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

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Example Questions

Example Question #1 : Equilateral Triangles

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

Possible Answers:

The triangle is scalene and obtuse.

The triangle is equilateral and acute.

The triangle is scalene and acute.

The triangle cannot exist.

The triangle is scalene and right.

Correct answer:

The triangle is equilateral and acute.

Explanation:

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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Example Question #102 : Triangles

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.

Possible Answers:

27.2

9.1

12.9

10.5

10.1

Correct answer:

9.1

Explanation:

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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Example Question #3 : How To Find The Length Of The Side Of An Equilateral Triangle

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

Possible Answers:

42.9

47.4

67.1

60.6

269.4

Correct answer:

42.9

Explanation:

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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Example Question #1 : How To Find The Length Of The Side Of An Equilateral Triangle

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 .

Possible Answers:

$\frac{ x} { \sqrt{3}}$

$\frac{x}{3}$

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

$\frac{ x} { \sqrt[4]{2}}$

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

Correct answer:

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

Explanation:

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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Example Question #3 : How To Find The Length Of The Side Of An Equilateral Triangle

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 .

Possible Answers:

$\frac{x }{3}$

$\frac{x }{2}$

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

$\frac{x\sqrt{6} }{2}$

$\frac{x\sqrt{3} }{3}$

Correct answer:

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

Explanation:

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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Example Question #2 : Equilateral Triangles

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 .

Possible Answers:

$\frac{ 4x}{ \sqrt{3}}$

$2x \sqrt{3}$

$\frac{ 2x}{ \sqrt{3}}$

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

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

Correct answer:

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

Explanation:

The area of a square is A=s^2 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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Example Question #1 : Equilateral Triangles

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 .

Possible Answers:

$x\sqrt{3}$

$x\sqrt{6}$

Correct answer:

$x\sqrt{6}$

Explanation:

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

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #1 : Equilateral Triangles

Example Question #102 : Triangles

Example Question #3 : How To Find The Length Of The Side Of An Equilateral Triangle

Example Question #1 : How To Find The Length Of The Side Of An Equilateral Triangle

Example Question #3 : How To Find The Length Of The Side Of An Equilateral Triangle

Example Question #2 : Equilateral Triangles

Example Question #1 : Equilateral Triangles

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