How to find the length of the side of an equilateral triangle - PSAT Math

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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.

Compare your answer with the correct one above

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 :

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

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

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

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

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

Compare your answer with the correct one above

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 :

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

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

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

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

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

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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 :

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

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

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

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

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

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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 :

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

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

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

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

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

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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 :

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

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

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

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

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

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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 :

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

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

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

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

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

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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