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PSAT Math : Geometry

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Example Questions

Example Question #531 : Geometry

Which of the following sets of line-segment lengths can form a triangle?

Possible Answers:

$2,3,5$

$7,9,16$

$5,12,17$

$10,15,25$

$5,12,13$

Correct answer:

$5,12,13$

Explanation:

In any given triangle, the sum of any two sides is greater than the third. The incorrect answers have the sum of two sides equal to the third.

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Example Question #463 : Geometry

In right $\Delta ABC$ , $\angle ABC$ $=$ $2x$ and $\angle BCA= \frac{x}{2}$ .

What is the value of $x$ ?

Possible Answers:

Correct answer:

Explanation:

There are 180 degrees in every triangle. Since this triangle is a right triangle, one of the angles measures 90 degrees.

Therefore, $90 + 2x + \frac{x}{2}= 180$ .

$90=2.5x$

$x=36$

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Example Question #1 : How To Find An Angle In A Right Triangle

Thingy_5

Refer to the above diagram. Which of the following gives a valid alternative name for $\Delta AGF$ ?

Possible Answers:

$\Delta FGA$

$\Delta G FA$

$\Delta G AF$

All four of the other choices gives a valid alternative name for the triangle.

$\Delta AFG$

Correct answer:

All four of the other choices gives a valid alternative name for the triangle.

Explanation:

A triangle can be named after its three vertices in any order, so all of the choices given are valid.

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

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

Possible Answers:

The triangle cannot exist.

The triangle is scalene and obtuse.

The triangle is scalene and right.

The triangle is scalene and acute.

The triangle is equilateral and acute.

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 #1 : Equilateral 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:

10.1

10.5

9.1

27.2

12.9

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 #2 : Equilateral Triangles

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:

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

$\frac{x}{3}$

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

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

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

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 : Geometry

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #531 : Geometry

Example Question #463 : Geometry

Example Question #1 : How To Find An Angle In A Right Triangle

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

Example Question #1 : Equilateral Triangles

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

Example Question #2 : Equilateral Triangles

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

All PSAT Math Resources

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