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PSAT Math : Plane Geometry

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Example Questions

Example Question #1 : Triangles

Triangle

In the figure above, what is the positive difference, in degrees, between the measures of angle ACB and angle CBD?

Possible Answers:

Correct answer:

Explanation:

In the figure above, angle ADB is a right angle. Because side AC is a straight line, angle CDB must also be a right angle.

Let’s examine triangle ADB. The sum of the measures of the three angles must be 180 degrees, and we know that angle ADB must be 90 degrees, since it is a right angle. We can now set up the following equation.

x + y + 90 = 180

Subtract 90 from both sides.

x + y = 90

Next, we will look at triangle CDB. We know that angle CDB is also 90 degrees, so we will write the following equation:

y – 10 + 2x – 20 + 90 = 180

y + 2x + 60 = 180

Subtract 60 from both sides.

y + 2x = 120

We have a system of equations consisting of x + y = 90 and y + 2x = 120. We can solve this system by solving one equation in terms of x and then substituting this value into the second equation. Let’s solve for y in the equation x + y = 90.

x + y = 90

Subtract x from both sides.

y = 90 – x

Next, we can substitute 90 – x into the equation y + 2x = 120.

(90 – x) + 2x = 120

90 + x = 120

x = 120 – 90 = 30

x = 30

Since y = 90 – x, y = 90 – 30 = 60.

The question ultimately asks us to find the positive difference between the measures of ACB and CBD. The measure of ACB = 2x – 20 = 2(30) – 20 = 40 degrees. The measure of CBD = y – 10 = 60 – 10 = 50 degrees. The positive difference between 50 degrees and 40 degrees is 10.

The answer is 10.

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Example Question #1 : Triangles

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

Possible Answers:

$10,15,25$

$7,9,16$

$2,3,5$

$5,12,17$

$5,12,13$

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 #31 : 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 #211 : Plane Geometry

Thingy_5

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

Possible Answers:

$\Delta G AF$

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

$\Delta G FA$

$\Delta FGA$

$\Delta AFG$

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 #212 : Plane Geometry

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

Possible Answers:

The triangle is equilateral and acute.

The triangle is scalene and obtuse.

The triangle cannot exist.

The triangle is scalene and right.

The triangle is scalene 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 #2 : 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 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

27.2

9.1

12.9

10.5

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 : 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:

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

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

$\frac{x}{3}$

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

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

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 #5 : How To Find The Length Of The Side Of An Equilateral Triangle

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:

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

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

$2x \sqrt{3}$

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

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

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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PSAT Math : Plane Geometry

Study concepts, example questions & explanations for PSAT Math

All PSAT Math Resources

Example Questions

Example Question #1 : Triangles

Example Question #1 : Triangles

Example Question #31 : Geometry

Example Question #211 : Plane Geometry

Example Question #212 : Plane Geometry

Example Question #2 : 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 : 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 #5 : How To Find The Length Of The Side Of An Equilateral Triangle

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