a short leg half as long as its hypotenuse , which is . Its long leg is times as long as its short leg, which will be . Its area is half the product of its legs, so the area will be

The area of an equilateral triangle is given by the formula

,

so set and solve for :

The circle with circumference 100 has radius

Its area is

We can substitute this for in the equation for the area of an equilateral triangle, and solve for :

The correct response is 42.9.

An isosceles right triangle is also a triangle, whose legs each measure the length of the hypotenuse divided by . Therefore, since the hypotenuse measures , each leg measures .

The area of a right triangle is half the product of its legs, so this right triangle has area

The area of an equilateral triangle is given by the formula

,

so set and solve for :

A right triangle with hypotenuse 12 and leg 8 also has leg

The area of a right triangle is half the product of its legs, so this right triangle has area

,

which is also the area of the given equilateral triangle.

The area of an equilateral triangle is given by the formula

so if we set , we can solve for :

The correct choice is 9.1.

A regular hexagon can be divided by its three diameters into six congruent equilateral triangles. Since each triangle will have sidelength , each will have area equal to

Multiply by 6 to get the area of the hexagon:

We can substitute this for in the equation for the area of an equilateral triangle, and solve for :

, the correct response.

The area of a square is where represents the side length. In our case the side length is therefore, the area of the square is ; this will also be the area of the equilateral triangle.

The formula for the area of an equilateral triangle with sidelength is

If we let , we can solve for in the equation:

which is the correct response.

The area of a triangle is (1/2)*base*height. We know that the area = 15 cm, and the base is 5 cm, so:

15 = 1/2 * 5 * height

3 = 1/2 * height

6 = height