Equilateral Triangles - PSAT Math

Card 0 of 16

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 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

$\textup{What is the area of an equilateral triangle with sides each length 10?}$

Answer

$\textup{Vertical height line will divide the triangle into two congruent}$

$\textup{30-60-90 triangles with hyp = 10 and short leg = 5. Use reference table}$

$\textup{or Pythagorean theorem to find height }=5\sqrt{3}$

$A=\frac{1}{2}bh,:b=10,:h=5\sqrt{3}: : : : : : A=\frac{1}{2}(10)5\sqrt{3}=25\sqrt{3}$

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Question

A triangle has a base of 5 cm and an area of 15 cm. What is the height of the triangle?

Answer

The area of a triangle is (1/2)baseheight. 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

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Question

In the figure above, AB = AD = AE = BD = BC = CD = DE = 1. What is the distance from A to C?

Answer

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Question

A triangles has sides of 5, 9, and $x$ . Which of the folowing CANNOT be a possible value of $x$ ?

Answer

The sum of the lengths of the shortest sides of a triangle cannot be less than the third side.

3 + 5 = 8 < 9, so 3 can't be a value of x.

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Question

Which of the following describes a triangle with sides of length one meter, 100 centimeters, and 10 decimeters?

Answer

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

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Question

Two triangles have the same area. One is an equilateral triangle. The other is a $30^{\circ }- 60^{\circ }-90^{\circ }$ right triangle with hypotenuse . Give the sidelength of the equilateral triangle in terms of .

Answer

A $30^{\circ }- 60^{\circ }-90^{\circ }$ right triangle has a short leg half as long as its hypotenuse , which is $\frac{x}{2}$ . Its long leg is $\sqrt{3}$ times as long as its short leg, which will be $\frac{x\sqrt{3}}{2}$ . Its area is half the product of its legs, so the area will be

$A = \frac{1}{2} \cdot \frac{x}{2} \cdot \frac{x\sqrt{3}}{2} = \frac{x^{2}\sqrt{3}}{8}$

The area of an equilateral triangle is given by the formula

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

so set $A = \frac{x^{2}\sqrt{3}}{8}$ and solve for :

$\frac{s^{2}\sqrt{3}}{4} = \frac{x^{2}\sqrt{3}}{8}$

$\frac{s^{2}\sqrt{3}}{4} \cdot \frac{4} {\sqrt{3}} = \frac{x^{2}\sqrt{3}}{8} \cdot \frac{4} {\sqrt{3}}$

$s^{2} = \frac{x^{2} }{2}$

$s =\sqrt{ \frac{x^{2} }{2}}$

$s = \frac{x }{\sqrt{2}} = \frac{x\sqrt{2}}{2}$

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Question

An equilateral triangle has the same area as a circle with circumference 100. To the nearest tenth, give the sidelength of the triangle.

Answer

The circle with circumference 100 has radius

$r = \frac{C}{2 \pi} = \frac{100 }{2 \pi} = \frac{50}{\pi}$

Its area is

$A = \pi r^{2}$

$A = \pi \cdot \left ( \frac{50}{\pi} \right )^{2}$

$A = \pi \cdot \frac{2,500}{\pi ^{2}}$

$A = \frac{2,500}{\pi }$

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

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

$\frac{s^{2}\sqrt{3}}{4} = \frac{2,500}{\pi}$

$\frac{s^{2}\sqrt{3}}{4} \cdot \frac{4}{\sqrt{3}} = \frac{2,500}{\pi} \cdot \frac{4}{\sqrt{3}}$

$s^{2} = \frac{2,500}{\pi} \cdot \frac{4}{\sqrt{3}} \approx \frac{10,000}{3.14159\cdot 1.732} \approx 1,837.76$

$s \approx \sqrt{1,837.76} \approx 42.9$

The correct response is 42.9.

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Question

Two triangles have the same area. One is an equilateral triangle. The other is an isosceles right triangle with hypotenuse . Give the sidelength of the equilateral triangle in terms of .

Answer

An isosceles right triangle is also a $45^{\circ } - 45^{\circ } - 90^{\circ }$ triangle, whose legs each measure the length of the hypotenuse divided by $\sqrt{2}$ . Therefore, since the hypotenuse measures , each leg measures $\frac{x}{\sqrt{2}}$ .

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

$A = \frac{1}{2} \cdot \frac{x}{\sqrt{2}} \cdot \frac{x}{\sqrt{2}} = \frac{x^{2}}{4}$

The area of an equilateral triangle is given by the formula

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

so set $A = \frac{x^{2}}{4}$ and solve for :

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

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

$s^{2} = \frac{x^{2}} { \sqrt{3}}$

$s =\sqrt{ \frac{x^{2}} { \sqrt{3}}}$

$s =\frac{ x} { \sqrt[4]{3}}$

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Question

Two triangles have the same area. One is an equilateral triangle. The other is a right triangle with hypotenuse 12 and one leg of length 8. Give the sidelength of the equilateral triangle to the nearest tenth.

Answer

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

$a = \sqrt{12^{2}-8^{2}} = \sqrt{144-64} = \sqrt{80} \approx 8.944$

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

$A \approx \frac{1}{2} \cdot 8 \cdot 8.944 \approx 35.777$ ,

which is also the area of the given equilateral triangle.

The area of an equilateral triangle is given by the formula

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

so if we set $A \approx 35.777$ , we can solve for :

$\frac{s^{2}\sqrt{3}}{4} \approx35.777$

$s^{2} \approx35.777 \cdot \frac{4}{\sqrt{3}} \approx 35.777 \cdot \frac{4}{1.732} \approx82.623$

$s \approx \sqrt{82.623} \approx 9.090$

The correct choice is 9.1.

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Question

A regular hexagon and an equilateral triangle have the same area. Call the side length of the hexagon . Give the side length of the equilateral triangle in terms of .

Answer

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

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

Multiply by 6 to get the area of the hexagon:

$A = 6A' = 6 \cdot \frac{x^{2}\sqrt{3}}{4} = \frac{3x^{2}\sqrt{3}}{2}$

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

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

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

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

$s^{2}= 6x^{2}$

$s = \sqrt{6x^{2}} = x\sqrt{6}$ , the correct response.

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Question

A square and an equilateral triangle have the same area. Call the side length of the square . Give the side length of the equilateral triangle in terms of .

Answer

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 $x^{2}$ ; this will also be the area of the equilateral triangle.

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

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

If we let $A = x^{2}$ , we can solve for in the equation:

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

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

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

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

$s =\frac{ 2x}{ \sqrt[4]{3}}$

which is the correct response.

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Question

A square rug border consists of a continuous pattern of equilateral triangles, with isosceles triangles as corners, one of which is shown above. If the length of each equilateral triangle side is 5 inches, and there are 40 triangles in total, what is the total perimeter of the rug?

The inner angles of the corner triangles is 30°.

Answer

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.

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Question

The height of an equilateral triangle is $\dpi{100} \small 2\sqrt{3}$

What is the triangle's perimeter?

Answer

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 $\dpi{100} \small 2:\sqrt{3}$

The length of the longer leg of the 30-60-90 triangle in this problem is $\dpi{100} \small 2\sqrt{3}$

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.

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